Question

Establish each identity.

${(a\sin \theta +b\cos \theta )}^2+{(a\cos \theta -b\sin \theta )}^2=a^2+b^2$

Short answer

The left side of the given equation has been simplified to bring the right side.

Step-by-step solution

Step 1. Given Information

The given equation is${(a\sin \theta +b\cos \theta )}^2+{(a\cos \theta -b\sin \theta )}^2=a^2+b^2$

Step 2. Calculation 

Consider the left side of the given equation and simplify to bring the right side in order to establish the equation.

$$\begin{gathered} {(a\sin \theta +b\cos \theta )}^2+{(a\cos \theta -b\sin \theta )}^2=a^2{\sin }^2\theta +b^2{\cos }^2\theta +2ab\sin \theta \cos \theta +a^2{\cos }^2\theta +b^2{\sin }^2\theta -2ab\sin \theta \cos \theta \\ =a^2({\sin }^2\theta +{\cos }^2\theta )+b^2({\cos }^2\theta +{\sin }^2\theta ) \\ =a^2+b^2 \end{gathered}$$