Show that the equation is not an identity by finding a value of x for which both sides are defined but not equal.
-$\sin ( x + \pi ) = \sin x$
A) $\frac { \pi } { 2 }$
B) 0
C) $2 \pi$
D) $\pi$

Question

Quizplus · Accepted Answer

For $x = \frac{\pi}{2}$, the left side becomes $\sin(\frac{\pi}{2} + \pi) = \sin(\frac{3\pi}{2}) = -1$, while the right side is $\sin(\frac{\pi}{2}) = 1$, showing the equation is not an identity.

Show That the Equation Is Not an Identity by Finding sin⁡(x+π)=sin⁡x\sin ( x + \pi ) = \sin xsin(x+π)=sinx

Show That the Equation Is Not an Identity by Finding $\sin ( x + \pi ) = \sin x$