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Theorem List for Metamath Proof Explorer - 20101-20200   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremangrtmuld 20101* Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  Y  e.  CC )   &    |-  ( ph  ->  Z  e.  CC )   &    |-  ( ph  ->  X  =/=  0 )   &    |-  ( ph  ->  Y  =/=  0
 )   &    |-  ( ph  ->  Z  =/=  0 )   &    |-  ( ph  ->  ( Z  /  Y )  e.  RR )   =>    |-  ( ph  ->  ( ( X F Y )  e.  { ( pi  /  2 ) ,  -u ( pi  /  2
 ) }  <->  ( X F Z )  e.  { ( pi  /  2 ) ,  -u ( pi  /  2
 ) } ) )
 
Theoremang180lem1 20102* Lemma for ang180 20107. Show that the "revolution number"  N is an integer, using efeq1 19886 to show that since the product of the three arguments  A ,  1  / 
( 1  -  A
) ,  ( A  -  1 )  /  A is  -u 1, the sum of the logarithms must be an integer multiple of  2
pi _i away from  pi _i  =  log ( -u 1 ). (Contributed by Mario Carneiro, 23-Sep-2014.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  T  =  ( ( ( log `  (
 1  /  ( 1  -  A ) ) )  +  ( log `  (
 ( A  -  1
 )  /  A )
 ) )  +  ( log `  A ) )   &    |-  N  =  ( (
 ( T  /  _i )  /  ( 2  x.  pi ) )  -  ( 1  /  2
 ) )   =>    |-  ( ( A  e.  CC  /\  A  =/=  0  /\  A  =/=  1 ) 
 ->  ( N  e.  ZZ  /\  ( T  /  _i )  e.  RR )
 )
 
Theoremang180lem2 20103* Lemma for ang180 20107. Show that the revolution number  N is strictly between  -u 2 and  1. Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 19922, but the resulting bound gives only  N  <_ 
1 for the upper bound. The case  N  =  1 is not ruled out here, but it is in some sense an "edge case" that can only happen under very specific conditions; in particular we show that all the angle arguments  A ,  1  /  ( 1  -  A ) ,  ( A  -  1 )  /  A must lie on the negative real axis, which is a contradiction because clearly if  A is negative then the other two are positive real. (Contributed by Mario Carneiro, 23-Sep-2014.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  T  =  ( ( ( log `  (
 1  /  ( 1  -  A ) ) )  +  ( log `  (
 ( A  -  1
 )  /  A )
 ) )  +  ( log `  A ) )   &    |-  N  =  ( (
 ( T  /  _i )  /  ( 2  x.  pi ) )  -  ( 1  /  2
 ) )   =>    |-  ( ( A  e.  CC  /\  A  =/=  0  /\  A  =/=  1 ) 
 ->  ( -u 2  <  N  /\  N  <  1 ) )
 
Theoremang180lem3 20104* Lemma for ang180 20107. Since ang180lem1 20102 shows that  N is an integer and ang180lem2 20103 shows that  N is strictly between  -u 2 and  1, it follows that  N  e.  { -u 1 ,  0 }, and these two cases correspond to the two possible values for  T. (Contributed by Mario Carneiro, 23-Sep-2014.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  T  =  ( ( ( log `  (
 1  /  ( 1  -  A ) ) )  +  ( log `  (
 ( A  -  1
 )  /  A )
 ) )  +  ( log `  A ) )   &    |-  N  =  ( (
 ( T  /  _i )  /  ( 2  x.  pi ) )  -  ( 1  /  2
 ) )   =>    |-  ( ( A  e.  CC  /\  A  =/=  0  /\  A  =/=  1 ) 
 ->  T  e.  { -u ( _i  x.  pi ) ,  ( _i  x.  pi ) } )
 
Theoremang180lem4 20105* Lemma for ang180 20107. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( A  e.  CC  /\  A  =/=  0  /\  A  =/=  1 ) 
 ->  ( ( ( ( 1  -  A ) F 1 )  +  ( A F ( A  -  1 ) ) )  +  ( 1 F A ) )  e.  { -u pi ,  pi } )
 
Theoremang180lem5 20106* Lemma for ang180 20107: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( ( A  e.  CC  /\  A  =/=  0 )  /\  ( B  e.  CC  /\  B  =/=  0 )  /\  A  =/=  B )  ->  (
 ( ( ( A  -  B ) F A )  +  ( B F ( B  -  A ) ) )  +  ( A F B ) )  e. 
 { -u pi ,  pi } )
 
Theoremang180 20107* The sum of angles  m A B C  +  m B C A  +  m C A B in a triangle adds up to either  pi or  -u pi, i.e. 180 degrees. (The sign is due to the two possible orientations of vertex arrangement and our signed notion of angle). (Contributed by Mario Carneiro, 23-Sep-2014.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  B  /\  B  =/=  C  /\  A  =/=  C ) )  ->  (
 ( ( ( C  -  B ) F ( A  -  B ) )  +  (
 ( A  -  C ) F ( B  -  C ) ) )  +  ( ( B  -  A ) F ( C  -  A ) ) )  e. 
 { -u pi ,  pi } )
 
Theoremlawcoslem1 20108 Lemma for Law of Cosines lawcos 20109. Here we prove the law for a point at the origin and two distinct points U and V, using an expanded version of the signed angle expression on the complex plane. (Contributed by David A. Wheeler, 11-Jun-2015.)
 |-  ( ph  ->  U  e.  CC )   &    |-  ( ph  ->  V  e.  CC )   &    |-  ( ph  ->  U  =/=  0
 )   &    |-  ( ph  ->  V  =/=  0 )   =>    |-  ( ph  ->  (
 ( abs `  ( U  -  V ) ) ^
 2 )  =  ( ( ( ( abs `  U ) ^ 2
 )  +  ( ( abs `  V ) ^ 2 ) )  -  ( 2  x.  ( ( ( abs `  U )  x.  ( abs `  V ) )  x.  ( ( Re
 `  ( U  /  V ) )  /  ( abs `  ( U  /  V ) ) ) ) ) ) )
 
Theoremlawcos 20109* Law of Cosines. Given three distinct points A, B, and C, prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where  F is the signed angle construct (as used in ang180 20107),  X is the distance of line segment BC,  Y is the distance of line segment AC,  Z is the distance of line segment AB, and  O is the distinguished (signed) angle m/_ BCA on the complex plane. We translate triangle ABC to move C to the origin (C-C), B to U=(B-C), and A to V=(A-C), then use lemma lawcoslem1 20108 to prove this algebraically simpler case. The metamath convention is to use a signed angle; in this case the sign doesn't matter because we use the cosine of the angle (see cosneg 12422). The Pythagorean Theorem pythag 20110 is a special case of the law of cosines. The theorem's expression and approach were suggested by Mario Carneiro. (Contributed by David A. Wheeler, 12-Jun-2015.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  X  =  ( abs `  ( B  -  C ) )   &    |-  Y  =  ( abs `  ( A  -  C ) )   &    |-  Z  =  ( abs `  ( A  -  B ) )   &    |-  O  =  ( ( B  -  C ) F ( A  -  C ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  C  /\  B  =/=  C ) )  ->  ( Z ^ 2 )  =  ( ( ( X ^ 2 )  +  ( Y ^
 2 ) )  -  ( 2  x.  (
 ( X  x.  Y )  x.  ( cos `  O ) ) ) ) )
 
Theorempythag 20110* Pythagorean Theorem. Given three distinct points A, B, and C that form a right triangle (with the right angle at C), prove a relationship between their segment lengths. This theorem is expressed using the complex number plane as a plane, where  F is the signed angle construct (as used in ang180 20107),  X is the distance of line segment BC,  Y is the distance of line segment AC,  Z is the distance of line segment AB (the hypotenuse), and  O is the distinguished (signed) right angle m/_ BCA. We use the law of cosines lawcos 20109 to prove this, along with simple trig facts like coshalfpi 19832 and cosneg 12422. (Contributed by David A. Wheeler, 13-Jun-2015.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  X  =  ( abs `  ( B  -  C ) )   &    |-  Y  =  ( abs `  ( A  -  C ) )   &    |-  Z  =  ( abs `  ( A  -  B ) )   &    |-  O  =  ( ( B  -  C ) F ( A  -  C ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  C  /\  B  =/=  C )  /\  O  e.  { ( pi  / 
 2 ) ,  -u ( pi  /  2 ) }
 )  ->  ( Z ^ 2 )  =  ( ( X ^
 2 )  +  ( Y ^ 2 ) ) )
 
Theoremlogreclem 20111 Symmetry of the natural logarithm range by negation. Lemma for logrec 20112. (Contributed by Saveliy Skresanov, 27-Dec-2016.)
 |-  ( ( A  e.  ran 
 log  /\  -.  ( Im
 `  A )  =  pi )  ->  -u A  e.  ran  log )
 
Theoremlogrec 20112 Logarithm of a reciprocal changes sign. (Contributed by Saveliy Skresanov, 28-Dec-2016.)
 |-  ( ( A  e.  CC  /\  A  =/=  0  /\  ( Im `  ( log `  A ) )  =/=  pi )  ->  ( log `  A )  =  -u ( log `  (
 1  /  A )
 ) )
 
Theoremisosctrlem1 20113 Lemma for isosctr 20116. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
 |-  ( ( A  e.  CC  /\  ( abs `  A )  =  1  /\  -.  1  =  A ) 
 ->  ( Im `  ( log `  ( 1  -  A ) ) )  =/=  pi )
 
Theoremisosctrlem2 20114 Lemma for isosctr 20116. Corresponds to the case where one vertex is at 0, another at 1 and the third lies on the unit circle. (Contributed by Saveliy Skresanov, 31-Dec-2016.)
 |-  ( ( A  e.  CC  /\  ( abs `  A )  =  1  /\  -.  1  =  A ) 
 ->  ( Im `  ( log `  ( 1  -  A ) ) )  =  ( Im `  ( log `  ( -u A  /  ( 1  -  A ) ) ) ) )
 
Theoremisosctrlem3 20115* Lemma for isosctr 20116. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC )  /\  ( A  =/=  0  /\  B  =/=  0  /\  A  =/=  B )  /\  ( abs `  A )  =  ( abs `  B )
 )  ->  ( -u A F ( B  -  A ) )  =  ( ( A  -  B ) F -u B ) )
 
Theoremisosctr 20116* Isosceles triangle theorem. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   =>    |-  ( ( ( A  e.  CC  /\  B  e.  CC  /\  C  e.  CC )  /\  ( A  =/=  C  /\  B  =/=  C  /\  A  =/=  B )  /\  ( abs `  ( A  -  C ) )  =  ( abs `  ( B  -  C ) ) ) 
 ->  ( ( C  -  A ) F ( B  -  A ) )  =  ( ( A  -  B ) F ( C  -  B ) ) )
 
Theoremssscongptld 20117* If two triangles have equal sides, one angle in one triangle has the same cosine as the corresponding angle in the other triangle. This is a partial form of the SSS congruence theorem.

This theorem is proven by using lawcos 20109 on both triangles to express one side in terms of the other two, and then equating these expressions and reducing this algebraically to get an equality of cosines of angles. (Contributed by David Moews, 28-Feb-2017.)

 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  E  e.  CC )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   &    |-  ( ph  ->  D  =/=  E )   &    |-  ( ph  ->  E  =/=  G )   &    |-  ( ph  ->  ( abs `  ( A  -  B ) )  =  ( abs `  ( D  -  E ) ) )   &    |-  ( ph  ->  ( abs `  ( B  -  C ) )  =  ( abs `  ( E  -  G ) ) )   &    |-  ( ph  ->  ( abs `  ( C  -  A ) )  =  ( abs `  ( G  -  D ) ) )   =>    |-  ( ph  ->  ( cos `  ( ( A  -  B ) F ( C  -  B ) ) )  =  ( cos `  (
 ( D  -  E ) F ( G  -  E ) ) ) )
 
Theoremaffineequiv 20118 Equivalence between two ways of expressing  B as an affine combination of  A and  C. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   =>    |-  ( ph  ->  ( B  =  ( ( D  x.  A )  +  ( ( 1  -  D )  x.  C ) )  <->  ( C  -  B )  =  ( D  x.  ( C  -  A ) ) ) )
 
Theoremaffineequiv2 20119 Equivalence between two ways of expressing  B as an affine combination of  A and  C. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   =>    |-  ( ph  ->  ( B  =  ( ( D  x.  A )  +  ( ( 1  -  D )  x.  C ) )  <->  ( B  -  A )  =  (
 ( 1  -  D )  x.  ( C  -  A ) ) ) )
 
Theoremangpieqvdlem 20120 Equivalence used in the proof of angpieqvd 20123. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  A  =/=  C )   =>    |-  ( ph  ->  (
 -u ( ( C  -  B )  /  ( A  -  B ) )  e.  RR+  <->  ( ( C  -  B )  /  ( C  -  A ) )  e.  (
 0 (,) 1 ) ) )
 
Theoremangpieqvdlem2 20121* Equivalence used in angpieqvd 20123. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  ( -u ( ( C  -  B )  /  ( A  -  B ) )  e.  RR+  <->  ( ( A  -  B ) F ( C  -  B ) )  =  pi ) )
 
Theoremangpined 20122* If the angle at ABC is  pi, then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  (
 ( ( A  -  B ) F ( C  -  B ) )  =  pi  ->  A  =/=  C ) )
 
Theoremangpieqvd 20123* The angle ABC is  pi iff B is a nontrivial convex combination of A and C, i.e., iff B is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  B  =/=  C )   =>    |-  ( ph  ->  (
 ( ( A  -  B ) F ( C  -  B ) )  =  pi  <->  E. w  e.  (
 0 (,) 1 ) B  =  ( ( w  x.  A )  +  ( ( 1  -  w )  x.  C ) ) ) )
 
Theoremchordthmlem 20124* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 20117 to observe that, since AMQ and BMQ have equal sides, the angles QMB and QMA must be equal. Since they are supplementary, both must be right. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  / 
 2 ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   &    |-  ( ph  ->  A  =/=  B )   &    |-  ( ph  ->  Q  =/=  M )   =>    |-  ( ph  ->  (
 ( Q  -  M ) F ( B  -  M ) )  e. 
 { ( pi  / 
 2 ) ,  -u ( pi  /  2 ) }
 )
 
Theoremchordthmlem2 20125* If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then QMP is a right angle. This is proven by reduction to the special case chordthmlem 20124, where P = B, and using angrtmuld 20101 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  RR )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  /  2 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  (
 ( 1  -  X )  x.  B ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   &    |-  ( ph  ->  P  =/=  M )   &    |-  ( ph  ->  Q  =/=  M )   =>    |-  ( ph  ->  (
 ( Q  -  M ) F ( P  -  M ) )  e. 
 { ( pi  / 
 2 ) ,  -u ( pi  /  2 ) }
 )
 
Theoremchordthmlem3 20126 If M is the midpoint of AB, AQ = BQ, and P is on the line AB, then PQ 2 = QM 2  + PM 2 . This follows from chordthmlem2 20125 and the Pythagorean theorem (pythag 20110) in the case where P and Q are unequal to M. If either P or Q equals M, the result is trivial. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  RR )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  / 
 2 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  ( (
 1  -  X )  x.  B ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   =>    |-  ( ph  ->  (
 ( abs `  ( P  -  Q ) ) ^
 2 )  =  ( ( ( abs `  ( Q  -  M ) ) ^ 2 )  +  ( ( abs `  ( P  -  M ) ) ^ 2 ) ) )
 
Theoremchordthmlem4 20127 If P is on the segment AB and M is the midpoint of AB, then PA  x. PB = BM 2  - PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity  X  x.  (
1  -  X )  =  ( 1  / 
2 ) 2  -  ( ( 1  /  2 )  -  X ) 2 . (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  X  e.  (
 0 [,] 1 ) )   &    |-  ( ph  ->  M  =  ( ( A  +  B )  /  2
 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  ( ( 1  -  X )  x.  B ) ) )   =>    |-  ( ph  ->  ( ( abs `  ( P  -  A ) )  x.  ( abs `  ( P  -  B ) ) )  =  ( ( ( abs `  ( B  -  M ) ) ^ 2 )  -  ( ( abs `  ( P  -  M ) ) ^ 2 ) ) )
 
Theoremchordthmlem5 20128 If P is on the segment AB and AQ = BQ, then PA  x. PB = BQ 2  - PQ 2 . This follows from two uses of chordthmlem3 20126 to show that PQ 2 = QM 2  + PM 2 and BQ 2 = QM 2  + BM 2 , so BQ 2  - PQ 2 = (QM 2  + BM 2 )  - (QM 2  + PM 2 ) = BM 2  - PM 2 , which equals PA  x. PB by chordthmlem4 20127. (Contributed by David Moews, 28-Feb-2017.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  ( 0 [,] 1
 ) )   &    |-  ( ph  ->  P  =  ( ( X  x.  A )  +  ( ( 1  -  X )  x.  B ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   =>    |-  ( ph  ->  (
 ( abs `  ( P  -  A ) )  x.  ( abs `  ( P  -  B ) ) )  =  ( ( ( abs `  ( B  -  Q ) ) ^ 2 )  -  ( ( abs `  ( P  -  Q ) ) ^ 2 ) ) )
 
Theoremchordthm 20129* The intersecting chords theorem. If points A, B, C, and D lie on a circle (with center Q, say), and the point P is on the interior of the segments AB and CD, then the two products of lengths PA  x. PB and PC  x. PD are equal. The Euclidean plane is identified with the complex plane, and the fact that P is on AB and on CD is expressed by the hypothesis that the angles APB and CPD are equal to  pi. The result is proven by using chordthmlem5 20128 twice to show that PA  x. PB and PC  x. PD both equal BQ 2  - PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. (Contributed by David Moews, 28-Feb-2017.)
 |-  F  =  ( x  e.  ( CC  \  { 0 } ) ,  y  e.  ( CC  \  { 0 } )  |->  ( Im `  ( log `  ( y  /  x ) ) ) )   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  A  =/=  P )   &    |-  ( ph  ->  B  =/=  P )   &    |-  ( ph  ->  C  =/=  P )   &    |-  ( ph  ->  D  =/=  P )   &    |-  ( ph  ->  ( ( A  -  P ) F ( B  -  P ) )  =  pi )   &    |-  ( ph  ->  ( ( C  -  P ) F ( D  -  P ) )  =  pi )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( B  -  Q ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( C  -  Q ) ) )   &    |-  ( ph  ->  ( abs `  ( A  -  Q ) )  =  ( abs `  ( D  -  Q ) ) )   =>    |-  ( ph  ->  (
 ( abs `  ( P  -  A ) )  x.  ( abs `  ( P  -  B ) ) )  =  ( ( abs `  ( P  -  C ) )  x.  ( abs `  ( P  -  D ) ) ) )
 
13.3.6  Solutions of quadratic, cubic, and quartic equations
 
Theoremquad2 20130 The quadratic equation, without specifying the particular branch  D to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  ( D ^ 2 )  =  ( ( B ^
 2 )  -  (
 4  x.  ( A  x.  C ) ) ) )   =>    |-  ( ph  ->  (
 ( ( A  x.  ( X ^ 2 ) )  +  ( ( B  x.  X )  +  C ) )  =  0  <->  ( X  =  ( ( -u B  +  D )  /  (
 2  x.  A ) )  \/  X  =  ( ( -u B  -  D )  /  (
 2  x.  A ) ) ) ) )
 
Theoremquad 20131 The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  D  =  ( ( B ^ 2
 )  -  ( 4  x.  ( A  x.  C ) ) ) )   =>    |-  ( ph  ->  (
 ( ( A  x.  ( X ^ 2 ) )  +  ( ( B  x.  X )  +  C ) )  =  0  <->  ( X  =  ( ( -u B  +  ( sqr `  D ) )  /  (
 2  x.  A ) )  \/  X  =  ( ( -u B  -  ( sqr `  D ) )  /  (
 2  x.  A ) ) ) ) )
 
Theorem1cubrlem 20132 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ( -u 1  ^ c  ( 2  /  3 ) )  =  ( ( -u 1  +  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 )  /\  (
 ( -u 1  ^ c  ( 2  /  3
 ) ) ^ 2
 )  =  ( (
 -u 1  -  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) )
 
Theorem1cubr 20133 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  R  =  { 1 ,  ( ( -u 1  +  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) ,  (
 ( -u 1  -  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) }   =>    |-  ( A  e.  R 
 <->  ( A  e.  CC  /\  ( A ^ 3
 )  =  1 ) )
 
Theoremdcubic1lem 20134 Lemma for dcubic1 20136 and dcubic2 20135: simplify the cubic equation under the substitution  X  =  U  -  M  /  U. (Contributed by Mario Carneiro, 26-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  U  e.  CC )   &    |-  ( ph  ->  U  =/=  0 )   &    |-  ( ph  ->  X  =  ( U  -  ( M  /  U ) ) )   =>    |-  ( ph  ->  (
 ( ( X ^
 3 )  +  (
 ( P  x.  X )  +  Q )
 )  =  0  <->  ( ( ( U ^ 3 ) ^ 2 )  +  ( ( Q  x.  ( U ^ 3 ) )  -  ( M ^ 3 ) ) )  =  0 ) )
 
Theoremdcubic2 20135* Reverse direction of dcubic 20137. Given a solution  U to the "substitution" quadratic equation  X  =  U  -  M  /  U, show that  X is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  U  e.  CC )   &    |-  ( ph  ->  U  =/=  0 )   &    |-  ( ph  ->  X  =  ( U  -  ( M  /  U ) ) )   &    |-  ( ph  ->  ( ( X ^ 3
 )  +  ( ( P  x.  X )  +  Q ) )  =  0 )   =>    |-  ( ph  ->  E. r  e.  CC  (
 ( r ^ 3
 )  =  1  /\  X  =  ( (
 r  x.  T )  -  ( M  /  ( r  x.  T ) ) ) ) )
 
Theoremdcubic1 20136 Forward direction of dcubic 20137: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   &    |-  ( ph  ->  X  =  ( T  -  ( M 
 /  T ) ) )   =>    |-  ( ph  ->  (
 ( X ^ 3
 )  +  ( ( P  x.  X )  +  Q ) )  =  0 )
 
Theoremdcubic 20137* Solutions to the depressed cubic, a special case of cubic 20140. (The definitions of  M ,  N ,  G ,  T here differ from mcubic 20138 by scale factors of  -u 9,  5 4,  5 4 and  -u 2
7 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( G  -  N ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  +  ( M ^ 3 ) ) )   &    |-  ( ph  ->  M  =  ( P  / 
 3 ) )   &    |-  ( ph  ->  N  =  ( Q  /  2 ) )   &    |-  ( ph  ->  T  =/=  0 )   =>    |-  ( ph  ->  ( ( ( X ^
 3 )  +  (
 ( P  x.  X )  +  Q )
 )  =  0  <->  E. r  e.  CC  ( ( r ^
 3 )  =  1 
 /\  X  =  ( ( r  x.  T )  -  ( M  /  ( r  x.  T ) ) ) ) ) )
 
Theoremmcubic 20138* Solutions to a monic cubic equation, a special case of cubic 20140. (Contributed by Mario Carneiro, 24-Apr-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^
 3 )  =  ( ( N  +  G )  /  2 ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^ 2 )  -  ( 4  x.  ( M ^ 3 ) ) ) )   &    |-  ( ph  ->  M  =  ( ( B ^ 2 )  -  ( 3  x.  C ) ) )   &    |-  ( ph  ->  N  =  ( ( ( 2  x.  ( B ^ 3
 ) )  -  (
 9  x.  ( B  x.  C ) ) )  +  (; 2 7  x.  D ) ) )   &    |-  ( ph  ->  T  =/=  0
 )   =>    |-  ( ph  ->  (
 ( ( ( X ^ 3 )  +  ( B  x.  ( X ^ 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  E. r  e.  CC  ( ( r ^
 3 )  =  1 
 /\  X  =  -u ( ( ( B  +  ( r  x.  T ) )  +  ( M  /  (
 r  x.  T ) ) )  /  3
 ) ) ) )
 
Theoremcubic2 20139* The solution to the general cubic equation, for arbitrary choices  G and  T of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  e.  CC )   &    |-  ( ph  ->  ( T ^ 3 )  =  ( ( N  +  G )  / 
 2 ) )   &    |-  ( ph  ->  G  e.  CC )   &    |-  ( ph  ->  ( G ^ 2 )  =  ( ( N ^
 2 )  -  (
 4  x.  ( M ^ 3 ) ) ) )   &    |-  ( ph  ->  M  =  ( ( B ^ 2 )  -  ( 3  x.  ( A  x.  C ) ) ) )   &    |-  ( ph  ->  N  =  ( ( ( 2  x.  ( B ^ 3 ) )  -  ( ( 9  x.  A )  x.  ( B  x.  C ) ) )  +  (; 2 7  x.  ( ( A ^ 2 )  x.  D ) ) ) )   &    |-  ( ph  ->  T  =/=  0 )   =>    |-  ( ph  ->  ( ( ( ( A  x.  ( X ^
 3 ) )  +  ( B  x.  ( X ^ 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  E. r  e.  CC  ( ( r ^
 3 )  =  1 
 /\  X  =  -u ( ( ( B  +  ( r  x.  T ) )  +  ( M  /  (
 r  x.  T ) ) )  /  (
 3  x.  A ) ) ) ) )
 
Theoremcubic 20140* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 3689 to convert the existential quantifier to a triple disjunction. (Contributed by Mario Carneiro, 26-Apr-2015.)
 |-  R  =  { 1 ,  ( ( -u 1  +  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) ,  (
 ( -u 1  -  ( _i  x.  ( sqr `  3
 ) ) )  / 
 2 ) }   &    |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  A  =/=  0 )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  T  =  ( ( ( N  +  ( sqr `  G )
 )  /  2 )  ^ c  ( 1  /  3 ) ) )   &    |-  ( ph  ->  G  =  ( ( N ^ 2 )  -  ( 4  x.  ( M ^ 3 ) ) ) )   &    |-  ( ph  ->  M  =  ( ( B ^ 2 )  -  ( 3  x.  ( A  x.  C ) ) ) )   &    |-  ( ph  ->  N  =  ( ( ( 2  x.  ( B ^ 3 ) )  -  ( ( 9  x.  A )  x.  ( B  x.  C ) ) )  +  (; 2 7  x.  ( ( A ^ 2 )  x.  D ) ) ) )   &    |-  ( ph  ->  M  =/=  0 )   =>    |-  ( ph  ->  ( ( ( ( A  x.  ( X ^
 3 ) )  +  ( B  x.  ( X ^ 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  E. r  e.  R  X  =  -u ( ( ( B  +  (
 r  x.  T ) )  +  ( M 
 /  ( r  x.  T ) ) ) 
 /  ( 3  x.  A ) ) ) )
 
Theorembinom4 20141 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 12283, but it turns out to be just as much work to put it into this form after clearing all the sums and calculating binomial coefficients.) (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ( A  e.  CC  /\  B  e.  CC )  ->  ( ( A  +  B ) ^
 4 )  =  ( ( ( A ^
 4 )  +  (
 4  x.  ( ( A ^ 3 )  x.  B ) ) )  +  ( ( 6  x.  ( ( A ^ 2 )  x.  ( B ^
 2 ) ) )  +  ( ( 4  x.  ( A  x.  ( B ^ 3 ) ) )  +  ( B ^ 4 ) ) ) ) )
 
Theoremdquartlem1 20142 Lemma for dquart 20144. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  S  e.  CC )   &    |-  ( ph  ->  M  =  ( ( 2  x.  S ) ^
 2 ) )   &    |-  ( ph  ->  M  =/=  0
 )   &    |-  ( ph  ->  I  e.  CC )   &    |-  ( ph  ->  ( I ^ 2 )  =  ( ( -u ( S ^ 2 )  -  ( B  / 
 2 ) )  +  ( ( C  / 
 4 )  /  S ) ) )   =>    |-  ( ph  ->  ( ( ( ( X ^ 2 )  +  ( ( M  +  B )  /  2
 ) )  +  (
 ( ( ( M 
 /  2 )  x.  X )  -  ( C  /  4 ) ) 
 /  S ) )  =  0  <->  ( X  =  ( -u S  +  I
 )  \/  X  =  ( -u S  -  I
 ) ) ) )
 
Theoremdquartlem2 20143 Lemma for dquart 20144. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  S  e.  CC )   &    |-  ( ph  ->  M  =  ( ( 2  x.  S ) ^
 2 ) )   &    |-  ( ph  ->  M  =/=  0
 )   &    |-  ( ph  ->  I  e.  CC )   &    |-  ( ph  ->  ( I ^ 2 )  =  ( ( -u ( S ^ 2 )  -  ( B  / 
 2 ) )  +  ( ( C  / 
 4 )  /  S ) ) )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  (
 ( ( M ^
 3 )  +  (
 ( 2  x.  B )  x.  ( M ^
 2 ) ) )  +  ( ( ( ( B ^ 2
 )  -  ( 4  x.  D ) )  x.  M )  +  -u ( C ^ 2
 ) ) )  =  0 )   =>    |-  ( ph  ->  (
 ( ( ( M  +  B )  / 
 2 ) ^ 2
 )  -  ( ( ( C ^ 2
 )  /  4 )  /  M ) )  =  D )
 
Theoremdquart 20144 Solve a depressed quartic equation. To eliminate  S, which is the square root of a solution  M to the resolvent cubic equation, apply cubic 20140 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  S  e.  CC )   &    |-  ( ph  ->  M  =  ( ( 2  x.  S ) ^
 2 ) )   &    |-  ( ph  ->  M  =/=  0
 )   &    |-  ( ph  ->  I  e.  CC )   &    |-  ( ph  ->  ( I ^ 2 )  =  ( ( -u ( S ^ 2 )  -  ( B  / 
 2 ) )  +  ( ( C  / 
 4 )  /  S ) ) )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  (
 ( ( M ^
 3 )  +  (
 ( 2  x.  B )  x.  ( M ^
 2 ) ) )  +  ( ( ( ( B ^ 2
 )  -  ( 4  x.  D ) )  x.  M )  +  -u ( C ^ 2
 ) ) )  =  0 )   &    |-  ( ph  ->  J  e.  CC )   &    |-  ( ph  ->  ( J ^
 2 )  =  ( ( -u ( S ^
 2 )  -  ( B  /  2 ) )  -  ( ( C 
 /  4 )  /  S ) ) )   =>    |-  ( ph  ->  ( (
 ( ( X ^
 4 )  +  ( B  x.  ( X ^
 2 ) ) )  +  ( ( C  x.  X )  +  D ) )  =  0  <->  ( ( X  =  ( -u S  +  I )  \/  X  =  ( -u S  -  I
 ) )  \/  ( X  =  ( S  +  J )  \/  X  =  ( S  -  J ) ) ) ) )
 
Theoremquart1cl 20145 Closure lemmas for quart 20152. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   =>    |-  ( ph  ->  ( P  e.  CC  /\  Q  e.  CC  /\  R  e.  CC ) )
 
Theoremquart1lem 20146 Lemma for quart1 20147. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  Y  =  ( X  +  ( A 
 /  4 ) ) )   =>    |-  ( ph  ->  D  =  ( ( ( ( A ^ 4 ) 
 / ;; 2 5 6 )  +  ( P  x.  ( ( A 
 /  4 ) ^
 2 ) ) )  +  ( ( Q  x.  ( A  / 
 4 ) )  +  R ) ) )
 
Theoremquart1 20147 Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  Y  =  ( X  +  ( A 
 /  4 ) ) )   =>    |-  ( ph  ->  (
 ( ( X ^
 4 )  +  ( A  x.  ( X ^
 3 ) ) )  +  ( ( B  x.  ( X ^
 2 ) )  +  ( ( C  x.  X )  +  D ) ) )  =  ( ( ( Y ^ 4 )  +  ( P  x.  ( Y ^ 2 ) ) )  +  ( ( Q  x.  Y )  +  R ) ) )
 
Theoremquartlem1 20148 Lemma for quart 20152. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  P  e.  CC )   &    |-  ( ph  ->  Q  e.  CC )   &    |-  ( ph  ->  R  e.  CC )   &    |-  ( ph  ->  U  =  ( ( P ^
 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   =>    |-  ( ph  ->  ( U  =  ( (
 ( 2  x.  P ) ^ 2 )  -  ( 3  x.  (
 ( P ^ 2
 )  -  ( 4  x.  R ) ) ) )  /\  V  =  ( ( ( 2  x.  ( ( 2  x.  P ) ^
 3 ) )  -  ( 9  x.  (
 ( 2  x.  P )  x.  ( ( P ^ 2 )  -  ( 4  x.  R ) ) ) ) )  +  (; 2 7  x.  -u ( Q ^ 2 ) ) ) ) )
 
Theoremquartlem2 20149 Closure lemmas for quart 20152. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   =>    |-  ( ph  ->  ( U  e.  CC  /\  V  e.  CC  /\  W  e.  CC ) )
 
Theoremquartlem3 20150 Closure lemmas for quart 20152. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   &    |-  ( ph  ->  S  =  ( ( sqr `  M )  /  2
 ) )   &    |-  ( ph  ->  M  =  -u ( ( ( ( 2  x.  P )  +  T )  +  ( U  /  T ) )  /  3
 ) )   &    |-  ( ph  ->  T  =  ( ( ( V  +  W ) 
 /  2 )  ^ c  ( 1  /  3
 ) ) )   &    |-  ( ph  ->  T  =/=  0
 )   =>    |-  ( ph  ->  ( S  e.  CC  /\  M  e.  CC  /\  T  e.  CC ) )
 
Theoremquartlem4 20151 Closure lemmas for quart 20152. (Contributed by Mario Carneiro, 7-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   &    |-  ( ph  ->  S  =  ( ( sqr `  M )  /  2
 ) )   &    |-  ( ph  ->  M  =  -u ( ( ( ( 2  x.  P )  +  T )  +  ( U  /  T ) )  /  3
 ) )   &    |-  ( ph  ->  T  =  ( ( ( V  +  W ) 
 /  2 )  ^ c  ( 1  /  3
 ) ) )   &    |-  ( ph  ->  T  =/=  0
 )   &    |-  ( ph  ->  M  =/=  0 )   &    |-  ( ph  ->  I  =  ( sqr `  (
 ( -u ( S ^
 2 )  -  ( P  /  2 ) )  +  ( ( Q 
 /  4 )  /  S ) ) ) )   &    |-  ( ph  ->  J  =  ( sqr `  (
 ( -u ( S ^
 2 )  -  ( P  /  2 ) )  -  ( ( Q 
 /  4 )  /  S ) ) ) )   =>    |-  ( ph  ->  ( S  =/=  0  /\  I  e.  CC  /\  J  e.  CC ) )
 
Theoremquart 20152 The quartic equation, writing out all roots using square and cube root functions so that only direct substitutions remain, and we can actually claim to have a "quartic equation". Naturally, this theorem is ridiculously long (see quartfull 23092) if all the substitutions are performed. (Contributed by Mario Carneiro, 6-May-2015.)
 |-  ( ph  ->  A  e.  CC )   &    |-  ( ph  ->  B  e.  CC )   &    |-  ( ph  ->  C  e.  CC )   &    |-  ( ph  ->  D  e.  CC )   &    |-  ( ph  ->  X  e.  CC )   &    |-  ( ph  ->  E  =  -u ( A  /  4
 ) )   &    |-  ( ph  ->  P  =  ( B  -  ( ( 3  / 
 8 )  x.  ( A ^ 2 ) ) ) )   &    |-  ( ph  ->  Q  =  ( ( C  -  ( ( A  x.  B )  / 
 2 ) )  +  ( ( A ^
 3 )  /  8
 ) ) )   &    |-  ( ph  ->  R  =  ( ( D  -  (
 ( C  x.  A )  /  4 ) )  +  ( ( ( ( A ^ 2
 )  x.  B ) 
 / ; 1 6 )  -  ( ( 3  / ;; 2 5 6 )  x.  ( A ^
 4 ) ) ) ) )   &    |-  ( ph  ->  U  =  ( ( P ^ 2 )  +  (; 1 2  x.  R ) ) )   &    |-  ( ph  ->  V  =  ( ( -u ( 2  x.  ( P ^ 3 ) )  -  (; 2 7  x.  ( Q ^ 2 ) ) )  +  (; 7 2  x.  ( P  x.  R ) ) ) )   &    |-  ( ph  ->  W  =  ( sqr `  (
 ( V ^ 2
 )  -  ( 4  x.  ( U ^
 3 ) ) ) ) )   &    |-  ( ph  ->  S  =  ( ( sqr `  M )  /  2
 ) )   &    |-  ( ph  ->  M  =  -u ( ( ( ( 2  x.  P )  +  T )  +  ( U  /  T ) )  /  3
 ) )   &    |-  ( ph  ->  T  =  ( ( ( V  +  W ) 
 /  2 )  ^ c  ( 1  /  3
 ) ) )   &    |-  ( ph  ->  T  =/=  0
 )   &    |-  ( ph  ->  M  =/=  0 )   &    |-  ( ph  ->  I  =  ( sqr `  (
 ( -u ( S ^
 2 )  -  ( P  /  2 ) )  +  ( ( Q 
 /  4 )  /  S ) ) ) )   &    |-  ( ph  ->  J  =  ( sqr `  (
 ( -u ( S ^
 2 )  -  ( P  /  2 ) )  -  ( ( Q 
 /  4 )  /  S ) ) ) )   =>    |-  ( ph  ->  (
 ( ( ( X ^ 4 )  +  ( A  x.  ( X ^ 3 ) ) )  +  ( ( B  x.  ( X ^ 2 ) )  +  ( ( C  x.  X )  +  D ) ) )  =  0  <->  ( ( X  =  ( ( E  -  S )  +  I )  \/  X  =  ( ( E  -  S )  -  I
 ) )  \/  ( X  =  ( ( E  +  S )  +  J )  \/  X  =  ( ( E  +  S )  -  J ) ) ) ) )
 
13.3.7  Inverse trigonometric functions
 
Syntaxcasin 20153 The arcsine function.
 class arcsin
 
Syntaxcacos 20154 The arccosine function.
 class arccos
 
Syntaxcatan 20155 The arctangent function.
 class arctan
 
Definitiondf-asin 20156 Define the arcsine function. Because  sin is not a one-to-one function, the literal inverse  `' sin is not a function. Rather than attempt to find the right domain on which to restrict  sin in order to get a total function, we just define it in terms of  log, which we already know is total (except at  0). There are branch points at  -u 1 and  1 (at which the function is defined), and branch cuts along the real line not between  -u
1 and  1, which is to say  (  -oo ,  -u 1 )  u.  (
1 ,  +oo ). (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arcsin  =  ( x  e.  CC  |->  ( -u _i  x.  ( log `  ( ( _i 
 x.  x )  +  ( sqr `  ( 1  -  ( x ^ 2
 ) ) ) ) ) ) )
 
Definitiondf-acos 20157 Define the arccosine function. See also remarks for df-asin 20156. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely  (  -oo ,  -u
1 )  u.  (
1 ,  +oo ). (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arccos  =  ( x  e.  CC  |->  ( ( pi  / 
 2 )  -  (arcsin `  x ) ) )
 
Definitiondf-atan 20158 Define the arctangent function. See also remarks for df-asin 20156. Unlike arcsin and arccos, this function is not defined everywhere, because  tan ( z )  =/=  pm _i for all  z  e.  CC. For all other  z, there is a formula for arctan ( z ) in terms of  log, and we take that as the definition. Branch points are at  pm _i; branch cuts are on the pure imaginary axis not between  -u _i and  _i, which is to say  { z  e.  CC  |  ( _i  x.  z )  e.  (  -oo ,  -u
1 )  u.  (
1 ,  +oo ) }. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arctan  =  ( x  e.  ( CC  \  { -u _i ,  _i } )  |->  ( ( _i  /  2
 )  x.  ( ( log `  ( 1  -  ( _i  x.  x ) ) )  -  ( log `  ( 1  +  ( _i  x.  x ) ) ) ) ) )
 
Theoremasinlem 20159 The argument to the logarithm in df-asin 20156 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  ( ( _i  x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2 ) ) ) )  =/=  0 )
 
Theoremasinlem2 20160 The argument to the logarithm in df-asin 20156 has the property that replacing  A with  -u A in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  ( ( ( _i 
 x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2
 ) ) ) )  x.  ( ( _i 
 x.  -u A )  +  ( sqr `  ( 1  -  ( -u A ^ 2
 ) ) ) ) )  =  1 )
 
Theoremasinlem3a 20161 Lemma for asinlem3 20162. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Im `  A )  <_  0 ) 
 ->  0  <_  ( Re
 `  ( ( _i 
 x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2
 ) ) ) ) ) )
 
Theoremasinlem3 20162 The argument to the logarithm in df-asin 20156 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  0  <_  ( Re `  ( ( _i  x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2 ) ) ) ) ) )
 
Theoremasinf 20163 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arcsin : CC --> CC
 
Theoremasincl 20164 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arcsin `  A )  e.  CC )
 
Theoremacosf 20165 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arccos : CC --> CC
 
Theoremacoscl 20166 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arccos `  A )  e.  CC )
 
Theorematandm 20167 Since the property is a little lengthy, we abbreviate  A  e.  CC  /\  A  =/=  -u _i  /\  A  =/=  _i as  A  e.  dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  A  =/=  -u _i  /\  A  =/=  _i ) )
 
Theorematandm2 20168 This form of atandm 20167 is a bit more useful for showing that the logarithms in df-atan 20158 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  ( 1  -  ( _i  x.  A ) )  =/=  0  /\  ( 1  +  ( _i  x.  A ) )  =/=  0 ) )
 
Theorematandm3 20169 A compact form of atandm 20167. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  ( A ^
 2 )  =/=  -u 1
 ) )
 
Theorematandm4 20170 A compact form of atandm 20167. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  <->  ( A  e.  CC  /\  ( 1  +  ( A ^ 2
 ) )  =/=  0
 ) )
 
Theorematanf 20171 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |- arctan : ( CC  \  { -u _i ,  _i } ) --> CC
 
Theorematancl 20172 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  ->  (arctan `  A )  e. 
 CC )
 
Theoremasinval 20173 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arcsin `  A )  =  ( -u _i  x.  ( log `  ( ( _i 
 x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2
 ) ) ) ) ) ) )
 
Theoremacosval 20174 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  (arccos `  A )  =  ( ( pi  / 
 2 )  -  (arcsin `  A ) ) )
 
Theorematanval 20175 Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  ->  (arctan `  A )  =  ( ( _i  / 
 2 )  x.  (
 ( log `  ( 1  -  ( _i  x.  A ) ) )  -  ( log `  ( 1  +  ( _i  x.  A ) ) ) ) ) )
 
Theorematanre 20176 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  RR  ->  A  e.  dom arctan )
 
Theoremasinneg 20177 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  (arcsin `  -u A )  =  -u (arcsin `  A ) )
 
Theoremacosneg 20178 The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  (arccos `  -u A )  =  ( pi  -  (arccos `  A ) ) )
 
Theoremefiasin 20179 The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  CC  ->  ( exp `  ( _i  x.  (arcsin `  A ) ) )  =  ( ( _i  x.  A )  +  ( sqr `  ( 1  -  ( A ^ 2 ) ) ) ) )
 
Theoremsinasin 20180 The arcsine function is an inverse to  sin. This is the main property that justifies the notation arcsin or  sin
^ -u 1. Because  sin is not an injection, the other converse identity asinsin 20183 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  ( sin `  (arcsin `  A ) )  =  A )
 
Theoremcosacos 20181 The arccosine function is an inverse to  cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
 |-  ( A  e.  CC  ->  ( cos `  (arccos `  A ) )  =  A )
 
Theoremasinsinlem 20182 Lemma for asinsin 20183. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  e.  ( -u ( pi  /  2
 ) (,) ( pi  / 
 2 ) ) ) 
 ->  0  <  ( Re
 `  ( exp `  ( _i  x.  A ) ) ) )
 
Theoremasinsin 20183 The arcsine function composed with 
sin is equal to the identity. This plus sinasin 20180 allow us to view  sin and arcsin as inverse operations to each other. For ease of use, we have not defined precisely the correct domain of correctness of this identity; in addition to the main region described here it is also true for some points on the branch cuts, namely when  A  =  ( pi 
/  2 )  -  _i y for non-negative real  y and also symmetrically at  A  =  _i y  -  ( pi  / 
2 ). In particular, when restricted to reals this identity extends to the closed interval  [ -u (
pi  /  2 ) ,  ( pi  / 
2 ) ], not just the open interval (see reasinsin 20187). (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  e.  ( -u ( pi  /  2
 ) (,) ( pi  / 
 2 ) ) ) 
 ->  (arcsin `  ( sin `  A ) )  =  A )
 
Theoremacoscos 20184 The arccosine function is an inverse to  cos. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  ( Re `  A )  e.  (
 0 (,) pi ) ) 
 ->  (arccos `  ( cos `  A ) )  =  A )
 
Theoremasin1 20185 The arcsine of  1 is  pi  / 
2. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  (arcsin `  1 )  =  ( pi  /  2
 )
 
Theoremacos1 20186 The arcsine of  1 is  pi  / 
2. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  (arccos `  1 )  =  0
 
Theoremreasinsin 20187 The arcsine function composed with 
sin is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u ( pi  /  2
 ) [,] ( pi  / 
 2 ) )  ->  (arcsin `  ( sin `  A ) )  =  A )
 
Theoremasinsinb 20188 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  B  e.  CC  /\  ( Re `  B )  e.  ( -u ( pi  /  2 ) (,) ( pi  /  2
 ) ) )  ->  ( (arcsin `  A )  =  B  <->  ( sin `  B )  =  A )
 )
 
Theoremacoscosb 20189 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( ( A  e.  CC  /\  B  e.  CC  /\  ( Re `  B )  e.  ( 0 (,) pi ) )  ->  ( (arccos `  A )  =  B  <->  ( cos `  B )  =  A )
 )
 
Theoremasinbnd 20190 The arcsine function has range within a vertical strip of the complex plane with real part between  -u pi  /  2 and  pi  /  2. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( Re `  (arcsin `  A ) )  e.  ( -u ( pi  / 
 2 ) [,] ( pi  /  2 ) ) )
 
Theoremacosbnd 20191 The arccosine function has range within a vertical strip of the complex plane with real part between  0 and  pi. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( Re `  (arccos `  A ) )  e.  ( 0 [,] pi ) )
 
Theoremasinrebnd 20192 Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u 1 [,] 1 ) 
 ->  (arcsin `  A )  e.  ( -u ( pi  / 
 2 ) [,] ( pi  /  2 ) ) )
 
Theoremasinrecl 20193 The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u 1 [,] 1 ) 
 ->  (arcsin `  A )  e.  RR )
 
Theoremacosrecl 20194 The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  ( -u 1 [,] 1 ) 
 ->  (arccos `  A )  e.  RR )
 
Theoremcosasin 20195 The cosine of the arcsine of  A is  sqr ( 1  -  A ^ 2 ). (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( cos `  (arcsin `  A ) )  =  ( sqr `  (
 1  -  ( A ^ 2 ) ) ) )
 
Theoremsinacos 20196 The sine of the arccosine of  A is  sqr ( 1  -  A ^ 2 ). (Contributed by Mario Carneiro, 2-Apr-2015.)
 |-  ( A  e.  CC  ->  ( sin `  (arccos `  A ) )  =  ( sqr `  (
 1  -  ( A ^ 2 ) ) ) )
 
Theorematandmneg 20197 The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  -u A  e.  dom arctan )
 
Theorematanneg 20198 The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.)
 |-  ( A  e.  dom arctan  ->  (arctan `  -u A )  =  -u (arctan `  A )
 )
 
Theorematan0 20199 The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  (arctan `  0 )  =  0
 
Theorematandmcj 20200 The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.)
 |-  ( A  e.  dom arctan  ->  ( * `  A )  e.  dom arctan )
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