Question

The average value of $\sin 2^o,\sin 4^o,\sin 6^o,.........\sin {180}^o$ is :-

• A. $\frac{1}{90}\cos 1^o$
• B. $\frac{1}{90}\sin 1^o$
• C. $\frac{1}{90}\cot 1^o$
• D. None of these

Answer 1

The correct option is C $\frac{1}{90}\cot 1^o$

$\frac{\sin 2+\sin 4+...\sin 180}{180}=\frac{1}{90}(\sin 2+\sin 4+...+\sin 88+1)$
$=\frac{1}{90}\left(\frac{\sin \frac{44.2}{2}}{\sin \frac{2}{2}}\sin (2+\frac{43.2}{2})+1\right)=\frac{1}{90}(\frac{\sin 44\sin 45}{\sin 1}+1)$
$=\frac{1}{90}\frac{\cos 1}{\sin 1}=\frac{1}{90}\cot 1$