[tex]\bf \textit{Cofunction Identities} \\\\ sin\left(90^o-\theta\right)=cos(\theta) \qquad cos\left(90^o-\theta\right)=sin(\theta) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ cos(157)\implies cos(90^o+67^o)\implies cos[90^o-(-67^o)] \\\\\\ sin(-67^o)\implies -sin(67^o)[/tex]
you can check the example in the picture below, why the cosine of the complementary angle is equals to the sine of its sibling.
