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Type | Label | Description |
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Statement | ||
Theorem | angpieqvdlem2 26101* | Equivalence used in angpieqvd 26103. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) β β’ (π β (-((πΆ β π΅) / (π΄ β π΅)) β β+ β ((π΄ β π΅)πΉ(πΆ β π΅)) = Ο)) | ||
Theorem | angpined 26102* | If the angle at ABC is Ο, then π΄ is not equal to πΆ. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) β β’ (π β (((π΄ β π΅)πΉ(πΆ β π΅)) = Ο β π΄ β πΆ)) | ||
Theorem | angpieqvd 26103* | The angle ABC is Ο iff π΅ is a nontrivial convex combination of π΄ and πΆ, i.e., iff π΅ is in the interior of the segment AC. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β π΅) & β’ (π β π΅ β πΆ) β β’ (π β (((π΄ β π΅)πΉ(πΆ β π΅)) = Ο β βπ€ β (0(,)1)π΅ = ((π€ Β· π΄) + ((1 β π€) Β· πΆ)))) | ||
Theorem | chordthmlem 26104* | If π is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 26094 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 angles. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) & β’ (π β π΄ β π΅) & β’ (π β π β π) β β’ (π β ((π β π)πΉ(π΅ β π)) β {(Ο / 2), -(Ο / 2)}) | ||
Theorem | chordthmlem2 26105* | 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 26104, where P = B, and using angrtmuld 26080 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) & β’ (π β π β π) & β’ (π β π β π) β β’ (π β ((π β π)πΉ(π β π)) β {(Ο / 2), -(Ο / 2)}) | ||
Theorem | chordthmlem3 26106 | 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 26105 and the Pythagorean theorem (pythag 26089) 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.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) β β’ (π β ((absβ(π β π))β2) = (((absβ(π β π))β2) + ((absβ(π β π))β2))) | ||
Theorem | chordthmlem4 26107 | If P is on the segment AB and M is the midpoint of AB, then PA Β· PB = BM 2 β PM 2 . If all lengths are reexpressed as fractions of AB, this reduces to the identity π Β· (1 β π) = (1 / 2) 2 β ((1 / 2) β π) 2 . (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β (0[,]1)) & β’ (π β π = ((π΄ + π΅) / 2)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) β β’ (π β ((absβ(π β π΄)) Β· (absβ(π β π΅))) = (((absβ(π΅ β π))β2) β ((absβ(π β π))β2))) | ||
Theorem | chordthmlem5 26108 | If P is on the segment AB and AQ = BQ, then PA Β· PB = BQ 2 β PQ 2 . This follows from two uses of chordthmlem3 26106 to show that PQ 2 = QM 2 + PM 2 and BQ 2 = QM 2 + BM 2 , so BQ 2 β PQ 2 = (QM 2 + BM 2 ) β (QM 2 + PM 2 ) = BM 2 β PM 2 , which equals PA Β· PB by chordthmlem4 26107. (Contributed by David Moews, 28-Feb-2017.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β π β β) & β’ (π β π β (0[,]1)) & β’ (π β π = ((π Β· π΄) + ((1 β π) Β· π΅))) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) β β’ (π β ((absβ(π β π΄)) Β· (absβ(π β π΅))) = (((absβ(π΅ β π))β2) β ((absβ(π β π))β2))) | ||
Theorem | chordthm 26109* | 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 Β· PB and PC Β· 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 Ο. The result is proven by using chordthmlem5 26108 twice to show that PA Β· PB and PC Β· PD both equal BQ 2 β PQ 2 . This is similar to the proof of the theorem given in Euclid's Elements, where it is Proposition III.35. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π΄ β π) & β’ (π β π΅ β π) & β’ (π β πΆ β π) & β’ (π β π· β π) & β’ (π β ((π΄ β π)πΉ(π΅ β π)) = Ο) & β’ (π β ((πΆ β π)πΉ(π· β π)) = Ο) & β’ (π β π β β) & β’ (π β (absβ(π΄ β π)) = (absβ(π΅ β π))) & β’ (π β (absβ(π΄ β π)) = (absβ(πΆ β π))) & β’ (π β (absβ(π΄ β π)) = (absβ(π· β π))) β β’ (π β ((absβ(π β π΄)) Β· (absβ(π β π΅))) = ((absβ(π β πΆ)) Β· (absβ(π β π·)))) | ||
Theorem | heron 26110* | Heron's formula gives the area of a triangle given only the side lengths. If points A, B, C form a triangle, then the area of the triangle, represented here as (1 / 2) Β· π Β· π Β· abs(sinπ), is equal to the square root of π Β· (π β π) Β· (π β π) Β· (π β π), where π = (π + π + π) / 2 is half the perimeter of the triangle. Based on work by Jon Pennant. This is Metamath 100 proof #57. (Contributed by Mario Carneiro, 10-Mar-2019.) |
β’ πΉ = (π₯ β (β β {0}), π¦ β (β β {0}) β¦ (ββ(logβ(π¦ / π₯)))) & β’ π = (absβ(π΅ β πΆ)) & β’ π = (absβ(π΄ β πΆ)) & β’ π = (absβ(π΄ β π΅)) & β’ π = ((π΅ β πΆ)πΉ(π΄ β πΆ)) & β’ π = (((π + π) + π) / 2) & β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π΄ β πΆ) & β’ (π β π΅ β πΆ) β β’ (π β (((1 / 2) Β· (π Β· π)) Β· (absβ(sinβπ))) = (ββ((π Β· (π β π)) Β· ((π β π) Β· (π β π))))) | ||
Theorem | quad2 26111 | The quadratic equation, without specifying the particular branch π· to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π· β β) & β’ (π β (π·β2) = ((π΅β2) β (4 Β· (π΄ Β· πΆ)))) β β’ (π β (((π΄ Β· (πβ2)) + ((π΅ Β· π) + πΆ)) = 0 β (π = ((-π΅ + π·) / (2 Β· π΄)) β¨ π = ((-π΅ β π·) / (2 Β· π΄))))) | ||
Theorem | quad 26112 | The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π· = ((π΅β2) β (4 Β· (π΄ Β· πΆ)))) β β’ (π β (((π΄ Β· (πβ2)) + ((π΅ Β· π) + πΆ)) = 0 β (π = ((-π΅ + (ββπ·)) / (2 Β· π΄)) β¨ π = ((-π΅ β (ββπ·)) / (2 Β· π΄))))) | ||
Theorem | 1cubrlem 26113 | The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ ((-1βπ(2 / 3)) = ((-1 + (i Β· (ββ3))) / 2) β§ ((-1βπ(2 / 3))β2) = ((-1 β (i Β· (ββ3))) / 2)) | ||
Theorem | 1cubr 26114 | The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ π = {1, ((-1 + (i Β· (ββ3))) / 2), ((-1 β (i Β· (ββ3))) / 2)} β β’ (π΄ β π β (π΄ β β β§ (π΄β3) = 1)) | ||
Theorem | dcubic1lem 26115 | Lemma for dcubic1 26117 and dcubic2 26116: simplify the cubic equation under the substitution π = π β π / π. (Contributed by Mario Carneiro, 26-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π = (π β (π / π))) β β’ (π β (((πβ3) + ((π Β· π) + π)) = 0 β (((πβ3)β2) + ((π Β· (πβ3)) β (πβ3))) = 0)) | ||
Theorem | dcubic2 26116* | Reverse direction of dcubic 26118. Given a solution π to the "substitution" quadratic equation π = π β π / π, show that π is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) & β’ (π β π β β) & β’ (π β π β 0) & β’ (π β π = (π β (π / π))) & β’ (π β ((πβ3) + ((π Β· π) + π)) = 0) β β’ (π β βπ β β ((πβ3) = 1 β§ π = ((π Β· π) β (π / (π Β· π))))) | ||
Theorem | dcubic1 26117 | Forward direction of dcubic 26118: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) & β’ (π β π = (π β (π / π))) β β’ (π β ((πβ3) + ((π Β· π) + π)) = 0) | ||
Theorem | dcubic 26118* | Solutions to the depressed cubic, a special case of cubic 26121. (The definitions of π, π, πΊ, π here differ from mcubic 26119 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = (πΊ β π)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) + (πβ3))) & β’ (π β π = (π / 3)) & β’ (π β π = (π / 2)) & β’ (π β π β 0) β β’ (π β (((πβ3) + ((π Β· π) + π)) = 0 β βπ β β ((πβ3) = 1 β§ π = ((π Β· π) β (π / (π Β· π)))))) | ||
Theorem | mcubic 26119* | Solutions to a monic cubic equation, a special case of cubic 26121. (Contributed by Mario Carneiro, 24-Apr-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = ((π + πΊ) / 2)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) β (4 Β· (πβ3)))) & β’ (π β π = ((π΅β2) β (3 Β· πΆ))) & β’ (π β π = (((2 Β· (π΅β3)) β (9 Β· (π΅ Β· πΆ))) + (;27 Β· π·))) & β’ (π β π β 0) β β’ (π β ((((πβ3) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β βπ β β ((πβ3) = 1 β§ π = -(((π΅ + (π Β· π)) + (π / (π Β· π))) / 3)))) | ||
Theorem | cubic2 26120* | The solution to the general cubic equation, for arbitrary choices πΊ and π of the square and cube roots. (Contributed by Mario Carneiro, 23-Apr-2015.) |
β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β (πβ3) = ((π + πΊ) / 2)) & β’ (π β πΊ β β) & β’ (π β (πΊβ2) = ((πβ2) β (4 Β· (πβ3)))) & β’ (π β π = ((π΅β2) β (3 Β· (π΄ Β· πΆ)))) & β’ (π β π = (((2 Β· (π΅β3)) β ((9 Β· π΄) Β· (π΅ Β· πΆ))) + (;27 Β· ((π΄β2) Β· π·)))) & β’ (π β π β 0) β β’ (π β ((((π΄ Β· (πβ3)) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β βπ β β ((πβ3) = 1 β§ π = -(((π΅ + (π Β· π)) + (π / (π Β· π))) / (3 Β· π΄))))) | ||
Theorem | cubic 26121* | The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4666 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.) |
β’ π = {1, ((-1 + (i Β· (ββ3))) / 2), ((-1 β (i Β· (ββ3))) / 2)} & β’ (π β π΄ β β) & β’ (π β π΄ β 0) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β π = (((π + (ββπΊ)) / 2)βπ(1 / 3))) & β’ (π β πΊ = ((πβ2) β (4 Β· (πβ3)))) & β’ (π β π = ((π΅β2) β (3 Β· (π΄ Β· πΆ)))) & β’ (π β π = (((2 Β· (π΅β3)) β ((9 Β· π΄) Β· (π΅ Β· πΆ))) + (;27 Β· ((π΄β2) Β· π·)))) & β’ (π β π β 0) β β’ (π β ((((π΄ Β· (πβ3)) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β βπ β π π = -(((π΅ + (π Β· π)) + (π / (π Β· π))) / (3 Β· π΄)))) | ||
Theorem | binom4 26122 | Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15650, 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.) |
β’ ((π΄ β β β§ π΅ β β) β ((π΄ + π΅)β4) = (((π΄β4) + (4 Β· ((π΄β3) Β· π΅))) + ((6 Β· ((π΄β2) Β· (π΅β2))) + ((4 Β· (π΄ Β· (π΅β3))) + (π΅β4))))) | ||
Theorem | dquartlem1 26123 | Lemma for dquart 26125. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((2 Β· π)β2)) & β’ (π β π β 0) & β’ (π β πΌ β β) & β’ (π β (πΌβ2) = ((-(πβ2) β (π΅ / 2)) + ((πΆ / 4) / π))) β β’ (π β ((((πβ2) + ((π + π΅) / 2)) + ((((π / 2) Β· π) β (πΆ / 4)) / π)) = 0 β (π = (-π + πΌ) β¨ π = (-π β πΌ)))) | ||
Theorem | dquartlem2 26124 | Lemma for dquart 26125. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((2 Β· π)β2)) & β’ (π β π β 0) & β’ (π β πΌ β β) & β’ (π β (πΌβ2) = ((-(πβ2) β (π΅ / 2)) + ((πΆ / 4) / π))) & β’ (π β π· β β) & β’ (π β (((πβ3) + ((2 Β· π΅) Β· (πβ2))) + ((((π΅β2) β (4 Β· π·)) Β· π) + -(πΆβ2))) = 0) β β’ (π β ((((π + π΅) / 2)β2) β (((πΆβ2) / 4) / π)) = π·) | ||
Theorem | dquart 26125 | Solve a depressed quartic equation. To eliminate π, which is the square root of a solution π to the resolvent cubic equation, apply cubic 26121 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((2 Β· π)β2)) & β’ (π β π β 0) & β’ (π β πΌ β β) & β’ (π β (πΌβ2) = ((-(πβ2) β (π΅ / 2)) + ((πΆ / 4) / π))) & β’ (π β π· β β) & β’ (π β (((πβ3) + ((2 Β· π΅) Β· (πβ2))) + ((((π΅β2) β (4 Β· π·)) Β· π) + -(πΆβ2))) = 0) & β’ (π β π½ β β) & β’ (π β (π½β2) = ((-(πβ2) β (π΅ / 2)) β ((πΆ / 4) / π))) β β’ (π β ((((πβ4) + (π΅ Β· (πβ2))) + ((πΆ Β· π) + π·)) = 0 β ((π = (-π + πΌ) β¨ π = (-π β πΌ)) β¨ (π = (π + π½) β¨ π = (π β π½))))) | ||
Theorem | quart1cl 26126 | Closure lemmas for quart 26133. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) β β’ (π β (π β β β§ π β β β§ π β β)) | ||
Theorem | quart1lem 26127 | Lemma for quart1 26128. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π β β) & β’ (π β π = (π + (π΄ / 4))) β β’ (π β π· = ((((π΄β4) / ;;256) + (π Β· ((π΄ / 4)β2))) + ((π Β· (π΄ / 4)) + π ))) | ||
Theorem | quart1 26128 | Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π β β) & β’ (π β π = (π + (π΄ / 4))) β β’ (π β (((πβ4) + (π΄ Β· (πβ3))) + ((π΅ Β· (πβ2)) + ((πΆ Β· π) + π·))) = (((πβ4) + (π Β· (πβ2))) + ((π Β· π) + π ))) | ||
Theorem | quartlem1 26129 | Lemma for quart 26133. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π β β) & β’ (π β π β β) & β’ (π β π β β) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) β β’ (π β (π = (((2 Β· π)β2) β (3 Β· ((πβ2) β (4 Β· π )))) β§ π = (((2 Β· ((2 Β· π)β3)) β (9 Β· ((2 Β· π) Β· ((πβ2) β (4 Β· π ))))) + (;27 Β· -(πβ2))))) | ||
Theorem | quartlem2 26130 | Closure lemmas for quart 26133. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) β β’ (π β (π β β β§ π β β β§ π β β)) | ||
Theorem | quartlem3 26131 | Closure lemmas for quart 26133. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) & β’ (π β π = ((ββπ) / 2)) & β’ (π β π = -((((2 Β· π) + π) + (π / π)) / 3)) & β’ (π β π = (((π + π) / 2)βπ(1 / 3))) & β’ (π β π β 0) β β’ (π β (π β β β§ π β β β§ π β β)) | ||
Theorem | quartlem4 26132 | Closure lemmas for quart 26133. (Contributed by Mario Carneiro, 7-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) & β’ (π β π = ((ββπ) / 2)) & β’ (π β π = -((((2 Β· π) + π) + (π / π)) / 3)) & β’ (π β π = (((π + π) / 2)βπ(1 / 3))) & β’ (π β π β 0) & β’ (π β π β 0) & β’ (π β πΌ = (ββ((-(πβ2) β (π / 2)) + ((π / 4) / π)))) & β’ (π β π½ = (ββ((-(πβ2) β (π / 2)) β ((π / 4) / π)))) β β’ (π β (π β 0 β§ πΌ β β β§ π½ β β)) | ||
Theorem | quart 26133 | 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 33520) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.) |
β’ (π β π΄ β β) & β’ (π β π΅ β β) & β’ (π β πΆ β β) & β’ (π β π· β β) & β’ (π β π β β) & β’ (π β πΈ = -(π΄ / 4)) & β’ (π β π = (π΅ β ((3 / 8) Β· (π΄β2)))) & β’ (π β π = ((πΆ β ((π΄ Β· π΅) / 2)) + ((π΄β3) / 8))) & β’ (π β π = ((π· β ((πΆ Β· π΄) / 4)) + ((((π΄β2) Β· π΅) / ;16) β ((3 / ;;256) Β· (π΄β4))))) & β’ (π β π = ((πβ2) + (;12 Β· π ))) & β’ (π β π = ((-(2 Β· (πβ3)) β (;27 Β· (πβ2))) + (;72 Β· (π Β· π )))) & β’ (π β π = (ββ((πβ2) β (4 Β· (πβ3))))) & β’ (π β π = ((ββπ) / 2)) & β’ (π β π = -((((2 Β· π) + π) + (π / π)) / 3)) & β’ (π β π = (((π + π) / 2)βπ(1 / 3))) & β’ (π β π β 0) & β’ (π β π β 0) & β’ (π β πΌ = (ββ((-(πβ2) β (π / 2)) + ((π / 4) / π)))) & β’ (π β π½ = (ββ((-(πβ2) β (π / 2)) β ((π / 4) / π)))) β β’ (π β ((((πβ4) + (π΄ Β· (πβ3))) + ((π΅ Β· (πβ2)) + ((πΆ Β· π) + π·))) = 0 β ((π = ((πΈ β π) + πΌ) β¨ π = ((πΈ β π) β πΌ)) β¨ (π = ((πΈ + π) + π½) β¨ π = ((πΈ + π) β π½))))) | ||
Syntax | casin 26134 | The arcsine function. |
class arcsin | ||
Syntax | cacos 26135 | The arccosine function. |
class arccos | ||
Syntax | catan 26136 | The arctangent function. |
class arctan | ||
Definition | df-asin 26137 | Define the arcsine function. Because sin is not a one-to-one function, the literal inverse β‘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 -1 and 1 (at which the function is defined), and branch cuts along the real line not between -1 and 1, which is to say (-β, -1) βͺ (1, +β). (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arcsin = (π₯ β β β¦ (-i Β· (logβ((i Β· π₯) + (ββ(1 β (π₯β2))))))) | ||
Definition | df-acos 26138 | Define the arccosine function. See also remarks for df-asin 26137. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely (-β, -1) βͺ (1, +β). (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arccos = (π₯ β β β¦ ((Ο / 2) β (arcsinβπ₯))) | ||
Definition | df-atan 26139 | Define the arctangent function. See also remarks for df-asin 26137. Unlike arcsin and arccos, this function is not defined everywhere, because tan(π§) β Β±i for all π§ β β. For all other π§, there is a formula for arctan(π§) in terms of log, and we take that as the definition. Branch points are at Β±i; branch cuts are on the pure imaginary axis not between -i and i, which is to say {π§ β β β£ (i Β· π§) β (-β, -1) βͺ (1, +β)}. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arctan = (π₯ β (β β {-i, i}) β¦ ((i / 2) Β· ((logβ(1 β (i Β· π₯))) β (logβ(1 + (i Β· π₯)))))) | ||
Theorem | asinlem 26140 | The argument to the logarithm in df-asin 26137 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β ((i Β· π΄) + (ββ(1 β (π΄β2)))) β 0) | ||
Theorem | asinlem2 26141 | The argument to the logarithm in df-asin 26137 has the property that replacing π΄ with -π΄ in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β (((i Β· π΄) + (ββ(1 β (π΄β2)))) Β· ((i Β· -π΄) + (ββ(1 β (-π΄β2))))) = 1) | ||
Theorem | asinlem3a 26142 | Lemma for asinlem3 26143. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β€ 0) β 0 β€ (ββ((i Β· π΄) + (ββ(1 β (π΄β2)))))) | ||
Theorem | asinlem3 26143 | The argument to the logarithm in df-asin 26137 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β 0 β€ (ββ((i Β· π΄) + (ββ(1 β (π΄β2)))))) | ||
Theorem | asinf 26144 | Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arcsin:ββΆβ | ||
Theorem | asincl 26145 | Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arcsinβπ΄) β β) | ||
Theorem | acosf 26146 | Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arccos:ββΆβ | ||
Theorem | acoscl 26147 | Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arccosβπ΄) β β) | ||
Theorem | atandm 26148 | Since the property is a little lengthy, we abbreviate π΄ β β β§ π΄ β -i β§ π΄ β i as π΄ β dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ π΄ β -i β§ π΄ β i)) | ||
Theorem | atandm2 26149 | This form of atandm 26148 is a bit more useful for showing that the logarithms in df-atan 26139 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ (1 β (i Β· π΄)) β 0 β§ (1 + (i Β· π΄)) β 0)) | ||
Theorem | atandm3 26150 | A compact form of atandm 26148. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ (π΄β2) β -1)) | ||
Theorem | atandm4 26151 | A compact form of atandm 26148. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β (π΄ β β β§ (1 + (π΄β2)) β 0)) | ||
Theorem | atanf 26152 | Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ arctan:(β β {-i, i})βΆβ | ||
Theorem | atancl 26153 | Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (arctanβπ΄) β β) | ||
Theorem | asinval 26154 | Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arcsinβπ΄) = (-i Β· (logβ((i Β· π΄) + (ββ(1 β (π΄β2))))))) | ||
Theorem | acosval 26155 | Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arccosβπ΄) = ((Ο / 2) β (arcsinβπ΄))) | ||
Theorem | atanval 26156 | Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (arctanβπ΄) = ((i / 2) Β· ((logβ(1 β (i Β· π΄))) β (logβ(1 + (i Β· π΄)))))) | ||
Theorem | atanre 26157 | A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β π΄ β dom arctan) | ||
Theorem | asinneg 26158 | The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β (arcsinβ-π΄) = -(arcsinβπ΄)) | ||
Theorem | acosneg 26159 | The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β β β (arccosβ-π΄) = (Ο β (arccosβπ΄))) | ||
Theorem | efiasin 26160 | The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (expβ(i Β· (arcsinβπ΄))) = ((i Β· π΄) + (ββ(1 β (π΄β2))))) | ||
Theorem | sinasin 26161 | The arcsine function is an inverse to sin. This is the main property that justifies the notation arcsin or sinβ-1. Because sin is not an injection, the other converse identity asinsin 26164 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β (sinβ(arcsinβπ΄)) = π΄) | ||
Theorem | cosacos 26162 | The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.) |
β’ (π΄ β β β (cosβ(arccosβπ΄)) = π΄) | ||
Theorem | asinsinlem 26163 | Lemma for asinsin 26164. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β (-(Ο / 2)(,)(Ο / 2))) β 0 < (ββ(expβ(i Β· π΄)))) | ||
Theorem | asinsin 26164 | The arcsine function composed with sin is equal to the identity. This plus sinasin 26161 allow 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 π΄ = (Ο / 2) β iπ¦ for nonnegative real π¦ and also symmetrically at π΄ = iπ¦ β (Ο / 2). In particular, when restricted to reals this identity extends to the closed interval [-(Ο / 2), (Ο / 2)], not just the open interval (see reasinsin 26168). (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β (-(Ο / 2)(,)(Ο / 2))) β (arcsinβ(sinβπ΄)) = π΄) | ||
Theorem | acoscos 26165 | The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β (0(,)Ο)) β (arccosβ(cosβπ΄)) = π΄) | ||
Theorem | asin1 26166 | The arcsine of 1 is Ο / 2. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (arcsinβ1) = (Ο / 2) | ||
Theorem | acos1 26167 | The arccosine of 1 is 0. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (arccosβ1) = 0 | ||
Theorem | reasinsin 26168 | The arcsine function composed with sin is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β (-(Ο / 2)[,](Ο / 2)) β (arcsinβ(sinβπ΄)) = π΄) | ||
Theorem | asinsinb 26169 | Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ ((π΄ β β β§ π΅ β β β§ (ββπ΅) β (-(Ο / 2)(,)(Ο / 2))) β ((arcsinβπ΄) = π΅ β (sinβπ΅) = π΄)) | ||
Theorem | acoscosb 26170 | Relationship between cosine and arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ ((π΄ β β β§ π΅ β β β§ (ββπ΅) β (0(,)Ο)) β ((arccosβπ΄) = π΅ β (cosβπ΅) = π΄)) | ||
Theorem | asinbnd 26171 | The arcsine function has range within a vertical strip of the complex plane with real part between -Ο / 2 and Ο / 2. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β β β (ββ(arcsinβπ΄)) β (-(Ο / 2)[,](Ο / 2))) | ||
Theorem | acosbnd 26172 | The arccosine function has range within a vertical strip of the complex plane with real part between 0 and Ο. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β β β (ββ(arccosβπ΄)) β (0[,]Ο)) | ||
Theorem | asinrebnd 26173 | Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β (-1[,]1) β (arcsinβπ΄) β (-(Ο / 2)[,](Ο / 2))) | ||
Theorem | asinrecl 26174 | The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β (-1[,]1) β (arcsinβπ΄) β β) | ||
Theorem | acosrecl 26175 | The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β (-1[,]1) β (arccosβπ΄) β β) | ||
Theorem | cosasin 26176 | The cosine of the arcsine of π΄ is β(1 β π΄β2). (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β β β (cosβ(arcsinβπ΄)) = (ββ(1 β (π΄β2)))) | ||
Theorem | sinacos 26177 | The sine of the arccosine of π΄ is β(1 β π΄β2). (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β β β (sinβ(arccosβπ΄)) = (ββ(1 β (π΄β2)))) | ||
Theorem | atandmneg 26178 | The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β -π΄ β dom arctan) | ||
Theorem | atanneg 26179 | The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β (arctanβ-π΄) = -(arctanβπ΄)) | ||
Theorem | atan0 26180 | The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (arctanβ0) = 0 | ||
Theorem | atandmcj 26181 | The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β dom arctan β (ββπ΄) β dom arctan) | ||
Theorem | atancj 26182 | The arctangent function distributes under conjugation. (The condition that β(π΄) β 0 is necessary because the branch cuts are chosen so that the negative imaginary line "agrees with" neighboring values with negative real part, while the positive imaginary line agrees with values with positive real part. This makes atanneg 26179 true unconditionally but messes up conjugation symmetry, and it is impossible to have both in a single-valued function. The claim is true on the imaginary line between -1 and 1, though.) (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β 0) β (π΄ β dom arctan β§ (ββ(arctanβπ΄)) = (arctanβ(ββπ΄)))) | ||
Theorem | atanrecl 26183 | The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ (π΄ β β β (arctanβπ΄) β β) | ||
Theorem | efiatan 26184 | Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β dom arctan β (expβ(i Β· (arctanβπ΄))) = ((ββ(1 + (i Β· π΄))) / (ββ(1 β (i Β· π΄))))) | ||
Theorem | atanlogaddlem 26185 | Lemma for atanlogadd 26186. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ ((π΄ β dom arctan β§ 0 β€ (ββπ΄)) β ((logβ(1 + (i Β· π΄))) + (logβ(1 β (i Β· π΄)))) β ran log) | ||
Theorem | atanlogadd 26186 | The rule β(π§π€) = (βπ§)(βπ€) is not always true on the complex numbers, but it is true when the arguments of π§ and π€ sum to within the interval (-Ο, Ο], so there are some cases such as this one with π§ = 1 + iπ΄ and π€ = 1 β iπ΄ which are true unconditionally. This result can also be stated as "β(1 + π§) + β(1 β π§) is analytic". (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β ((logβ(1 + (i Β· π΄))) + (logβ(1 β (i Β· π΄)))) β ran log) | ||
Theorem | atanlogsublem 26187 | Lemma for atanlogsub 26188. (Contributed by Mario Carneiro, 4-Apr-2015.) |
β’ ((π΄ β dom arctan β§ 0 < (ββπ΄)) β (ββ((logβ(1 + (i Β· π΄))) β (logβ(1 β (i Β· π΄))))) β (-Ο(,)Ο)) | ||
Theorem | atanlogsub 26188 | A variation on atanlogadd 26186, to show that β(1 + iπ§) / β(1 β iπ§) = β((1 + iπ§) / (1 β iπ§)) under more limited conditions. (Contributed by Mario Carneiro, 4-Apr-2015.) |
β’ ((π΄ β dom arctan β§ (ββπ΄) β 0) β ((logβ(1 + (i Β· π΄))) β (logβ(1 β (i Β· π΄)))) β ran log) | ||
Theorem | efiatan2 26189 | Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β (expβ(i Β· (arctanβπ΄))) = ((1 + (i Β· π΄)) / (ββ(1 + (π΄β2))))) | ||
Theorem | 2efiatan 26190 | Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β dom arctan β (expβ(2 Β· (i Β· (arctanβπ΄)))) = (((2 Β· i) / (π΄ + i)) β 1)) | ||
Theorem | tanatan 26191 | The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (π΄ β dom arctan β (tanβ(arctanβπ΄)) = π΄) | ||
Theorem | atandmtan 26192 | The tangent function has range contained in the domain of the arctangent. (Contributed by Mario Carneiro, 31-Mar-2015.) |
β’ ((π΄ β β β§ (cosβπ΄) β 0) β (tanβπ΄) β dom arctan) | ||
Theorem | cosatan 26193 | The cosine of an arctangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β (cosβ(arctanβπ΄)) = (1 / (ββ(1 + (π΄β2))))) | ||
Theorem | cosatanne0 26194 | The arctangent function has range contained in the domain of the tangent. (Contributed by Mario Carneiro, 3-Apr-2015.) |
β’ (π΄ β dom arctan β (cosβ(arctanβπ΄)) β 0) | ||
Theorem | atantan 26195 | The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 5-Apr-2015.) |
β’ ((π΄ β β β§ (ββπ΄) β (-(Ο / 2)(,)(Ο / 2))) β (arctanβ(tanβπ΄)) = π΄) | ||
Theorem | atantanb 26196 | Relationship between tangent and arctangent. (Contributed by Mario Carneiro, 5-Apr-2015.) |
β’ ((π΄ β dom arctan β§ π΅ β β β§ (ββπ΅) β (-(Ο / 2)(,)(Ο / 2))) β ((arctanβπ΄) = π΅ β (tanβπ΅) = π΄)) | ||
Theorem | atanbndlem 26197 | Lemma for atanbnd 26198. (Contributed by Mario Carneiro, 5-Apr-2015.) |
β’ (π΄ β β+ β (arctanβπ΄) β (-(Ο / 2)(,)(Ο / 2))) | ||
Theorem | atanbnd 26198 | The arctangent function is bounded by Ο / 2 on the reals. (Contributed by Mario Carneiro, 5-Apr-2015.) |
β’ (π΄ β β β (arctanβπ΄) β (-(Ο / 2)(,)(Ο / 2))) | ||
Theorem | atanord 26199 | The arctangent function is strictly increasing. (Contributed by Mario Carneiro, 5-Apr-2015.) |
β’ ((π΄ β β β§ π΅ β β) β (π΄ < π΅ β (arctanβπ΄) < (arctanβπ΅))) | ||
Theorem | atan1 26200 | The arctangent of 1 is Ο / 4. (Contributed by Mario Carneiro, 2-Apr-2015.) |
β’ (arctanβ1) = (Ο / 4) |
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