Question

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

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

Answer 1

Answer: (A)

$\sqrt{2}$

$k=\frac{2\left(\sin 1^{}+\sin 2^{}+\sin 3^{}+....+\sin {89}^{}\right)}{2(\cos 1^{}+\cos 2^{}+.....+\cos {44}^{})+1}$
$\Rightarrow k=\frac{2\cos {45}^{}(\sin ({90}^{}-{89}^{})+\sin ({90}^{}-{88}^{})+\sin ({90}^{}-{87}^{})+.....+\sin ({90}^{}-{89}^{}))}{\left(2\cos 1^{}\cos {45}^{}\right)+\left(2\cos 2^{}\cos {45}^{}\right)+....+\left(2\cos {44}^{}\cos {45}^{}\right)+\cos {45}^{}}$
We know that $2\cos A\cos B=\cos (A+B)+\cos (A-B)$
$\therefore k=\frac{\sqrt{2}(\cos 1^{}+\cos 2^{}+\cos 3^{}+.....+\cos {89}^{})}{(\cos {46}^{}+\cos {44}^{})+(\cos {47}^{}+\cos {43}^{})+....+(\cos {89}^{}+\cos 1^{})+\cos {45}^{}}\Rightarrow k=\frac{\sqrt{2}(\cos 1^{}+\cos 2^{}+\cos 3^{}+.....+\cos {89}^{})}{(\cos 1^{}+\cos 2^{}+\cos 3^{}+.....+\cos {89}^{})}$
$\Rightarrow k=\frac{\sqrt{2}(\cos 1^{}+\cos 2^{}+\cos 3^{}+.....+\cos {89}^{})}{(\cos 1^{}+\cos 2^{}+\cos 3^{}+.....+\cos {89}^{})}$
$\therefore k=\sqrt{2}$
Ans: A