Question

tanho
10. Express cosh^7theta? in terms of hyperbolic cosines of multiples of theta​

Answer 1

Answer:

$\cos ^7 \theta=\frac{1}{64}[\cos 7 \theta+7 \cos 5 \theta+21 \cos 3 \theta+35 \cos \theta]$

Step-by-step explanation:

The trigonometric or circular functions have analogues in the hyperbolic functions. The hyperbolic function may be found in Laplace's equations in cartesian coordinates, solutions to linear differential equations, and calculations of distance and angles in hyperbolic geometry. The hyperbolic function, often known as the hyperbolic angle, typically occurs in the real argument. The fundamental hyperbolic operations are:

Overextended sine (sinh)

distorted sine (cosh)

hyperbolic digression (tanh)

$\begin{aligned} & \text { Let, } x=\cos \theta+i \sin \theta \quad \frac{1}{x}=\cos \theta-i \sin \theta \\ & \therefore x^n=(\cos \theta+i \sin \theta)^n=\cos n \theta+i \sin \theta \\ & \text { Also } \frac{1}{x^n}=(\cos \theta-i \sin \theta)^n=\cos n \theta-i \sin n \theta \\ & \{\because \text { DeMoivre s Theorem }\}\end{aligned}$

$x^n+\frac{1}{x^n}=2 \cos n \theta$

$\cos ^7 \theta=(\cos \theta)^7=\left[\frac{1}{2}\left(x+\frac{1}{x}\right)\right]^7=\frac{1}{2^7}\left(x+\frac{1}{x}\right)^7 \quad$ from (i)

$\begin{aligned} & (\cos \theta)^7= \\ & \frac{1}{2^7}\left[x^7+7 x^6 \cdot \frac{1}{x}+21 x^5 \cdot \frac{1}{x^2}+35 x^4 \cdot \frac{1}{x^3}+35 x^3 \cdot \frac{1}{x^4}+21 x^2 \cdot \frac{1}{x^5}+7 x \cdot \frac{1}{x^6}+\frac{1}{x^7}\right] \\ & \ldots \text { Expanding binomially }\end{aligned}$

$\begin{aligned} & =\frac{1}{128}\left[x^7+7\left(x^5+\frac{1}{x^5}\right)+21\left(x^3+\frac{1}{x^3}\right)+35\left(x+\frac{1}{x}\right)+\frac{1}{x^7}\right] \\ & =\frac{1}{128}[2 \cos 7 \theta+7(2 \cos 5 \theta)+21(2 \cos 3 \theta)+35(2 \cos \theta)] \quad \text { from }(i) \\ & \therefore \cos ^7 \theta=\frac{1}{64}[\cos 7 \theta+7 \cos 5 \theta+21 \cos 3 \theta+35 \cos \theta]\end{aligned}$

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