## Trigonometry 7th Edition

Counterexample: $\displaystyle \theta=-\frac{\pi}{2}$ (sample answer)
The problem that leads us to search for a counterexample is that on the RHS, we have a POSITIVE square root of a number. So, we find a $\theta$ for which $\sin\theta$ is negative. Sine is negative in quadrants III and IV, so we can select $\displaystyle \theta=-\frac{\pi}{2}$ (sample) LHS$=-1$ RHS=$\sqrt{1-0}=+1$ LHS$\neq$RHS so the identity is not valid.