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Theorem List for Metamath Proof Explorer - 24601-24700   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremangpieqvdlem2 24601* Equivalence used in angpieqvd 24603. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (-((𝐶𝐵) / (𝐴𝐵)) ∈ ℝ+ ↔ ((𝐴𝐵)𝐹(𝐶𝐵)) = π))

Theoremangpined 24602* If the angle at ABC is π, then A is not equal to C. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (((𝐴𝐵)𝐹(𝐶𝐵)) = π → 𝐴𝐶))

Theoremangpieqvd 24603* The angle ABC is π 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)       (𝜑 → (((𝐴𝐵)𝐹(𝐶𝐵)) = π ↔ ∃𝑤 ∈ (0(,)1)𝐵 = ((𝑤 · 𝐴) + ((1 − 𝑤) · 𝐶))))

Theoremchordthmlem 24604* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 24597 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑𝐴𝐵)    &   (𝜑𝑄𝑀)       (𝜑 → ((𝑄𝑀)𝐹(𝐵𝑀)) ∈ {(π / 2), -(π / 2)})

Theoremchordthmlem2 24605* 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 24604, where P = B, and using angrtmuld 24583 to observe that QMP is right iff QMB is. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℝ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑𝑃𝑀)    &   (𝜑𝑄𝑀)       (𝜑 → ((𝑄𝑀)𝐹(𝑃𝑀)) ∈ {(π / 2), -(π / 2)})

Theoremchordthmlem3 24606 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 24605 and the Pythagorean theorem (pythag 24592) 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℝ)    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))       (𝜑 → ((abs‘(𝑃𝑄))↑2) = (((abs‘(𝑄𝑀))↑2) + ((abs‘(𝑃𝑀))↑2)))

Theoremchordthmlem4 24607 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 𝑋 · (1 − 𝑋) = (1 / 2) 2 − ((1 / 2) − 𝑋) 2 . (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑋 ∈ (0[,]1))    &   (𝜑𝑀 = ((𝐴 + 𝐵) / 2))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = (((abs‘(𝐵𝑀))↑2) − ((abs‘(𝑃𝑀))↑2)))

Theoremchordthmlem5 24608 If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 PQ 2 . This follows from two uses of chordthmlem3 24606 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 24607. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ (0[,]1))    &   (𝜑𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = (((abs‘(𝐵𝑄))↑2) − ((abs‘(𝑃𝑄))↑2)))

Theoremchordthm 24609* 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 π. The result is proven by using chordthmlem5 24608 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. This is Metamath 100 proof #55. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 ∈ ℂ)    &   (𝜑𝐴𝑃)    &   (𝜑𝐵𝑃)    &   (𝜑𝐶𝑃)    &   (𝜑𝐷𝑃)    &   (𝜑 → ((𝐴𝑃)𝐹(𝐵𝑃)) = π)    &   (𝜑 → ((𝐶𝑃)𝐹(𝐷𝑃)) = π)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐵𝑄)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐶𝑄)))    &   (𝜑 → (abs‘(𝐴𝑄)) = (abs‘(𝐷𝑄)))       (𝜑 → ((abs‘(𝑃𝐴)) · (abs‘(𝑃𝐵))) = ((abs‘(𝑃𝐶)) · (abs‘(𝑃𝐷))))

Theoremheron 24610* 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) · 𝑋 · 𝑌 · abs(sin𝑂), is equal to the square root of 𝑆 · (𝑆𝑋) · (𝑆𝑌) · (𝑆𝑍), where 𝑆 = (𝑋 + 𝑌 + 𝑍) / 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.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑋 = (abs‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))    &   𝑆 = (((𝑋 + 𝑌) + 𝑍) / 2)    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐶)    &   (𝜑𝐵𝐶)       (𝜑 → (((1 / 2) · (𝑋 · 𝑌)) · (abs‘(sin‘𝑂))) = (√‘((𝑆 · (𝑆𝑋)) · ((𝑆𝑌) · (𝑆𝑍)))))

14.3.7  Solutions of quadratic, cubic, and quartic equations

Theoremquad2 24611 The quadratic equation, without specifying the particular branch 𝐷 to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑 → (𝐷↑2) = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))       (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + 𝐷) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵𝐷) / (2 · 𝐴)))))

(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐷 = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))       (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + (√‘𝐷)) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − (√‘𝐷)) / (2 · 𝐴)))))

Theorem1cubrlem 24613 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
((-1↑𝑐(2 / 3)) = ((-1 + (i · (√‘3))) / 2) ∧ ((-1↑𝑐(2 / 3))↑2) = ((-1 − (i · (√‘3))) / 2))

Theorem1cubr 24614 The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}       (𝐴𝑅 ↔ (𝐴 ∈ ℂ ∧ (𝐴↑3) = 1))

Theoremdcubic1lem 24615 Lemma for dcubic1 24617 and dcubic2 24616: simplify the cubic equation under the substitution 𝑋 = 𝑈𝑀 / 𝑈. (Contributed by Mario Carneiro, 26-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑈 ∈ ℂ)    &   (𝜑𝑈 ≠ 0)    &   (𝜑𝑋 = (𝑈 − (𝑀 / 𝑈)))       (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ (((𝑈↑3)↑2) + ((𝑄 · (𝑈↑3)) − (𝑀↑3))) = 0))

Theoremdcubic2 24616* Reverse direction of dcubic 24618. Given a solution 𝑈 to the "substitution" quadratic equation 𝑋 = 𝑈𝑀 / 𝑈, show that 𝑋 is in the desired form. (Contributed by Mario Carneiro, 25-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑈 ∈ ℂ)    &   (𝜑𝑈 ≠ 0)    &   (𝜑𝑋 = (𝑈 − (𝑀 / 𝑈)))    &   (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)       (𝜑 → ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇)))))

Theoremdcubic1 24617 Forward direction of dcubic 24618: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑋 = (𝑇 − (𝑀 / 𝑇)))       (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)

Theoremdcubic 24618* Solutions to the depressed cubic, a special case of cubic 24621. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 24619 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = (𝐺𝑁))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   (𝜑𝑀 = (𝑃 / 3))    &   (𝜑𝑁 = (𝑄 / 2))    &   (𝜑𝑇 ≠ 0)       (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇))))))

Theoremmcubic 24619* Solutions to a monic cubic equation, a special case of cubic 24621. (Contributed by Mario Carneiro, 24-Apr-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   (𝜑𝑀 = ((𝐵↑2) − (3 · 𝐶)))    &   (𝜑𝑁 = (((2 · (𝐵↑3)) − (9 · (𝐵 · 𝐶))) + (27 · 𝐷)))    &   (𝜑𝑇 ≠ 0)       (𝜑 → ((((𝑋↑3) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / 3))))

Theoremcubic2 24620* 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.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 ∈ ℂ)    &   (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   (𝜑𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   (𝜑𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (27 · ((𝐴↑2) · 𝐷))))    &   (𝜑𝑇 ≠ 0)       (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴)))))

Theoremcubic 24621* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4273 to convert the existential quantifier to a triple disjunction. This is Metamath 100 proof #37. (Contributed by Mario Carneiro, 26-Apr-2015.)
𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐴 ≠ 0)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑇 = (((𝑁 + (√‘𝐺)) / 2)↑𝑐(1 / 3)))    &   (𝜑𝐺 = ((𝑁↑2) − (4 · (𝑀↑3))))    &   (𝜑𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   (𝜑𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (27 · ((𝐴↑2) · 𝐷))))    &   (𝜑𝑀 ≠ 0)       (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟𝑅 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴))))

Theorembinom4 24622 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 14606, 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 ∈ ℂ ∧ 𝐵 ∈ ℂ) → ((𝐴 + 𝐵)↑4) = (((𝐴↑4) + (4 · ((𝐴↑3) · 𝐵))) + ((6 · ((𝐴↑2) · (𝐵↑2))) + ((4 · (𝐴 · (𝐵↑3))) + (𝐵↑4)))))

Theoremdquartlem1 24623 Lemma for dquart 24625. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑆 ∈ ℂ)    &   (𝜑𝑀 = ((2 · 𝑆)↑2))    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 ∈ ℂ)    &   (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))       (𝜑 → ((((𝑋↑2) + ((𝑀 + 𝐵) / 2)) + ((((𝑀 / 2) · 𝑋) − (𝐶 / 4)) / 𝑆)) = 0 ↔ (𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆𝐼))))

Theoremdquartlem2 24624 Lemma for dquart 24625. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑆 ∈ ℂ)    &   (𝜑𝑀 = ((2 · 𝑆)↑2))    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 ∈ ℂ)    &   (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)       (𝜑 → ((((𝑀 + 𝐵) / 2)↑2) − (((𝐶↑2) / 4) / 𝑀)) = 𝐷)

Theoremdquart 24625 Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 24621 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑆 ∈ ℂ)    &   (𝜑𝑀 = ((2 · 𝑆)↑2))    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 ∈ ℂ)    &   (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)    &   (𝜑𝐽 ∈ ℂ)    &   (𝜑 → (𝐽↑2) = ((-(𝑆↑2) − (𝐵 / 2)) − ((𝐶 / 4) / 𝑆)))       (𝜑 → ((((𝑋↑4) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ((𝑋 = (-𝑆 + 𝐼) ∨ 𝑋 = (-𝑆𝐼)) ∨ (𝑋 = (𝑆 + 𝐽) ∨ 𝑋 = (𝑆𝐽)))))

Theoremquart1cl 24626 Closure lemmas for quart 24633. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))       (𝜑 → (𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ ∧ 𝑅 ∈ ℂ))

Theoremquart1lem 24627 Lemma for quart1 24628. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 = (𝑋 + (𝐴 / 4)))       (𝜑𝐷 = ((((𝐴↑4) / 256) + (𝑃 · ((𝐴 / 4)↑2))) + ((𝑄 · (𝐴 / 4)) + 𝑅)))

Theoremquart1 24628 Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 = (𝑋 + (𝐴 / 4)))       (𝜑 → (((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = (((𝑌↑4) + (𝑃 · (𝑌↑2))) + ((𝑄 · 𝑌) + 𝑅)))

Theoremquartlem1 24629 Lemma for quart 24633. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝑃 ∈ ℂ)    &   (𝜑𝑄 ∈ ℂ)    &   (𝜑𝑅 ∈ ℂ)    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))       (𝜑 → (𝑈 = (((2 · 𝑃)↑2) − (3 · ((𝑃↑2) − (4 · 𝑅)))) ∧ 𝑉 = (((2 · ((2 · 𝑃)↑3)) − (9 · ((2 · 𝑃) · ((𝑃↑2) − (4 · 𝑅))))) + (27 · -(𝑄↑2)))))

Theoremquartlem2 24630 Closure lemmas for quart 24633. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))       (𝜑 → (𝑈 ∈ ℂ ∧ 𝑉 ∈ ℂ ∧ 𝑊 ∈ ℂ))

Theoremquartlem3 24631 Closure lemmas for quart 24633. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   (𝜑𝑆 = ((√‘𝑀) / 2))    &   (𝜑𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   (𝜑𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   (𝜑𝑇 ≠ 0)       (𝜑 → (𝑆 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ 𝑇 ∈ ℂ))

Theoremquartlem4 24632 Closure lemmas for quart 24633. (Contributed by Mario Carneiro, 7-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   (𝜑𝑆 = ((√‘𝑀) / 2))    &   (𝜑𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   (𝜑𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   (𝜑𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))       (𝜑 → (𝑆 ≠ 0 ∧ 𝐼 ∈ ℂ ∧ 𝐽 ∈ ℂ))

Theoremquart 24633 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 31273) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝐸 = -(𝐴 / 4))    &   (𝜑𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   (𝜑𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   (𝜑𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / 16) − ((3 / 256) · (𝐴↑4)))))    &   (𝜑𝑈 = ((𝑃↑2) + (12 · 𝑅)))    &   (𝜑𝑉 = ((-(2 · (𝑃↑3)) − (27 · (𝑄↑2))) + (72 · (𝑃 · 𝑅))))    &   (𝜑𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   (𝜑𝑆 = ((√‘𝑀) / 2))    &   (𝜑𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   (𝜑𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   (𝜑𝑇 ≠ 0)    &   (𝜑𝑀 ≠ 0)    &   (𝜑𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   (𝜑𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))       (𝜑 → ((((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = 0 ↔ ((𝑋 = ((𝐸𝑆) + 𝐼) ∨ 𝑋 = ((𝐸𝑆) − 𝐼)) ∨ (𝑋 = ((𝐸 + 𝑆) + 𝐽) ∨ 𝑋 = ((𝐸 + 𝑆) − 𝐽)))))

14.3.8  Inverse trigonometric functions

Syntaxcasin 24634 The arcsine function.
class arcsin

Syntaxcacos 24635 The arccosine function.
class arccos

Syntaxcatan 24636 The arctangent function.
class arctan

Definitiondf-asin 24637 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 -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 (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin = (𝑥 ∈ ℂ ↦ (-i · (log‘((i · 𝑥) + (√‘(1 − (𝑥↑2)))))))

Definitiondf-acos 24638 Define the arccosine function. See also remarks for df-asin 24637. Since we define arccos in terms of arcsin, it shares the same branch points and cuts, namely (-∞, -1) ∪ (1, +∞). (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos = (𝑥 ∈ ℂ ↦ ((π / 2) − (arcsin‘𝑥)))

Definitiondf-atan 24639 Define the arctangent function. See also remarks for df-asin 24637. Unlike arcsin and arccos, this function is not defined everywhere, because tan(𝑧) ≠ ±i for all 𝑧 ∈ ℂ. 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 ±i; branch cuts are on the pure imaginary axis not between -i and i, which is to say {𝑧 ∈ ℂ ∣ (i · 𝑧) ∈ (-∞, -1) ∪ (1, +∞)}. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan = (𝑥 ∈ (ℂ ∖ {-i, i}) ↦ ((i / 2) · ((log‘(1 − (i · 𝑥))) − (log‘(1 + (i · 𝑥))))))

Theoremasinlem 24640 The argument to the logarithm in df-asin 24637 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)

Theoremasinlem2 24641 The argument to the logarithm in df-asin 24637 has the property that replacing 𝐴 with -𝐴 in the expression gives the reciprocal. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (((i · 𝐴) + (√‘(1 − (𝐴↑2)))) · ((i · -𝐴) + (√‘(1 − (-𝐴↑2))))) = 1)

Theoremasinlem3a 24642 Lemma for asinlem3 24643. (Contributed by Mario Carneiro, 1-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))

Theoremasinlem3 24643 The argument to the logarithm in df-asin 24637 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))

Theoremasinf 24644 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin:ℂ⟶ℂ

Theoremasincl 24645 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ)

Theoremacosf 24646 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos:ℂ⟶ℂ

Theoremacoscl 24647 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ)

Theorematandm 24648 Since the property is a little lengthy, we abbreviate 𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i as 𝐴 ∈ dom arctan. This is the necessary precondition for the definition of arctan to make sense. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ 𝐴 ≠ -i ∧ 𝐴 ≠ i))

Theorematandm2 24649 This form of atandm 24648 is a bit more useful for showing that the logarithms in df-atan 24639 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0))

Theorematandm3 24650 A compact form of atandm 24648. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1))

Theorematandm4 24651 A compact form of atandm 24648. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0))

Theorematanf 24652 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan:(ℂ ∖ {-i, i})⟶ℂ

Theorematancl 24653 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ)

Theoremasinval 24654 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))))

Theoremacosval 24655 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴)))

Theorematanval 24656 Value of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (arctan‘𝐴) = ((i / 2) · ((log‘(1 − (i · 𝐴))) − (log‘(1 + (i · 𝐴))))))

Theorematanre 24657 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℝ → 𝐴 ∈ dom arctan)

Theoremasinneg 24658 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴))

Theoremacosneg 24659 The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴)))

Theoremefiasin 24660 The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (exp‘(i · (arcsin‘𝐴))) = ((i · 𝐴) + (√‘(1 − (𝐴↑2)))))

Theoremsinasin 24661 The arcsine function is an inverse to sin. This is the main property that justifies the notation arcsin or sin↑-1. Because sin is not an injection, the other converse identity asinsin 24664 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴)

Theoremcosacos 24662 The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴)

Theoremasinsinlem 24663 Lemma for asinsin 24664. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴))))

Theoremasinsin 24664 The arcsine function composed with sin is equal to the identity. This plus sinasin 24661 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 = (π / 2) − i𝑦 for nonnegative real 𝑦 and also symmetrically at 𝐴 = i𝑦 − (π / 2). In particular, when restricted to reals this identity extends to the closed interval [-(π / 2), (π / 2)], not just the open interval (see reasinsin 24668). (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴)

Theoremacoscos 24665 The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (0(,)π)) → (arccos‘(cos‘𝐴)) = 𝐴)

Theoremasin1 24666 The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
(arcsin‘1) = (π / 2)

Theoremacos1 24667 The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
(arccos‘1) = 0

Theoremreasinsin 24668 The arcsine function composed with sin is equal to the identity. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-(π / 2)[,](π / 2)) → (arcsin‘(sin‘𝐴)) = 𝐴)

Theoremasinsinb 24669 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (-(π / 2)(,)(π / 2))) → ((arcsin‘𝐴) = 𝐵 ↔ (sin‘𝐵) = 𝐴))

Theoremacoscosb 24670 Relationship between sine and arcsine. (Contributed by Mario Carneiro, 2-Apr-2015.)
((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (0(,)π)) → ((arccos‘𝐴) = 𝐵 ↔ (cos‘𝐵) = 𝐴))

Theoremasinbnd 24671 The arcsine function has range within a vertical strip of the complex plane with real part between -π / 2 and π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (ℜ‘(arcsin‘𝐴)) ∈ (-(π / 2)[,](π / 2)))

Theoremacosbnd 24672 The arccosine function has range within a vertical strip of the complex plane with real part between 0 and π. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (ℜ‘(arccos‘𝐴)) ∈ (0[,]π))

Theoremasinrebnd 24673 Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2)))

Theoremasinrecl 24674 The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ)

Theoremacosrecl 24675 The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ)

Theoremcosasin 24676 The cosine of the arcsine of 𝐴 is √(1 − 𝐴↑2). (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (cos‘(arcsin‘𝐴)) = (√‘(1 − (𝐴↑2))))

Theoremsinacos 24677 The sine of the arccosine of 𝐴 is √(1 − 𝐴↑2). (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ ℂ → (sin‘(arccos‘𝐴)) = (√‘(1 − (𝐴↑2))))

Theorematandmneg 24678 The domain of the arctangent function is closed under negatives. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → -𝐴 ∈ dom arctan)

Theorematanneg 24679 The arctangent function is odd. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → (arctan‘-𝐴) = -(arctan‘𝐴))

Theorematan0 24680 The arctangent of zero is zero. (Contributed by Mario Carneiro, 31-Mar-2015.)
(arctan‘0) = 0

Theorematandmcj 24681 The arctangent function distributes under conjugation. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (∗‘𝐴) ∈ dom arctan)

Theorematancj 24682 The arctangent function distributes under conjugation. (The condition that ℜ(𝐴) ≠ 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 24679 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.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ≠ 0) → (𝐴 ∈ dom arctan ∧ (∗‘(arctan‘𝐴)) = (arctan‘(∗‘𝐴))))

Theorematanrecl 24683 The arctangent function is real for all real inputs. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℝ → (arctan‘𝐴) ∈ ℝ)

Theoremefiatan 24684 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ dom arctan → (exp‘(i · (arctan‘𝐴))) = ((√‘(1 + (i · 𝐴))) / (√‘(1 − (i · 𝐴)))))

((𝐴 ∈ dom arctan ∧ 0 ≤ (ℜ‘𝐴)) → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log)

Theorematanlogadd 24686 The rule √(𝑧𝑤) = (√𝑧)(√𝑤) is not always true on the complex numbers, but it is true when the arguments of 𝑧 and 𝑤 sum to within the interval (-π, π], so there are some cases such as this one with 𝑧 = 1 + i𝐴 and 𝑤 = 1 − i𝐴 which are true unconditionally. This result can also be stated as "√(1 + 𝑧) + √(1 − 𝑧) is analytic". (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → ((log‘(1 + (i · 𝐴))) + (log‘(1 − (i · 𝐴)))) ∈ ran log)

Theorematanlogsublem 24687 Lemma for atanlogsub 24688. (Contributed by Mario Carneiro, 4-Apr-2015.)
((𝐴 ∈ dom arctan ∧ 0 < (ℜ‘𝐴)) → (ℑ‘((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴))))) ∈ (-π(,)π))

Theorematanlogsub 24688 A variation on atanlogadd 24686, to show that √(1 + i𝑧) / √(1 − i𝑧) = √((1 + i𝑧) / (1 − i𝑧)) under more limited conditions. (Contributed by Mario Carneiro, 4-Apr-2015.)
((𝐴 ∈ dom arctan ∧ (ℜ‘𝐴) ≠ 0) → ((log‘(1 + (i · 𝐴))) − (log‘(1 − (i · 𝐴)))) ∈ ran log)

Theoremefiatan2 24689 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → (exp‘(i · (arctan‘𝐴))) = ((1 + (i · 𝐴)) / (√‘(1 + (𝐴↑2)))))

Theorem2efiatan 24690 Value of the exponential of an artcangent. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ dom arctan → (exp‘(2 · (i · (arctan‘𝐴)))) = (((2 · i) / (𝐴 + i)) − 1))

Theoremtanatan 24691 The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 2-Apr-2015.)
(𝐴 ∈ dom arctan → (tan‘(arctan‘𝐴)) = 𝐴)

Theorematandmtan 24692 The tangent function has range contained in the domain of the arctangent. (Contributed by Mario Carneiro, 31-Mar-2015.)
((𝐴 ∈ ℂ ∧ (cos‘𝐴) ≠ 0) → (tan‘𝐴) ∈ dom arctan)

Theoremcosatan 24693 The cosine of an arctangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) = (1 / (√‘(1 + (𝐴↑2)))))

Theoremcosatanne0 24694 The arctangent function has range contained in the domain of the tangent. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan → (cos‘(arctan‘𝐴)) ≠ 0)

Theorematantan 24695 The arctangent function is an inverse to tan. (Contributed by Mario Carneiro, 5-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arctan‘(tan‘𝐴)) = 𝐴)

Theorematantanb 24696 Relationship between tangent and arctangent. (Contributed by Mario Carneiro, 5-Apr-2015.)
((𝐴 ∈ dom arctan ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (-(π / 2)(,)(π / 2))) → ((arctan‘𝐴) = 𝐵 ↔ (tan‘𝐵) = 𝐴))

Theorematanbndlem 24697 Lemma for atanbnd 24698. (Contributed by Mario Carneiro, 5-Apr-2015.)
(𝐴 ∈ ℝ+ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2)))

Theorematanbnd 24698 The arctangent function is bounded by π / 2 on the reals. (Contributed by Mario Carneiro, 5-Apr-2015.)
(𝐴 ∈ ℝ → (arctan‘𝐴) ∈ (-(π / 2)(,)(π / 2)))

Theorematanord 24699 The arctangent function is strictly increasing. (Contributed by Mario Carneiro, 5-Apr-2015.)
((𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ) → (𝐴 < 𝐵 ↔ (arctan‘𝐴) < (arctan‘𝐵)))

Theorematan1 24700 The arctangent of 1 is π / 4. (Contributed by Mario Carneiro, 2-Apr-2015.)
(arctan‘1) = (π / 4)

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