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Theorem List for Metamath Proof Explorer - 20101-20200   *Has distinct variable group(s)
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Statement

13.3.5  Theorems of Pythagoras, isosceles triangles, and intersecting chords

Theoremangval 20101* Define the angle function, which takes two complex numbers, treated as vectors from the origin, and returns the angle between them, in the range . To convert from the geometry notation, , the measure of the angle with legs , where is more counterclockwise for positive angles, is represented by . (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremangcan 20102* Cancel a constant multiplier in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremangneg 20103* Cancel a negative sign in the angle function. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremangvald 20104* The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 20101. (Contributed by David Moews, 28-Feb-2017.)

Theoremangcld 20105* The (signed) angle between two vectors is in . Deduction form. (Contributed by David Moews, 28-Feb-2017.)

Theoremangrteqvd 20106* Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017.)

Theoremcosangneg2d 20107* The cosine of the angle between and is the negative of that between and . If A, B and C are collinear points, this implies that the cosines of DBA and DBC sum to zero, i.e., that DBA and DBC are supplementary. (Contributed by David Moews, 28-Feb-2017.)

Theoremangrtmuld 20108* Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.)

Theoremang180lem1 20109* Lemma for ang180 20114. Show that the "revolution number" is an integer, using efeq1 19893 to show that since the product of the three arguments is , the sum of the logarithms must be an integer multiple of away from . (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem2 20110* Lemma for ang180 20114. Show that the revolution number is strictly between and . Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 19929, but the resulting bound gives only for the upper bound. The case 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 must lie on the negative real axis, which is a contradiction because clearly if is negative then the other two are positive real. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem3 20111* Lemma for ang180 20114. Since ang180lem1 20109 shows that is an integer and ang180lem2 20110 shows that is strictly between and , it follows that , and these two cases correspond to the two possible values for . (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem4 20112* Lemma for ang180 20114. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180lem5 20113* Lemma for ang180 20114: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.)

Theoremang180 20114* The sum of angles in a triangle adds up to either or , 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.)

Theoremlawcoslem1 20115 Lemma for Law of Cosines lawcos 20116. 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.)

Theoremlawcos 20116* 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 is the signed angle construct (as used in ang180 20114), is the distance of line segment BC, is the distance of line segment AC, is the distance of line segment AB, and 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 20115 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 12429). The Pythagorean Theorem pythag 20117 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.)

Theorempythag 20117* 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 is the signed angle construct (as used in ang180 20114), is the distance of line segment BC, is the distance of line segment AC, is the distance of line segment AB (the hypotenuse), and is the distinguished (signed) right angle m/_ BCA. We use the law of cosines lawcos 20116 to prove this, along with simple trig facts like coshalfpi 19839 and cosneg 12429. (Contributed by David A. Wheeler, 13-Jun-2015.)

Theoremlogreclem 20118 Symmetry of the natural logarithm range by negation. Lemma for logrec 20119. (Contributed by Saveliy Skresanov, 27-Dec-2016.)

Theoremlogrec 20119 Logarithm of a reciprocal changes sign. (Contributed by Saveliy Skresanov, 28-Dec-2016.)

Theoremisosctrlem1 20120 Lemma for isosctr 20123. (Contributed by Saveliy Skresanov, 30-Dec-2016.)

Theoremisosctrlem2 20121 Lemma for isosctr 20123. 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.)

Theoremisosctrlem3 20122* Lemma for isosctr 20123. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)

Theoremisosctr 20123* Isosceles triangle theorem. (Contributed by Saveliy Skresanov, 1-Jan-2017.)

Theoremssscongptld 20124* 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 20116 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.)

Theoremaffineequiv 20125 Equivalence between two ways of expressing as an affine combination of and . (Contributed by David Moews, 28-Feb-2017.)

Theoremaffineequiv2 20126 Equivalence between two ways of expressing as an affine combination of and . (Contributed by David Moews, 28-Feb-2017.)

Theoremangpieqvdlem 20127 Equivalence used in the proof of angpieqvd 20130. (Contributed by David Moews, 28-Feb-2017.)

Theoremangpieqvdlem2 20128* Equivalence used in angpieqvd 20130. (Contributed by David Moews, 28-Feb-2017.)

Theoremangpined 20129* If the angle at ABC is , then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)

Theoremangpieqvd 20130* The angle ABC is 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.)

Theoremchordthmlem 20131* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 20124 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.)

Theoremchordthmlem2 20132* 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 20131, where P = B, and using angrtmuld 20108 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem3 20133 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 20132 and the Pythagorean theorem (pythag 20117) 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.)

Theoremchordthmlem4 20134 If P is on the segment AB and M is the midpoint of AB, then PA PB = BM 2 PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity 2 2 . (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthmlem5 20135 If P is on the segment AB and AQ = BQ, then PA PB = BQ 2 PQ 2 . This follows from two uses of chordthmlem3 20133 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 PB by chordthmlem4 20134. (Contributed by David Moews, 28-Feb-2017.)

Theoremchordthm 20136* 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 PB and PC 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 . The result is proven by using chordthmlem5 20135 twice to show that PA PB and PC 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.)

13.3.6  Solutions of quadratic, cubic, and quartic equations

Theoremquad2 20137 The quadratic equation, without specifying the particular branch to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theorem1cubrlem 20139 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theorem1cubr 20140 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)

Theoremdcubic1lem 20141 Lemma for dcubic1 20143 and dcubic2 20142: simplify the cubic equation under the substitution . (Contributed by Mario Carneiro, 26-Apr-2015.)

Theoremdcubic2 20142* Reverse direction of dcubic 20144. Given a solution to the "substitution" quadratic equation , show that is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)

Theoremdcubic1 20143 Forward direction of dcubic 20144: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)

Theoremdcubic 20144* Solutions to the depressed cubic, a special case of cubic 20147. (The definitions of here differ from mcubic 20145 by scale factors of , , and respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)

Theoremmcubic 20145* Solutions to a monic cubic equation, a special case of cubic 20147. (Contributed by Mario Carneiro, 24-Apr-2015.)
;

Theoremcubic2 20146* The solution to the general cubic equation, for arbitrary choices and of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.)
;

Theoremcubic 20147* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 3691 to convert the existential quantifier to a triple disjunction. (Contributed by Mario Carneiro, 26-Apr-2015.)
;

Theorembinom4 20148 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 12290, 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.)

Theoremdquartlem1 20149 Lemma for dquart 20151. (Contributed by Mario Carneiro, 6-May-2015.)

Theoremdquartlem2 20150 Lemma for dquart 20151. (Contributed by Mario Carneiro, 6-May-2015.)

Theoremdquart 20151 Solve a depressed quartic equation. To eliminate , which is the square root of a solution to the resolvent cubic equation, apply cubic 20147 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)

Theoremquart1cl 20152 Closure lemmas for quart 20159. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;

Theoremquart1lem 20153 Lemma for quart1 20154. (Contributed by Mario Carneiro, 6-May-2015.)
; ;;                      ;;

Theoremquart1 20154 Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
; ;;

Theoremquartlem1 20155 Lemma for quart 20159. (Contributed by Mario Carneiro, 6-May-2015.)
;        ; ;        ;

Theoremquartlem2 20156 Closure lemmas for quart 20159. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;        ;        ; ;

Theoremquartlem3 20157 Closure lemmas for quart 20159. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;        ;        ; ;

Theoremquartlem4 20158 Closure lemmas for quart 20159. (Contributed by Mario Carneiro, 7-May-2015.)
; ;;        ;        ; ;

Theoremquart 20159 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 23688) if all the substitutions are performed. (Contributed by Mario Carneiro, 6-May-2015.)
; ;;        ;        ; ;

13.3.7  Inverse trigonometric functions

Syntaxcasin 20160 The arcsine function.
arcsin

Syntaxcacos 20161 The arccosine function.
arccos

Syntaxcatan 20162 The arctangent function.
arctan

Definitiondf-asin 20163 Define the arcsine function. Because is not a one-to-one function, the literal inverse is not a function. Rather than attempt to find the right domain on which to restrict in order to get a total function, we just define it in terms of , which we already know is total (except at ). There are branch points at and (at which the function is defined), and branch cuts along the real line not between and , which is to say . (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Definitiondf-acos 20164 Define the arccosine function. See also remarks for df-asin 20163. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely . (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos arcsin

Definitiondf-atan 20165 Define the arctangent function. See also remarks for df-asin 20163. Unlike arcsin and arccos, this function is not defined everywhere, because for all . For all other , there is a formula for arctan in terms of , and we take that as the definition. Branch points are at ; branch cuts are on the pure imaginary axis not between and , which is to say . (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theoremasinlem 20166 The argument to the logarithm in df-asin 20163 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)

Theoremasinlem2 20167 The argument to the logarithm in df-asin 20163 has the property that replacing with in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)

Theoremasinlem3a 20168 Lemma for asinlem3 20169. (Contributed by Mario Carneiro, 1-Apr-2015.)

Theoremasinlem3 20169 The argument to the logarithm in df-asin 20163 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)

Theoremasinf 20170 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremasincl 20171 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremacosf 20172 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos

Theoremacoscl 20173 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos

Theorematandm 20174 Since the property is a little lengthy, we abbreviate as arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematandm2 20175 This form of atandm 20174 is a bit more useful for showing that the logarithms in df-atan 20165 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematandm3 20176 A compact form of atandm 20174. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematandm4 20177 A compact form of atandm 20174. (Contributed by Mario Carneiro, 3-Apr-2015.)
arctan

Theorematanf 20178 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theorematancl 20179 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan arctan

Theoremasinval 20180 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremacosval 20181 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos arcsin

Theorematanval 20182 Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan arctan

Theorematanre 20183 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan

Theoremasinneg 20184 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
arcsin arcsin

Theoremacosneg 20185 The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
arccos arccos

Theoremefiasin 20186 The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin

Theoremsinasin 20187 The arcsine function is an inverse to . This is the main property that justifies the notation arcsin or . Because is not an injection, the other converse identity asinsin 20190 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
arcsin

Theoremcosacos 20188 The arccosine function is an inverse to . (Contributed by Mario Carneiro, 1-Apr-2015.)
arccos

Theoremasinsinlem 20189 Lemma for asinsin 20190. (Contributed by Mario Carneiro, 2-Apr-2015.)

Theoremasinsin 20190 The arcsine function composed with is equal to the identity. This plus sinasin 20187 allow us to view 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 for non-negative real and also symmetrically at . In particular, when restricted to reals this identity extends to the closed interval , not just the open interval (see reasinsin 20194). (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

Theoremacoscos 20191 The arccosine function is an inverse to . (Contributed by Mario Carneiro, 2-Apr-2015.)
arccos

Theoremasin1 20192 The arcsine of is . (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

Theoremacos1 20193 The arcsine of is . (Contributed by Mario Carneiro, 2-Apr-2015.)
arccos

Theoremreasinsin 20194 The arcsine function composed with is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

Theoremasinsinb 20195 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

Theoremacoscosb 20196 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
arccos

Theoremasinbnd 20197 The arcsine function has range within a vertical strip of the complex plane with real part between and . (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

Theoremacosbnd 20198 The arccosine function has range within a vertical strip of the complex plane with real part between and . (Contributed by Mario Carneiro, 2-Apr-2015.)
arccos

Theoremasinrebnd 20199 Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

Theoremasinrecl 20200 The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
arcsin

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