## Calculus: Early Transcendentals 8th Edition

(a) $sin(-\theta)=-sin\theta$ (b) $cos(-\theta)=cos\theta$
(a) Need to prove $sin(-\theta)=-sin\theta$ For a right triangle as depicted below $sin\theta=\frac{y}{\sqrt {x^{2}+y^{2}}}$ If we interchange the sign for $\theta$, we will have $sin(-\theta)=-\frac{y}{\sqrt {x^{2}+y^{2}}}$ Hence, $sin(-\theta)=-sin\theta$ (b) (a) Need to prove $cos(-\theta)=cos\theta$ For a right triangle , $cos\theta=\frac{x}{\sqrt {x^{2}+y^{2}}}$ If we interchange the sign for $\theta$, we will have $cos(-\theta)=\frac{x}{\sqrt {x^{2}+y^{2}}}$ Hence, $cos(-\theta)=cos\theta$