Question

Pythagorean identities: Use Pythagorean identities to rewrite each of the following trigonometric expressions.

${\csc }^{8}x=\_\_\_\_$

Short answer

$\frac{1}{8}\left[1-\cos 2x\frac{\left(\cos 4x+1\right)}{2}\right]$

Step-by-step solution

Step 1. Given information.

Consider the given information,

${\csc }^{8}x$

Step 2. Calculation.

Simplify the expression with the Pythagorean theorem.

$\begin{array}{rcl}{\cos }^{6}x& =& {\left({\cos }^{2}x\right)}^3\\ & =& {\left(1-{\sin }^{2}x\right)}^3\\ & =& {\left(1-\frac{1-\cos 2x}{2}\right)}^3\\ & =& \frac{1}{8}{\left(1-\cos 2x\right)}^3\\ & =& \frac{1}{8}\left(1-\cos 2x\right)\left(1+\cos 2x+{\cos }^{2}2x\right)\\ & =& \frac{1}{8}\left[1+\cos 2x+{\cos }^{2}2x-\cos 2x\left(1+\cos 2x+{\cos }^{2}2x\right)\right]\\ & =& \frac{1}{8}\left[1+\cos 2x+\frac{\cos 4x+1}{2}-\left(\cos 2x+\frac{\cos 4x+1}{2}+\cos 2x\frac{\cos 4x+1}{2}\right)\right]\\ & =& \frac{1}{8}\left[1+\cos 2x+\frac{\left(\cos 4x+1\right)}{2}-\cos 2x-\frac{\left(\cos 4x+1\right)}{2}-\cos 2x\frac{\left(\cos 4x+1\right)}{2}\right]\\ & =& \frac{1}{8}\left[1-\cos 2x\frac{\left(\cos 4x+1\right)}{2}\right]\end{array}$