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Question

Mathematics Question on Trigonometric Functions

The value of cos⁡ 15∘−sin⁡ 15∘\cos\, 15^{\circ} - \sin \,15^{\circ} is

A

0

B

12\frac{1}{\sqrt{2}}

C

−12-\frac{1}{\sqrt{2}}

D

122\frac{1}{2\sqrt{2}}

Answer

12\frac{1}{\sqrt{2}}

Explanation

Solution

Now, cos⁡15∘−sin⁡15∘\cos 15^{\circ}-\sin 15^{\circ}
=2(12cos⁡15∘−12sin⁡15∘)=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos 15^{\circ}-\frac{1}{\sqrt{2}} \sin 15^{\circ}\right)
=2(sin⁡45∘cos⁡15∘−cos⁡45∘sin⁡15∘)= \sqrt{2}\left(\sin 45^{\circ} \cos 15^{\circ}-\cos 45^{\circ} \sin 15^{\circ}\right)
=2sin⁡(45∘−15∘)=\sqrt{2} \sin \left(45^{\circ}-15^{\circ}\right)
=2sin⁡30∘=2×12= \sqrt{2} \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}
=12= \frac{1}{\sqrt{2}}