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Question

Mathematics Question on Trigonometric Functions

The value of cos2 45∘−sin2 15∘cos^2 \, 45^{\circ} - sin^2 \, 15^{\circ} is

A

32\frac{\sqrt{3}}{2}

B

34\frac{\sqrt{3}}{4}

C

3+122\frac{\sqrt{3} + 1}{2 \sqrt{2}}

D

3−122\frac{\sqrt{3} - 1}{2 \sqrt{2}}

Answer

34\frac{\sqrt{3}}{4}

Explanation

Solution

cos⁡245∘−sin⁡215∘\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}
=cos⁡(45∘+15∘)cos⁡(45∘−15∘)= \cos \left(45^{\circ}+15^{\circ}\right) \cos \left(45^{\circ}-15^{\circ}\right)
[∵cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)]\left[\because \cos ^{2} A-\sin ^{2} B=\cos (A+B) \cos (A-B)\right]
=cos⁡60∘cos⁡30∘=12×32=34= \cos 60^{\circ} \cos 30^{\circ}=\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}