Question

The value of the expression $\frac{2(\sin 1^o+\sin 2^o+\sin 3^o+...+\sin {89}^o)}{2(\cos 1^o+\cos 2^o+...+\cos {44}^o)+1}$ equals

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is A $\sqrt{2}$

Consider the numerator

$N^r=2(sin{1}^0+sin{2}^0+sin{3}^0+......+sin{89}^0)$

$\Rightarrow N^r=2\left[\right(sin{1}^0+sin{89}^0)+(sin{2}^0+sin{88}^0)+.....+(sin{44}^0+sin{46}^0)+sin{45}^0]$

We know that $sinC+sinD=2sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)$

Using this result we get,

$N^r=2\left[sin{45}^0\right[2(cos{44}^0+cos{43}^0+cos{42}^0+....cos{1}^0)]+1]$

$\Rightarrow \frac{N^r}{D^r}=2sin{45}^0$

$\Rightarrow \frac{N^r}{D^r}=\sqrt{2}$