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UPSC NDA I and II 2020 Math Solved Paper

Question Numbers: 52-54

Direction: Read the following information and answer the three items that follow:

Let α = β = 15°.

Question : 53 of 120

Marks: +1, -0

What is the value of sin 7α - cos 7β ?

Solution:

Concept:

Use trigonometric identities:

\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}

and

\sin(90^\circ+\theta)=\cos\theta

\cos(90^\circ+\theta)=-\sin\theta

Explanation:

Given expression:

\sin 7\alpha - \cos 7\beta

Substitute

\alpha=15^\circ

\beta=15^\circ

, then

7\alpha=105^\circ

7\beta=105^\circ

\sin 105^\circ - \cos 105^\circ

Write

\sin 105^\circ = \sin(90^\circ+15^\circ) = \sin90^\circ\cos15^\circ+\cos90^\circ\sin15^\circ = \cos15^\circ

Write

\cos 105^\circ = \cos(90^\circ+15^\circ) = \cos90^\circ\cos15^\circ - \sin90^\circ\sin15^\circ = -\sin15^\circ

Thus

\sin 105^\circ - \cos 105^\circ = \cos15^\circ - (-\sin15^\circ) = \cos15^\circ + \sin15^\circ

Rewrite

\cos15^\circ = \sin75^\circ

, so we have

\sin15^\circ + \sin75^\circ

Apply sum-to-product:

2\sin\frac{75+15}{2}\cos\frac{75-15}{2} = 2\sin45^\circ\cos30^\circ

Compute:

2 \times \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{\sqrt{2}}

Answer:

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

(Option D).

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