Answer
(a) $sin(-\theta)=-sin\theta$
(b) $cos(-\theta)=cos\theta$
Work Step by Step
(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, $cos(-\theta)=cos\theta$
