Chapter 7.4, Problem 82AYU

establish each identity.

$\frac{{\sin }^3\theta +co{s}^3\theta }{1-2{\cos }^2\theta }=\frac{\sec \theta -\sin \theta }{\tan \theta -1}$

To determine

To find: The trigonometric expression $\frac{{\text{sin}}^3\theta +{\text{ cos}}^3\theta }{1-2 {\text{cos}}^2\theta }=\frac{\sec \theta -\sin \theta }{\tan \theta -1}$.

Answer to Problem 82AYU

Solution:

The trigonometric expression $\frac{{\text{sin}}^3\theta +{\text{ cos}}^3\theta }{1-2 {\text{cos}}^2\theta }=\frac{\sec \theta -\sin \theta }{\tan \theta -1}$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{{\text{sin}}^3\theta +{\text{ cos}}^3\theta }{1-2 {\text{cos}}^2\theta }=\frac{\sec \theta -\sin \theta }{\tan \theta -1}$.

Calculation:

Left hand side

$\frac{{\text{sin}}^3\theta +{\text{ cos}}^3\theta }{1-2 {\text{cos}}^2\theta }$

Expanding the numerator and by Pythagorean identity in the denominator

$\frac{\left(\sin \theta +\cos \theta \right)\left({\text{sin}}^3\theta -\sin \theta \cos \theta +{\text{ cos}}^3\theta \right)}{{\text{sin}}^2\theta +{\text{ cos}}^2\theta -2{\text{cos}}^2\theta }$

$\frac{\left(\sin \theta +\cos \theta \right)\left({\text{sin}}^3\theta -\sin \theta \cos \theta +{\text{ cos}}^3\theta \right)}{{\text{sin}}^2\theta -{\text{ cos}}^2\theta }$

Expanding the denominator

$\frac{\left(\sin \theta +\cos \theta \right)\left({\text{sin}}^3\theta -\sin \theta \cos \theta +{\text{ cos}}^3\theta \right)}{(\sin \theta +\cos \theta )(\sin \theta -\cos \theta )}$

Cancelling the common factor $\left(\sin \theta +\cos \theta \right)$

$\frac{\left({\text{sin}}^3\theta -\sin \theta \cos \theta +{\text{ cos}}^3\theta \right)}{(\sin \theta -\cos \theta )}$

By Pythagorean identity

$\frac{\left(1-\sin \theta \cos \theta \right)}{(\sin \theta -\cos \theta )}$

Right hand side

$\frac{\sec \theta -\sin \theta }{\tan \theta -1}$

By Quotient identity

$\frac{\frac{1}{\cos \theta }-\sin \theta }{\frac{\sin \theta }{\cos \theta }-1}$

Adding the quotients and cancelling the common factor $\cos \theta$

$\frac{1-\sin \theta \cos \theta }{\sin \theta -\cos \theta }$

$\text{LHS = RHS}$