Solveeit Logo
Login

Question

Mathematics Question on Trigonometric Functions

What is the value of sin⁡ 1950∘−cos⁡ 1950∘\sin \, 1950^{\circ} - \cos \, 1950^{\circ}?

A

0

B

(3+1)/2(\sqrt{3} + 1) / 2

C

(1−3)/2( 1 - \sqrt{3} ) / 2

D

(3−1)/2(\sqrt{3} - 1) / 2

Answer

(3+1)/2(\sqrt{3} + 1) / 2

Explanation

Solution

The given expression sin⁡ 1950∘−cos⁡ 1950∘\sin \, 1950^{\circ} - \cos \, 1950^{\circ} can also be written as =sin⁡(10×180∘+150∘)−cos⁡(10×180∘+150∘)= \sin (10 \times 180^{\circ} + 150^{\circ} ) - \cos (10 \times 180^{\circ} + 150^{\circ}) =sin⁡150∘−cos⁡150∘= \sin 150^{\circ}- \cos 150^{\circ} =sin⁡(90∘+60∘)−cos⁡(90∘+60∘)= \sin (90^{\circ} + 60^{\circ}) - \cos (90^{\circ} + 60^{\circ}) =cos⁡60∘+sin⁡60∘= \cos 60^{\circ} + \sin 60^{\circ} =12+32=3+12 = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{\sqrt{3} + 1}{2}