P. 271 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 27001-27100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	angrteqvd 27001*	Two vectors are at a right angle iff their quotient is purely imaginary. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    ⇒   ⊢ (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (ℜ‘(𝑌 / 𝑋)) = 0))
Theorem	cosangneg2d 27002*	The cosine of the angle between 𝑋 and -𝑌 is the negative of that between 𝑋 and 𝑌. 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.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    ⇒   ⊢ (𝜑 → (cos‘(𝑋𝐹-𝑌)) = -(cos‘(𝑋𝐹𝑌)))
Theorem	angrtmuld 27003*	Perpendicularity of two vectors does not change under rescaling the second. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 ∈ ℂ)    &   ⊢ (𝜑 → 𝑍 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ≠ 0)    &   ⊢ (𝜑 → 𝑌 ≠ 0)    &   ⊢ (𝜑 → 𝑍 ≠ 0)    &   ⊢ (𝜑 → (𝑍 / 𝑌) ∈ ℝ)    ⇒   ⊢ (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (𝑋𝐹𝑍) ∈ {(π / 2), -(π / 2)}))
Theorem	ang180lem1 27004*	Lemma for ang180 27009. Show that the "revolution number" 𝑁 is an integer, using efeq1 26723 to show that since the product of the three arguments 𝐴, 1 / (1 − 𝐴), (𝐴 − 1) / 𝐴 is -1, the sum of the logarithms must be an integer multiple of 2πi away from πi = log(-1). (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   ⊢ 𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (𝑁 ∈ ℤ ∧ (𝑇 / i) ∈ ℝ))
Theorem	ang180lem2 27005*	Lemma for ang180 27009. Show that the revolution number 𝑁 is strictly between -2 and 1. Both bounds are established by iterating using the bounds on the imaginary part of the logarithm, logimcl 26764, 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   ⊢ 𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → (-2 < 𝑁 ∧ 𝑁 < 1))
Theorem	ang180lem3 27006*	Lemma for ang180 27009. Since ang180lem1 27004 shows that 𝑁 is an integer and ang180lem2 27005 shows that 𝑁 is strictly between -2 and 1, it follows that 𝑁 ∈ {-1, 0}, and these two cases correspond to the two possible values for 𝑇. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑇 = (((log‘(1 / (1 − 𝐴))) + (log‘((𝐴 − 1) / 𝐴))) + (log‘𝐴))    &   ⊢ 𝑁 = (((𝑇 / i) / (2 · π)) − (1 / 2))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → 𝑇 ∈ {-(i · π), (i · π)})
Theorem	ang180lem4 27007*	Lemma for ang180 27009. Reduce the statement to one variable. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ ((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐴 ≠ 1) → ((((1 − 𝐴)𝐹1) + (𝐴𝐹(𝐴 − 1))) + (1𝐹𝐴)) ∈ {-π, π})
Theorem	ang180lem5 27008*	Lemma for ang180 27009: Reduce the statement to two variables. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐴 ≠ 0) ∧ (𝐵 ∈ ℂ ∧ 𝐵 ≠ 0) ∧ 𝐴 ≠ 𝐵) → ((((𝐴 − 𝐵)𝐹𝐴) + (𝐵𝐹(𝐵 − 𝐴))) + (𝐴𝐹𝐵)) ∈ {-π, π})
Theorem	ang180 27009*	The sum of angles 𝑚𝐴𝐵𝐶 + 𝑚𝐵𝐶𝐴 + 𝑚𝐶𝐴𝐵 in a triangle adds up to either π or -π, i.e. 180 degrees. (The sign is due to the two possible orientations of vertex arrangement and our signed notion of angle). This is Metamath 100 proof #27. (Contributed by Mario Carneiro, 23-Sep-2014.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐶)) → ((((𝐶 − 𝐵)𝐹(𝐴 − 𝐵)) + ((𝐴 − 𝐶)𝐹(𝐵 − 𝐶))) + ((𝐵 − 𝐴)𝐹(𝐶 − 𝐴))) ∈ {-π, π})
Theorem	lawcoslem1 27010	Lemma for lawcos 27011. 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.)
⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑉 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑉 ≠ 0)    ⇒   ⊢ (𝜑 → ((abs‘(𝑈 − 𝑉))↑2) = ((((abs‘𝑈)↑2) + ((abs‘𝑉)↑2)) − (2 · (((abs‘𝑈) · (abs‘𝑉)) · ((ℜ‘(𝑈 / 𝑉)) / (abs‘(𝑈 / 𝑉)))))))
Theorem	lawcos 27011*	Law of cosines (also known as the Al-Kashi theorem or the generalized Pythagorean theorem, or the cosine formula or cosine rule). 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 27009), 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 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 27010 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 16228). The Pythagorean theorem pythag 27012 is a special case of the law of cosines. The theorem's expression and approach were suggested by Mario Carneiro. This is Metamath 100 proof #94. (Contributed by David A. Wheeler, 12-Jun-2015.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑋 = (abs‘(𝐵 − 𝐶))    &   ⊢ 𝑌 = (abs‘(𝐴 − 𝐶))    &   ⊢ 𝑍 = (abs‘(𝐴 − 𝐵))    &   ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶)) → (𝑍↑2) = (((𝑋↑2) + (𝑌↑2)) − (2 · ((𝑋 · 𝑌) · (cos‘𝑂)))))
Theorem	pythag 27012*	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 27009), 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 signed right angle m/_ BCA. We use the law of cosines lawcos 27011 to prove this, along with simple trigonometry facts like coshalfpi 26664 and cosneg 16228. (Contributed by David A. Wheeler, 13-Jun-2015.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ 𝑋 = (abs‘(𝐵 − 𝐶))    &   ⊢ 𝑌 = (abs‘(𝐴 − 𝐶))    &   ⊢ 𝑍 = (abs‘(𝐴 − 𝐵))    &   ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶) ∧ 𝑂 ∈ {(π / 2), -(π / 2)}) → (𝑍↑2) = ((𝑋↑2) + (𝑌↑2)))
Theorem	isosctrlem1 27013	Lemma for isosctr 27016. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
⊢ ((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)
Theorem	isosctrlem2 27014	Lemma for isosctr 27016. 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 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) = (ℑ‘(log‘(-𝐴 / (1 − 𝐴)))))
Theorem	isosctrlem3 27015*	Lemma for isosctr 27016. Corresponds to the case where one vertex is at 0. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ) ∧ (𝐴 ≠ 0 ∧ 𝐵 ≠ 0 ∧ 𝐴 ≠ 𝐵) ∧ (abs‘𝐴) = (abs‘𝐵)) → (-𝐴𝐹(𝐵 − 𝐴)) = ((𝐴 − 𝐵)𝐹-𝐵))
Theorem	isosctr 27016*	Isosceles triangle theorem. This is Metamath 100 proof #65. (Contributed by Saveliy Skresanov, 1-Jan-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    ⇒   ⊢ (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴 ≠ 𝐶 ∧ 𝐵 ≠ 𝐶 ∧ 𝐴 ≠ 𝐵) ∧ (abs‘(𝐴 − 𝐶)) = (abs‘(𝐵 − 𝐶))) → ((𝐶 − 𝐴)𝐹(𝐵 − 𝐴)) = ((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)))
Theorem	ssscongptld 27017*	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 27011 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 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 ∈ ℂ)    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    &   ⊢ (𝜑 → 𝐷 ≠ 𝐸)    &   ⊢ (𝜑 → 𝐸 ≠ 𝐺)    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝐵)) = (abs‘(𝐷 − 𝐸)))    &   ⊢ (𝜑 → (abs‘(𝐵 − 𝐶)) = (abs‘(𝐸 − 𝐺)))    &   ⊢ (𝜑 → (abs‘(𝐶 − 𝐴)) = (abs‘(𝐺 − 𝐷)))    ⇒   ⊢ (𝜑 → (cos‘((𝐴 − 𝐵)𝐹(𝐶 − 𝐵))) = (cos‘((𝐷 − 𝐸)𝐹(𝐺 − 𝐸))))
Theorem	affineequiv 27018	Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐶 − 𝐵) = (𝐷 · (𝐶 − 𝐴))))
Theorem	affineequiv2 27019	Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐵 − 𝐴) = ((1 − 𝐷) · (𝐶 − 𝐴))))
Theorem	affineequiv3 27020	Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ (𝐴 − 𝐵) = (𝐷 · (𝐶 − 𝐵))))
Theorem	affineequiv4 27021	Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶. (Contributed by AV, 22-Jan-2023.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    ⇒   ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐴 = ((𝐷 · (𝐶 − 𝐵)) + 𝐵)))
Theorem	affineequivne 27022	Equivalence between two ways of expressing 𝐴 as an affine combination of 𝐵 and 𝐶 if 𝐵 and 𝐶 are not equal. (Contributed by AV, 22-Jan-2023.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (𝐴 = (((1 − 𝐷) · 𝐵) + (𝐷 · 𝐶)) ↔ 𝐷 = ((𝐴 − 𝐵) / (𝐶 − 𝐵))))
Theorem	angpieqvdlem 27023	Equivalence used in the proof of angpieqvd 27026. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (-((𝐶 − 𝐵) / (𝐴 − 𝐵)) ∈ ℝ+ ↔ ((𝐶 − 𝐵) / (𝐶 − 𝐴)) ∈ (0(,)1)))
Theorem	angpieqvdlem2 27024*	Equivalence used in angpieqvd 27026. (Contributed by David Moews, 28-Feb-2017.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (-((𝐶 − 𝐵) / (𝐴 − 𝐵)) ∈ ℝ+ ↔ ((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π))
Theorem	angpined 27025*	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))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π → 𝐴 ≠ 𝐶))
Theorem	angpieqvd 27026*	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))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (((𝐴 − 𝐵)𝐹(𝐶 − 𝐵)) = π ↔ ∃𝑤 ∈ (0(,)1)𝐵 = ((𝑤 · 𝐴) + ((1 − 𝑤) · 𝐶))))
Theorem	chordthmlem 27027*	If 𝑀 is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 27017 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.)
⊢ 𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    &   ⊢ (𝜑 → 𝐴 ≠ 𝐵)    &   ⊢ (𝜑 → 𝑄 ≠ 𝑀)    ⇒   ⊢ (𝜑 → ((𝑄 − 𝑀)𝐹(𝐵 − 𝑀)) ∈ {(π / 2), -(π / 2)})
Theorem	chordthmlem2 27028*	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 27027, where P = B, and using angrtmuld 27003 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))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    &   ⊢ (𝜑 → 𝑃 ≠ 𝑀)    &   ⊢ (𝜑 → 𝑄 ≠ 𝑀)    ⇒   ⊢ (𝜑 → ((𝑄 − 𝑀)𝐹(𝑃 − 𝑀)) ∈ {(π / 2), -(π / 2)})
Theorem	chordthmlem3 27029	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 27028 and the Pythagorean theorem (pythag 27012) 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.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℝ)    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝑄))↑2) = (((abs‘(𝑄 − 𝑀))↑2) + ((abs‘(𝑃 − 𝑀))↑2)))
Theorem	chordthmlem4 27030	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.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ (0[,]1))    &   ⊢ (𝜑 → 𝑀 = ((𝐴 + 𝐵) / 2))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = (((abs‘(𝐵 − 𝑀))↑2) − ((abs‘(𝑃 − 𝑀))↑2)))
Theorem	chordthmlem5 27031	If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 − PQ 2 . This follows from two uses of chordthmlem3 27029 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 27030. (Contributed by David Moews, 28-Feb-2017.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ (0[,]1))    &   ⊢ (𝜑 → 𝑃 = ((𝑋 · 𝐴) + ((1 − 𝑋) · 𝐵)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = (((abs‘(𝐵 − 𝑄))↑2) − ((abs‘(𝑃 − 𝑄))↑2)))
Theorem	chordthm 27032*	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 27031 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))))    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝑃)    &   ⊢ (𝜑 → 𝐵 ≠ 𝑃)    &   ⊢ (𝜑 → 𝐶 ≠ 𝑃)    &   ⊢ (𝜑 → 𝐷 ≠ 𝑃)    &   ⊢ (𝜑 → ((𝐴 − 𝑃)𝐹(𝐵 − 𝑃)) = π)    &   ⊢ (𝜑 → ((𝐶 − 𝑃)𝐹(𝐷 − 𝑃)) = π)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐵 − 𝑄)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐶 − 𝑄)))    &   ⊢ (𝜑 → (abs‘(𝐴 − 𝑄)) = (abs‘(𝐷 − 𝑄)))    ⇒   ⊢ (𝜑 → ((abs‘(𝑃 − 𝐴)) · (abs‘(𝑃 − 𝐵))) = ((abs‘(𝑃 − 𝐶)) · (abs‘(𝑃 − 𝐷))))
Theorem	heron 27033*	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 · (𝑆 − 𝑋) · (𝑆 − 𝑌) · (𝑆 − 𝑍), 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‘(𝐵 − 𝐶))    &   ⊢ 𝑌 = (abs‘(𝐴 − 𝐶))    &   ⊢ 𝑍 = (abs‘(𝐴 − 𝐵))    &   ⊢ 𝑂 = ((𝐵 − 𝐶)𝐹(𝐴 − 𝐶))    &   ⊢ 𝑆 = (((𝑋 + 𝑌) + 𝑍) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 𝐶)    &   ⊢ (𝜑 → 𝐵 ≠ 𝐶)    ⇒   ⊢ (𝜑 → (((1 / 2) · (𝑋 · 𝑌)) · (abs‘(sin‘𝑂))) = (√‘((𝑆 · (𝑆 − 𝑋)) · ((𝑆 − 𝑌) · (𝑆 − 𝑍)))))
14.3.7  Solutions of quadratic, cubic, and quartic equations
Theorem	quad2 27034	The quadratic equation, without specifying the particular branch 𝐷 to the square root. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → (𝐷↑2) = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))    ⇒   ⊢ (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + 𝐷) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − 𝐷) / (2 · 𝐴)))))
Theorem	quad 27035	The quadratic equation. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 = ((𝐵↑2) − (4 · (𝐴 · 𝐶))))    ⇒   ⊢ (𝜑 → (((𝐴 · (𝑋↑2)) + ((𝐵 · 𝑋) + 𝐶)) = 0 ↔ (𝑋 = ((-𝐵 + (√‘𝐷)) / (2 · 𝐴)) ∨ 𝑋 = ((-𝐵 − (√‘𝐷)) / (2 · 𝐴)))))
Theorem	1cubrlem 27036	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))
Theorem	1cubr 27037	The cube roots of unity. (Contributed by Mario Carneiro, 23-Apr-2015.)
⊢ 𝑅 = {1, ((-1 + (i · (√‘3))) / 2), ((-1 − (i · (√‘3))) / 2)}    ⇒   ⊢ (𝐴 ∈ 𝑅 ↔ (𝐴 ∈ ℂ ∧ (𝐴↑3) = 1))
Theorem	dcubic1lem 27038	Lemma for dcubic1 27040 and dcubic2 27039: simplify the cubic equation under the substitution 𝑋 = 𝑈 − 𝑀 / 𝑈. (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈)))    ⇒   ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ (((𝑈↑3)↑2) + ((𝑄 · (𝑈↑3)) − (𝑀↑3))) = 0))
Theorem	dcubic2 27039*	Reverse direction of dcubic 27041. 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.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑈 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑈 − (𝑀 / 𝑈)))    &   ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)    ⇒   ⊢ (𝜑 → ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇)))))
Theorem	dcubic1 27040	Forward direction of dcubic 27041: the claimed formula produces solutions to the cubic equation. (Contributed by Mario Carneiro, 25-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑋 = (𝑇 − (𝑀 / 𝑇)))    ⇒   ⊢ (𝜑 → ((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0)
Theorem	dcubic 27041*	Solutions to the depressed cubic, a special case of cubic 27044. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 27042 by scale factors of -9, 54, 54 and -27 respectively, to simplify the algebra and presentation.) (Contributed by Mario Carneiro, 26-Apr-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = (𝐺 − 𝑁))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) + (𝑀↑3)))    &   ⊢ (𝜑 → 𝑀 = (𝑃 / 3))    &   ⊢ (𝜑 → 𝑁 = (𝑄 / 2))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → (((𝑋↑3) + ((𝑃 · 𝑋) + 𝑄)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = ((𝑟 · 𝑇) − (𝑀 / (𝑟 · 𝑇))))))
Theorem	mcubic 27042*	Solutions to a monic cubic equation, a special case of cubic 27044. (Contributed by Mario Carneiro, 24-Apr-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · 𝐶)))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − (9 · (𝐵 · 𝐶))) + (;27 · 𝐷)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝑋↑3) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ ℂ ((𝑟↑3) = 1 ∧ 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / 3))))
Theorem	cubic2 27043*	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.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 ∈ ℂ)    &   ⊢ (𝜑 → (𝑇↑3) = ((𝑁 + 𝐺) / 2))    &   ⊢ (𝜑 → 𝐺 ∈ ℂ)    &   ⊢ (𝜑 → (𝐺↑2) = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷))))    &   ⊢ (𝜑 → 𝑇 ≠ 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)))))
Theorem	cubic 27044*	The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4677 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)}    &   ⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐴 ≠ 0)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑇 = (((𝑁 + (√‘𝐺)) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝐺 = ((𝑁↑2) − (4 · (𝑀↑3))))    &   ⊢ (𝜑 → 𝑀 = ((𝐵↑2) − (3 · (𝐴 · 𝐶))))    &   ⊢ (𝜑 → 𝑁 = (((2 · (𝐵↑3)) − ((9 · 𝐴) · (𝐵 · 𝐶))) + (;27 · ((𝐴↑2) · 𝐷))))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    ⇒   ⊢ (𝜑 → ((((𝐴 · (𝑋↑3)) + (𝐵 · (𝑋↑2))) + ((𝐶 · 𝑋) + 𝐷)) = 0 ↔ ∃𝑟 ∈ 𝑅 𝑋 = -(((𝐵 + (𝑟 · 𝑇)) + (𝑀 / (𝑟 · 𝑇))) / (3 · 𝐴))))
Theorem	binom4 27045	Work out a quartic binomial. (You would think that by this point it would be faster to use binom 15910, 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)))))
Theorem	dquartlem1 27046	Lemma for dquart 27048. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ 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 = (-𝑆 − 𝐼))))
Theorem	dquartlem2 27047	Lemma for dquart 27048. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   ⊢ (𝜑 → 𝐷 ∈ 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)
Theorem	dquart 27048	Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 27044 or one of its variants. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑆 ∈ ℂ)    &   ⊢ (𝜑 → 𝑀 = ((2 · 𝑆)↑2))    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 ∈ ℂ)    &   ⊢ (𝜑 → (𝐼↑2) = ((-(𝑆↑2) − (𝐵 / 2)) + ((𝐶 / 4) / 𝑆)))    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → (((𝑀↑3) + ((2 · 𝐵) · (𝑀↑2))) + ((((𝐵↑2) − (4 · 𝐷)) · 𝑀) + -(𝐶↑2))) = 0)    &   ⊢ (𝜑 → 𝐽 ∈ 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 = (-𝑆 − 𝐼)) ∨ (𝑋 = (𝑆 + 𝐽) ∨ 𝑋 = (𝑆 − 𝐽)))))
Theorem	quart1cl 27049	Closure lemmas for quart 27056. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    ⇒   ⊢ (𝜑 → (𝑃 ∈ ℂ ∧ 𝑄 ∈ ℂ ∧ 𝑅 ∈ ℂ))
Theorem	quart1lem 27050	Lemma for quart1 27051. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4)))    ⇒   ⊢ (𝜑 → 𝐷 = ((((𝐴↑4) / ;;256) + (𝑃 · ((𝐴 / 4)↑2))) + ((𝑄 · (𝐴 / 4)) + 𝑅)))
Theorem	quart1 27051	Depress a quartic equation. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝑌 = (𝑋 + (𝐴 / 4)))    ⇒   ⊢ (𝜑 → (((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = (((𝑌↑4) + (𝑃 · (𝑌↑2))) + ((𝑄 · 𝑌) + 𝑅)))
Theorem	quartlem1 27052	Lemma for quart 27056. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝑃 ∈ ℂ)    &   ⊢ (𝜑 → 𝑄 ∈ ℂ)    &   ⊢ (𝜑 → 𝑅 ∈ ℂ)    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(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)))))
Theorem	quartlem2 27053	Closure lemmas for quart 27056. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    ⇒   ⊢ (𝜑 → (𝑈 ∈ ℂ ∧ 𝑉 ∈ ℂ ∧ 𝑊 ∈ ℂ))
Theorem	quartlem3 27054	Closure lemmas for quart 27056. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    ⇒   ⊢ (𝜑 → (𝑆 ∈ ℂ ∧ 𝑀 ∈ ℂ ∧ 𝑇 ∈ ℂ))
Theorem	quartlem4 27055	Closure lemmas for quart 27056. (Contributed by Mario Carneiro, 7-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   ⊢ (𝜑 → 𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))    ⇒   ⊢ (𝜑 → (𝑆 ≠ 0 ∧ 𝐼 ∈ ℂ ∧ 𝐽 ∈ ℂ))
Theorem	quart 27056	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 35670) if all the substitutions are performed. This is Metamath 100 proof #46. (Contributed by Mario Carneiro, 6-May-2015.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝐵 ∈ ℂ)    &   ⊢ (𝜑 → 𝐶 ∈ ℂ)    &   ⊢ (𝜑 → 𝐷 ∈ ℂ)    &   ⊢ (𝜑 → 𝑋 ∈ ℂ)    &   ⊢ (𝜑 → 𝐸 = -(𝐴 / 4))    &   ⊢ (𝜑 → 𝑃 = (𝐵 − ((3 / 8) · (𝐴↑2))))    &   ⊢ (𝜑 → 𝑄 = ((𝐶 − ((𝐴 · 𝐵) / 2)) + ((𝐴↑3) / 8)))    &   ⊢ (𝜑 → 𝑅 = ((𝐷 − ((𝐶 · 𝐴) / 4)) + ((((𝐴↑2) · 𝐵) / ;16) − ((3 / ;;256) · (𝐴↑4)))))    &   ⊢ (𝜑 → 𝑈 = ((𝑃↑2) + (;12 · 𝑅)))    &   ⊢ (𝜑 → 𝑉 = ((-(2 · (𝑃↑3)) − (;27 · (𝑄↑2))) + (;72 · (𝑃 · 𝑅))))    &   ⊢ (𝜑 → 𝑊 = (√‘((𝑉↑2) − (4 · (𝑈↑3)))))    &   ⊢ (𝜑 → 𝑆 = ((√‘𝑀) / 2))    &   ⊢ (𝜑 → 𝑀 = -((((2 · 𝑃) + 𝑇) + (𝑈 / 𝑇)) / 3))    &   ⊢ (𝜑 → 𝑇 = (((𝑉 + 𝑊) / 2)↑𝑐(1 / 3)))    &   ⊢ (𝜑 → 𝑇 ≠ 0)    &   ⊢ (𝜑 → 𝑀 ≠ 0)    &   ⊢ (𝜑 → 𝐼 = (√‘((-(𝑆↑2) − (𝑃 / 2)) + ((𝑄 / 4) / 𝑆))))    &   ⊢ (𝜑 → 𝐽 = (√‘((-(𝑆↑2) − (𝑃 / 2)) − ((𝑄 / 4) / 𝑆))))    ⇒   ⊢ (𝜑 → ((((𝑋↑4) + (𝐴 · (𝑋↑3))) + ((𝐵 · (𝑋↑2)) + ((𝐶 · 𝑋) + 𝐷))) = 0 ↔ ((𝑋 = ((𝐸 − 𝑆) + 𝐼) ∨ 𝑋 = ((𝐸 − 𝑆) − 𝐼)) ∨ (𝑋 = ((𝐸 + 𝑆) + 𝐽) ∨ 𝑋 = ((𝐸 + 𝑆) − 𝐽)))))
14.3.8  Inverse trigonometric functions
Syntax	casin 27057	The arcsine function.
class arcsin
Syntax	cacos 27058	The arccosine function.
class arccos
Syntax	catan 27059	The arctangent function.
class arctan
Definition	df-asin 27060	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)))))))
Definition	df-acos 27061	Define the arccosine function. See also remarks for df-asin 27060. 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)))
Definition	df-atan 27062	Define the arctangent function. See also remarks for df-asin 27060. 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))))))
Theorem	asinlem 27063	The argument to the logarithm in df-asin 27060 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)
Theorem	asinlem2 27064	The argument to the logarithm in df-asin 27060 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)
Theorem	asinlem3a 27065	Lemma for asinlem3 27066. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinlem3 27066	The argument to the logarithm in df-asin 27060 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))
Theorem	asinf 27067	Domain and codomain of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arcsin:ℂ⟶ℂ
Theorem	asincl 27068	Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ)
Theorem	acosf 27069	Domain and codoamin of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arccos:ℂ⟶ℂ
Theorem	acoscl 27070	Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ)
Theorem	atandm 27071	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))
Theorem	atandm2 27072	This form of atandm 27071 is a bit more useful for showing that the logarithms in df-atan 27062 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0))
Theorem	atandm3 27073	A compact form of atandm 27071. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1))
Theorem	atandm4 27074	A compact form of atandm 27071. (Contributed by Mario Carneiro, 3-Apr-2015.)
⊢ (𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0))
Theorem	atanf 27075	Domain and codoamin of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ arctan:(ℂ ∖ {-i, i})⟶ℂ
Theorem	atancl 27076	Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ)
Theorem	asinval 27077	Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))))
Theorem	acosval 27078	Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴)))
Theorem	atanval 27079	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))))))
Theorem	atanre 27080	A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℝ → 𝐴 ∈ dom arctan)
Theorem	asinneg 27081	The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴))
Theorem	acosneg 27082	The negative symmetry relation of the arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (arccos‘-𝐴) = (π − (arccos‘𝐴)))
Theorem	efiasin 27083	The exponential of the arcsine function. (Contributed by Mario Carneiro, 31-Mar-2015.)
⊢ (𝐴 ∈ ℂ → (exp‘(i · (arcsin‘𝐴))) = ((i · 𝐴) + (√‘(1 − (𝐴↑2)))))
Theorem	sinasin 27084	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 27087 is only true under limited circumstances. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (sin‘(arcsin‘𝐴)) = 𝐴)
Theorem	cosacos 27085	The arccosine function is an inverse to cos. (Contributed by Mario Carneiro, 1-Apr-2015.)
⊢ (𝐴 ∈ ℂ → (cos‘(arccos‘𝐴)) = 𝐴)
Theorem	asinsinlem 27086	Lemma for asinsin 27087. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → 0 < (ℜ‘(exp‘(i · 𝐴))))
Theorem	asinsin 27087	The arcsine function composed with sin is equal to the identity. This plus sinasin 27084 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 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 27091). (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ (-(π / 2)(,)(π / 2))) → (arcsin‘(sin‘𝐴)) = 𝐴)
Theorem	acoscos 27088	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)
Theorem	asin1 27089	The arcsine of 1 is π / 2. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (arcsin‘1) = (π / 2)
Theorem	acos1 27090	The arccosine of 1 is 0. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (arccos‘1) = 0
Theorem	reasinsin 27091	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)
Theorem	asinsinb 27092	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))
Theorem	acoscosb 27093	Relationship between cosine and arccosine. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (ℜ‘𝐵) ∈ (0(,)π)) → ((arccos‘𝐴) = 𝐵 ↔ (cos‘𝐵) = 𝐴))
Theorem	asinbnd 27094	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)))
Theorem	acosbnd 27095	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))
Theorem	asinrebnd 27096	Bounds on the arcsine function. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ (-(π / 2)[,](π / 2)))
Theorem	asinrecl 27097	The arcsine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ (-1[,]1) → (arcsin‘𝐴) ∈ ℝ)
Theorem	acosrecl 27098	The arccosine function is real in its principal domain. (Contributed by Mario Carneiro, 2-Apr-2015.)
⊢ (𝐴 ∈ (-1[,]1) → (arccos‘𝐴) ∈ ℝ)
Theorem	cosasin 27099	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))))
Theorem	sinacos 27100	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))))