Question

Prove that $\sin {35}^o\sin {55}^o-\cos {35}^o\cos {55}^o=0$

Answer 1

Consider $\sin {35}^0$ and $\sin {55}^0$ which can be rewritten as follows:

$\sin {35}^0=\sin {({90}^0-55}^0)=\cos {55}^0$

$\sin {55}^0=\sin {({90}^0-35}^0)=\cos {35}^0$

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

Therefore, $\sin {35}^0\sin {55}^0-\cos {35}^0\cos {55}^0$ can be evaluated as shown below:

$\sin {35}^0\sin {55}^0-\cos {35}^0\cos {55}^0=\cos {35}^0\cos {55}^0-\cos {35}^0\cos {55}^0=0$

Hence, $\sin {35}^0\sin {55}^0-\cos {35}^0\cos {55}^0=0$.