Chapter 14, Problem 44E

Solution

To Verify: The identity ${\sin }^2\left(\theta -\frac{\pi }{2}\right)={\cos }^2\theta$ .

The given identity ${\sin }^2\left(\theta -\frac{\pi }{2}\right)={\cos }^2\theta$ is verified.

Given information:

The trigonimetric identity ${\sin }^2\left(\theta -\frac{\pi }{2}\right)={\cos }^2\theta$

Concept and Formula Used:

We use the trigonometric identity $\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta ,\sin (-\theta )=-\sin \theta$ .

Calculation:

The given identity is ${\sin }^2\left(\theta -\frac{\pi }{2}\right)={\cos }^2\theta$

Taking LHS

  $\begin{array}{l}{\sin }^2\left(\theta -\frac{\pi }{2}\right)={\left(\sin \left[-\left(\frac{\pi }{2}-\theta \right)\right]\right)}^2\\ {\sin }^2\left(\theta -\frac{\pi }{2}\right)={\left(-\sin \left[\left(\frac{\pi }{2}-\theta \right)\right]\right)}^2\left(\sin \left(-\theta \right)=-\sin \theta \right)\\ {\sin }^2\left(\theta -\frac{\pi }{2}\right)={\left(-\cos \theta \right)}^2\left(\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta \right)\\ {\sin }^2\left(\theta -\frac{\pi }{2}\right)={\cos }^2\theta \\ \text{=RHS}\end{array}$

Hence verified

Conclusion:

The given identity ${\sin }^2\left(\theta -\frac{\pi }{2}\right)={\cos }^2\theta$ is verified.