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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 77-78

Consider the following for the next items that follow:

Let sin x + sin y = √3 (cos y - cos x); x + y =

\frac{\pi }{2}

, 0 < x, y <

\frac{\pi }{2}

Question : 77 of 120

Marks: +1, -0

What is a value of sin 3x + sin 3y?

Solution:

Concept:

Use sum-to-product formulas:

\sin C + \sin D = 2 \sin\frac{C+D}{2}\cos\frac{C-D}{2}

and

\cos C - \cos D = -2 \sin\frac{C+D}{2}\sin\frac{C-D}{2}

Explanation:

Given:

\sin x + \sin y = \sqrt{3}(\cos y - \cos x)

and

x+y = \frac{\pi}{2}

, with

0 < x,y < \frac{\pi}{2}

Apply sum-to-product:

2\sin\frac{x+y}{2}\cos\frac{x-y}{2} = \sqrt{3}\left(-2\sin\frac{y+x}{2}\sin\frac{y-x}{2}\right)

Since

\sin\frac{x+y}{2} = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}

cancels on both sides, we get:

\cos\frac{x-y}{2} = -\sqrt{3}\sin\frac{y-x}{2}

Using

\sin\frac{y-x}{2} = -\sin\frac{x-y}{2}

, the equation becomes:

\cos\frac{x-y}{2} = \sqrt{3}\sin\frac{x-y}{2}

Thus

\tan\frac{x-y}{2} = \frac{1}{\sqrt{3}}

, so

\frac{x-y}{2} = 30^{\circ}

(or

\frac{\pi}{6}

), hence

x - y = 60^{\circ}

Combined with

x + y = 90^{\circ}

, solving gives

x = 75^{\circ}

y = 15^{\circ}

Now

\sin 3x + \sin 3y = 2\sin\frac{3x+3y}{2}\cos\frac{3x-3y}{2}

Compute

3x+3y = 3\cdot 90^{\circ} = 270^{\circ}

, so

\frac{3x+3y}{2} = 135^{\circ}

; and

3x-3y = 3\cdot 60^{\circ} = 180^{\circ}

, so

\frac{3x-3y}{2} = 90^{\circ}

Thus

\sin 3x + \sin 3y = 2\sin 135^{\circ} \cos 90^{\circ} = 2\cdot \frac{\sqrt{2}}{2} \cdot 0 = 0

Answer:

0

(Option B).

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