you have that cos cos is a holomorphic function, which coincides with the cosine defined on the real numbers. Therefore, by the uniqueness theorem about holomorphic functions that coincide on a set with an accumulation point, the identity

In mathematics, hyperbolic functions are analogues of the ordinary trigonometric functions, but defined using the hyperbola rather than the circle. Just as the points (cos t, sin t) form a circle with a unit radius, the points (cosh t, sinh t) form the right half of the unit hyperbola.

剩下三个等式证明显然 练手题（选自龚昇《简明 复分析》） 求 cos (x+iy),sin (x+iy) 的实部和虚部，其中 x,y\in R. 求解过程如下： cos (x+iy)=cosxcosiy-sinxsiniy=cosxcoshy-isinxsinhy sin (x+iy)=sinxcosiy+cosxsiniy=sinxcoshy+icosxsinhy

How to show $$\sin (x+iy)=\sin (x) \cosh (y) + i\cos (x) \sinh (y)$$ I begin with $$\sin (x+iy) = \frac {e^ {x+iy}-e^ {-x-iy}} {2i} = \frac {e^xe^ {iy}-e^ {-x}e^ {-iy}} {2i}$$ $$ = \frac {e^xe^ {iy}-e^ {-x}e^ {iy}+...

\displaystyle \text {cosh}\ x - \text {cosh}\ y = 2 \text {sinh}\ \frac12 (x + y)\ \text {sinh}\ \frac12 (x - y) cosh x−cosh y = 2sinh 21(x+y) sinh 21(x−y)

2016年4月10日 · In geometric terms, cosθ is the x -coordinate of the point on the unit circle that's a counterclockwise rotation of θ (radians) from the positive x -axis (“at an angle of θ ”).

2022年11月11日 · 证明： cosh (x+y)=coshx\cdot coshy+sinhx\cdot sinhy 欧拉公式法e^ {ix} = (cos x+isin x) （欧拉公式）sinh x = \frac { (e^x-e^ {-x})} {2} cosh x = \frac { (e^x+e^ {-x})} {2} 易得， cosh (ix) = cosx ; sinh (ix) = …

2020年4月14日 · Real part of cosh (x+iy) = cosh x cos y Imaginary part of cosh (x+iy) = sinh x sin y Given : The function cosh (x+iy) To find : Real part and Imaginary part Solution : Step 1 of 2 : …

For any z ∈ C, we define the hyperbolic sine function by ez − e−z sinh(z) = cosh(z) = . Proposition 21.1. For any z ∈ C, cosh(z) = sinh(z). Proof. We have. = = sinh(z). and cosh(z) = cos(iz) and sinh(z) = −i sin(iz). It follows that sinh(−z) = −i sin(−iz) = i sin(iz) = − sinh(z),

\ [\tanh (x-y) = \frac {\tanh x - \tanh y} {1-\tanh x\tanh y} \] 积化和差公式 \ [\sin x\cos y = \frac {1} {2} [\sin (x+y) + \sin (x-y)] \] \ [\cos x\sin y = \frac {1} {2} [\sin (x+y) - \sin (x-y)] \] \ [\cos x\cos y = \frac {1} {2} [\cos (x+y) + \cos (x-y)] \] \ [\sin x\sin y = -\frac {1} {2} [\cos (x+y) - \cos (x-y)] \] 和差化积公式