Question

$\sin {70}^{\circ }\sin {50}^{\circ }-\cos {85}^{\circ }\cos {65}^{\circ }=$

• A. $0$
• B. $1$
• C. $\frac{\sqrt{5}+1}{4}$
• D. $\frac{\sqrt{3}+1}{4}$

Answer 1

The correct option is D $\frac{\sqrt{3}+1}{4}$

Let $\sin 70\cdot \sin 50-\cos 85\cdot \cos 65=I$

Using $2\sin A\sin B=\cos (A+B)-\cos (A-B)$

$2\cos A\cos B=\cos (A-B)+\cos (A+B)$

We get,

$I=\frac{1}{2}(\cos 20-\cos 120-\cos 150-\cos 20)$

$=\frac{1}{2}(-\cos 120-\cos 150)$

$=\frac{1}{2}(\sin 30+\sin 60)$

$=\frac{1}{2}(\frac{1}{2}+\frac{\sqrt{3}}{2})$

$=\frac{\sqrt{3}+1}{4}$