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Theorem List for Metamath Proof Explorer - 24201-24300   *Has distinct variable group(s)
TypeLabelDescription
Statement

Theoremrelogbzexp 24201 Power law for the general logarithm for integer powers: The logarithm of a positive real number to the power of an integer is equal to the product of the exponent and the logarithm of the base of the power. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 9-Jun-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝐶 ∈ ℝ+𝑁 ∈ ℤ) → (𝐵 logb (𝐶𝑁)) = (𝑁 · (𝐵 logb 𝐶)))

Theoremrelogbmul 24202 The logarithm of the product of two positive real numbers is the sum of logarithms. Property 2 of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 29-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+𝐶 ∈ ℝ+)) → (𝐵 logb (𝐴 · 𝐶)) = ((𝐵 logb 𝐴) + (𝐵 logb 𝐶)))

Theoremrelogbmulexp 24203 The logarithm of the product of a positive real and a positive real number to the power of a real number is the sum of the logarithm of the first real number and the scaled logarithm of the second real number. (Contributed by AV, 29-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+𝐶 ∈ ℝ+𝐸 ∈ ℝ)) → (𝐵 logb (𝐴 · (𝐶𝑐𝐸))) = ((𝐵 logb 𝐴) + (𝐸 · (𝐵 logb 𝐶))))

Theoremrelogbdiv 24204 The logarithm of the quotient of two positive real numbers is the difference of logarithms. Property 3 of [Cohen4] p. 361. (Contributed by AV, 29-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ (𝐴 ∈ ℝ+𝐶 ∈ ℝ+)) → (𝐵 logb (𝐴 / 𝐶)) = ((𝐵 logb 𝐴) − (𝐵 logb 𝐶)))

Theoremrelogbexp 24205 Identity law for general logarithm: the logarithm of a power to the base is the exponent. Property 6 of [Cohen4] p. 361. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by AV, 9-Jun-2020.)
((𝐵 ∈ ℝ+𝐵 ≠ 1 ∧ 𝑀 ∈ ℤ) → (𝐵 logb (𝐵𝑀)) = 𝑀)

Theoremnnlogbexp 24206 Identity law for general logarithm with integer base. (Contributed by Stefan O'Rear, 19-Sep-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑀 ∈ ℤ) → (𝐵 logb (𝐵𝑀)) = 𝑀)

Theoremlogbrec 24207 Logarithm of a reciprocal changes sign. See logrec 24188. Particular case of Property 3 of [Cohen4] p. 361. (Contributed by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝐴 ∈ ℝ+) → (𝐵 logb (1 / 𝐴)) = -(𝐵 logb 𝐴))

Theoremlogbleb 24208 The general logarithm function is monotone/increasing. See logleb 24040. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by AV, 31-May-2020.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑋 ∈ ℝ+𝑌 ∈ ℝ+) → (𝑋𝑌 ↔ (𝐵 logb 𝑋) ≤ (𝐵 logb 𝑌)))

Theoremlogblt 24209 The general logarithm function is strictly monotone/increasing. Property 2 of [Cohen4] p. 377. See logltb 24037. (Contributed by Stefan O'Rear, 19-Oct-2014.) (Revised by Thierry Arnoux, 27-Sep-2017.)
((𝐵 ∈ (ℤ‘2) ∧ 𝑋 ∈ ℝ+𝑌 ∈ ℝ+) → (𝑋 < 𝑌 ↔ (𝐵 logb 𝑋) < (𝐵 logb 𝑌)))

Theoremrelogbcxp 24210 Identity law for the general logarithm for real numbers. (Contributed by AV, 22-May-2020.)
((𝐵 ∈ (ℝ+ ∖ {1}) ∧ 𝑋 ∈ ℝ) → (𝐵 logb (𝐵𝑐𝑋)) = 𝑋)

Theoremcxplogb 24211 Identity law for the general logarithm. (Contributed by AV, 22-May-2020.)
((𝐵 ∈ (ℂ ∖ {0, 1}) ∧ 𝑋 ∈ (ℂ ∖ {0})) → (𝐵𝑐(𝐵 logb 𝑋)) = 𝑋)

Theoremrelogbcxpb 24212 The logarithm is the inverse of the exponentiation. Observation in [Cohen4] p. 348. (Contributed by AV, 11-Jun-2020.)
(((𝐵 ∈ ℝ+𝐵 ≠ 1) ∧ 𝑋 ∈ ℝ+𝑌 ∈ ℝ) → ((𝐵 logb 𝑋) = 𝑌 ↔ (𝐵𝑐𝑌) = 𝑋))

Theoremlogbmpt 24213* The general logarithm to a fixed base regarded as mapping. (Contributed by AV, 11-Jun-2020.)
((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (curry logb𝐵) = (𝑦 ∈ (ℂ ∖ {0}) ↦ ((log‘𝑦) / (log‘𝐵))))

Theoremlogbf 24214 The general logarithm to a fixed base regarded as function. (Contributed by AV, 11-Jun-2020.)
((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) → (curry logb𝐵):(ℂ ∖ {0})⟶ℂ)

Theoremlogbfval 24215 The general logarithm of a complex number to a fixed base. (Contributed by AV, 11-Jun-2020.)
(((𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ∧ 𝐵 ≠ 1) ∧ 𝑋 ∈ (ℂ ∖ {0})) → ((curry logb𝐵)‘𝑋) = (𝐵 logb 𝑋))

Theoremrelogbf 24216 The general logarithm to a real base greater than 1 regarded as function restricted to the positive integers. Property in [Cohen4] p. 349. (Contributed by AV, 12-Jun-2020.)
((𝐵 ∈ ℝ+ ∧ 1 < 𝐵) → ((curry logb𝐵) ↾ ℝ+):ℝ+⟶ℝ)

Theoremlogblog 24217 The general logarithm to the base being Euler's constant regarded as function is the natural logarithm. (Contributed by AV, 12-Jun-2020.)
(curry logb ‘e) = log

14.3.6  Theorems of Pythagoras, isosceles triangles, and intersecting chords

Theoremangval 24218* Define the angle function, which takes two complex numbers, treated as vectors from the origin, and returns the angle between them, in the range ( − π, π]. To convert from the geometry notation, 𝑚𝐴𝐵𝐶, the measure of the angle with legs 𝐴𝐵, 𝐶𝐵 where 𝐶 is more counterclockwise for positive angles, is represented by ((𝐶𝐵)𝐹(𝐴𝐵)). (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)) → (𝐴𝐹𝐵) = (ℑ‘(log‘(𝐵 / 𝐴))))

Theoremangcan 24219* Cancel a constant multiplier in the angle function. (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) ∧ (𝐶 ∈ ℂ ∧ 𝐶 ≠ 0)) → ((𝐶 · 𝐴)𝐹(𝐶 · 𝐵)) = (𝐴𝐹𝐵))

Theoremangneg 24220* Cancel a negative sign in the angle function. (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)) → (-𝐴𝐹-𝐵) = (𝐴𝐹𝐵))

Theoremangvald 24221* The (signed) angle between two vectors is the argument of their quotient. Deduction form of angval 24218. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → (𝑋𝐹𝑌) = (ℑ‘(log‘(𝑌 / 𝑋))))

Theoremangcld 24222* The (signed) angle between two vectors is in (-π(,]π). Deduction form. (Contributed by David Moews, 28-Feb-2017.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → (𝑋𝐹𝑌) ∈ (-π(,]π))

Theoremangrteqvd 24223* 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))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (ℜ‘(𝑌 / 𝑋)) = 0))

Theoremcosangneg2d 24224* 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))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑌 ≠ 0)       (𝜑 → (cos‘(𝑋𝐹-𝑌)) = -(cos‘(𝑋𝐹𝑌)))

Theoremangrtmuld 24225* 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))))    &   (𝜑𝑋 ∈ ℂ)    &   (𝜑𝑌 ∈ ℂ)    &   (𝜑𝑍 ∈ ℂ)    &   (𝜑𝑋 ≠ 0)    &   (𝜑𝑌 ≠ 0)    &   (𝜑𝑍 ≠ 0)    &   (𝜑 → (𝑍 / 𝑌) ∈ ℝ)       (𝜑 → ((𝑋𝐹𝑌) ∈ {(π / 2), -(π / 2)} ↔ (𝑋𝐹𝑍) ∈ {(π / 2), -(π / 2)}))

Theoremang180lem1 24226* Lemma for ang180 24231. Show that the "revolution number" 𝑁 is an integer, using efeq1 23966 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))

Theoremang180lem2 24227* Lemma for ang180 24231. 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 24007, 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 < 𝑁𝑁 < 1))

Theoremang180lem3 24228* Lemma for ang180 24231. Since ang180lem1 24226 shows that 𝑁 is an integer and ang180lem2 24227 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)})

Theoremang180lem4 24229* Lemma for ang180 24231. 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})

Theoremang180lem5 24230* Lemma for ang180 24231: 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) ∧ 𝐴𝐵) → ((((𝐴𝐵)𝐹𝐴) + (𝐵𝐹(𝐵𝐴))) + (𝐴𝐹𝐵)) ∈ {-π, π})

Theoremang180 24231* The sum of angles 𝑚𝐴𝐵𝐶 + 𝑚𝐵𝐶𝐴 + 𝑚𝐶𝐴𝐵 in a triangle adds up to either π or , 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) ∧ (𝐴𝐵𝐵𝐶𝐴𝐶)) → ((((𝐶𝐵)𝐹(𝐴𝐵)) + ((𝐴𝐶)𝐹(𝐵𝐶))) + ((𝐵𝐴)𝐹(𝐶𝐴))) ∈ {-π, π})

Theoremlawcoslem1 24232 Lemma for lawcos 24233. 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.)
(𝜑𝑈 ∈ ℂ)    &   (𝜑𝑉 ∈ ℂ)    &   (𝜑𝑈 ≠ 0)    &   (𝜑𝑉 ≠ 0)       (𝜑 → ((abs‘(𝑈𝑉))↑2) = ((((abs‘𝑈)↑2) + ((abs‘𝑉)↑2)) − (2 · (((abs‘𝑈) · (abs‘𝑉)) · ((ℜ‘(𝑈 / 𝑉)) / (abs‘(𝑈 / 𝑉)))))))

Theoremlawcos 24233* 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 24231), 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 24232 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 14585). The Pythagorean theorem pythag 24234 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‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐶𝐵𝐶)) → (𝑍↑2) = (((𝑋↑2) + (𝑌↑2)) − (2 · ((𝑋 · 𝑌) · (cos‘𝑂)))))

Theorempythag 24234* 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 24231), 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 24233 to prove this, along with simple trigonometry facts like coshalfpi 23912 and cosneg 14585. (Contributed by David A. Wheeler, 13-Jun-2015.)
𝐹 = (𝑥 ∈ (ℂ ∖ {0}), 𝑦 ∈ (ℂ ∖ {0}) ↦ (ℑ‘(log‘(𝑦 / 𝑥))))    &   𝑋 = (abs‘(𝐵𝐶))    &   𝑌 = (abs‘(𝐴𝐶))    &   𝑍 = (abs‘(𝐴𝐵))    &   𝑂 = ((𝐵𝐶)𝐹(𝐴𝐶))       (((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ) ∧ (𝐴𝐶𝐵𝐶) ∧ 𝑂 ∈ {(π / 2), -(π / 2)}) → (𝑍↑2) = ((𝑋↑2) + (𝑌↑2)))

Theoremisosctrlem1 24235 Lemma for isosctr 24238. (Contributed by Saveliy Skresanov, 30-Dec-2016.)
((𝐴 ∈ ℂ ∧ (abs‘𝐴) = 1 ∧ ¬ 1 = 𝐴) → (ℑ‘(log‘(1 − 𝐴))) ≠ π)

Theoremisosctrlem2 24236 Lemma for isosctr 24238. 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)))))

Theoremisosctrlem3 24237* Lemma for isosctr 24238. 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 ∧ 𝐴𝐵) ∧ (abs‘𝐴) = (abs‘𝐵)) → (-𝐴𝐹(𝐵𝐴)) = ((𝐴𝐵)𝐹-𝐵))

Theoremisosctr 24238* 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) ∧ (𝐴𝐶𝐵𝐶𝐴𝐵) ∧ (abs‘(𝐴𝐶)) = (abs‘(𝐵𝐶))) → ((𝐶𝐴)𝐹(𝐵𝐴)) = ((𝐴𝐵)𝐹(𝐶𝐵)))

Theoremssscongptld 24239* 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 24233 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))))    &   (𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)    &   (𝜑𝐸 ∈ ℂ)    &   (𝜑𝐺 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐵𝐶)    &   (𝜑𝐷𝐸)    &   (𝜑𝐸𝐺)    &   (𝜑 → (abs‘(𝐴𝐵)) = (abs‘(𝐷𝐸)))    &   (𝜑 → (abs‘(𝐵𝐶)) = (abs‘(𝐸𝐺)))    &   (𝜑 → (abs‘(𝐶𝐴)) = (abs‘(𝐺𝐷)))       (𝜑 → (cos‘((𝐴𝐵)𝐹(𝐶𝐵))) = (cos‘((𝐷𝐸)𝐹(𝐺𝐸))))

Theoremaffineequiv 24240 Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐶𝐵) = (𝐷 · (𝐶𝐴))))

Theoremaffineequiv2 24241 Equivalence between two ways of expressing 𝐵 as an affine combination of 𝐴 and 𝐶. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐷 ∈ ℂ)       (𝜑 → (𝐵 = ((𝐷 · 𝐴) + ((1 − 𝐷) · 𝐶)) ↔ (𝐵𝐴) = ((1 − 𝐷) · (𝐶𝐴))))

Theoremangpieqvdlem 24242 Equivalence used in the proof of angpieqvd 24245. (Contributed by David Moews, 28-Feb-2017.)
(𝜑𝐴 ∈ ℂ)    &   (𝜑𝐵 ∈ ℂ)    &   (𝜑𝐶 ∈ ℂ)    &   (𝜑𝐴𝐵)    &   (𝜑𝐴𝐶)       (𝜑 → (-((𝐶𝐵) / (𝐴𝐵)) ∈ ℝ+ ↔ ((𝐶𝐵) / (𝐶𝐴)) ∈ (0(,)1)))

Theoremangpieqvdlem2 24243* Equivalence used in angpieqvd 24245. (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 24244* 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 24245* 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 24246* If M is the midpoint of AB and AQ = BQ, then QMB is a right angle. The proof uses ssscongptld 24239 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 24247* 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 24246, where P = B, and using angrtmuld 24225 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 24248 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 24247 and the Pythagorean theorem (pythag 24234) 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 24249 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 24250 If P is on the segment AB and AQ = BQ, then PA · PB = BQ 2 PQ 2 . This follows from two uses of chordthmlem3 24248 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 24249. (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 24251* 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 24250 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 24252* 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 24253 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 24255 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 24256 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 24257 Lemma for dcubic1 24259 and dcubic2 24258: 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 24258* Reverse direction of dcubic 24260. 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 24259 Forward direction of dcubic 24260: 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 24260* Solutions to the depressed cubic, a special case of cubic 24263. (The definitions of 𝑀, 𝑁, 𝐺, 𝑇 here differ from mcubic 24261 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 24261* Solutions to a monic cubic equation, a special case of cubic 24263. (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 24262* 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 24263* The cubic equation, which gives the roots of an arbitrary (nondegenerate) cubic function. Use rextp 4091 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 24264 Work out a quartic binomial. (You would think that by this point it would be faster to use binom 14270, 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 24265 Lemma for dquart 24267. (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 24266 Lemma for dquart 24267. (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 24267 Solve a depressed quartic equation. To eliminate 𝑆, which is the square root of a solution 𝑀 to the resolvent cubic equation, apply cubic 24263 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 24268 Closure lemmas for quart 24275. (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 24269 Lemma for quart1 24270. (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 24270 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 24271 Lemma for quart 24275. (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 24272 Closure lemmas for quart 24275. (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 24273 Closure lemmas for quart 24275. (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 24274 Closure lemmas for quart 24275. (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 24275 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 30246) 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 24276 The arcsine function.
class arcsin

Syntaxcacos 24277 The arccosine function.
class arccos

Syntaxcatan 24278 The arctangent function.
class arctan

Definitiondf-asin 24279 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 24280 Define the arccosine function. See also remarks for df-asin 24279. 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 24281 Define the arctangent function. See also remarks for df-asin 24279. 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 24282 The argument to the logarithm in df-asin 24279 is always nonzero. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → ((i · 𝐴) + (√‘(1 − (𝐴↑2)))) ≠ 0)

Theoremasinlem2 24283 The argument to the logarithm in df-asin 24279 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 24284 Lemma for asinlem3 24285. (Contributed by Mario Carneiro, 1-Apr-2015.)
((𝐴 ∈ ℂ ∧ (ℑ‘𝐴) ≤ 0) → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))

Theoremasinlem3 24285 The argument to the logarithm in df-asin 24279 has nonnegative real part. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → 0 ≤ (ℜ‘((i · 𝐴) + (√‘(1 − (𝐴↑2))))))

Theoremasinf 24286 Domain and range of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arcsin:ℂ⟶ℂ

Theoremasincl 24287 Closure for the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arcsin‘𝐴) ∈ ℂ)

Theoremacosf 24288 Domain and range of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arccos:ℂ⟶ℂ

Theoremacoscl 24289 Closure for the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arccos‘𝐴) ∈ ℂ)

Theorematandm 24290 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 24291 This form of atandm 24290 is a bit more useful for showing that the logarithms in df-atan 24281 are well-defined. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 − (i · 𝐴)) ≠ 0 ∧ (1 + (i · 𝐴)) ≠ 0))

Theorematandm3 24292 A compact form of atandm 24290. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (𝐴↑2) ≠ -1))

Theorematandm4 24293 A compact form of atandm 24290. (Contributed by Mario Carneiro, 3-Apr-2015.)
(𝐴 ∈ dom arctan ↔ (𝐴 ∈ ℂ ∧ (1 + (𝐴↑2)) ≠ 0))

Theorematanf 24294 Domain and range of the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
arctan:(ℂ ∖ {-i, i})⟶ℂ

Theorematancl 24295 Closure for the arctan function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ dom arctan → (arctan‘𝐴) ∈ ℂ)

Theoremasinval 24296 Value of the arcsin function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arcsin‘𝐴) = (-i · (log‘((i · 𝐴) + (√‘(1 − (𝐴↑2)))))))

Theoremacosval 24297 Value of the arccos function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℂ → (arccos‘𝐴) = ((π / 2) − (arcsin‘𝐴)))

Theorematanval 24298 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 24299 A real number is in the domain of the arctangent function. (Contributed by Mario Carneiro, 31-Mar-2015.)
(𝐴 ∈ ℝ → 𝐴 ∈ dom arctan)

Theoremasinneg 24300 The arcsine function is odd. (Contributed by Mario Carneiro, 1-Apr-2015.)
(𝐴 ∈ ℂ → (arcsin‘-𝐴) = -(arcsin‘𝐴))

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