Question

Prove that $\cos {38}^o\cos {52}^o-\sin {38}^o\sin {52}^o=0$

Answer 1

Consider $\cos {38}^0$ and $\cos {52}^0$ which can be rewritten as follows:

$\cos {38}^0=\cos {({90}^0-52}^0)=\sin {52}^0$

$\cos {52}^0=\cos {({90}^0-38}^0)=\sin {38}^0$

$(\because \cos ({90}^0-x)=\sin x)$

Therefore, $\cos {38}^0\cos {52}^0-\sin {38}^0\sin {52}^0$ can be evaluated as shown below:

$\cos {38}^0\cos {52}^0-\sin {38}^0\sin {52}^0=\sin {38}^0\sin {52}^0-\sin {38}^0\sin {52}^0=0$

Hence, $\cos {38}^0\cos {52}^0-\sin {38}^0\sin {52}^0=0$.