What is a value of ³ x + ³ y?

Correct Answer is : rac368rac{3{6}}{8}rac368

Correct Answer is : rac368rac{3{6}}{8}rac368

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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 : 78 of 120

Marks: +1, -0

What is a value of

\cos^{3} x + \cos^{3} y

Solution:

Concept:

Use trigonometric sum-to-product identities and algebraic expansion to simplify and compute the required expression.

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 standard identities:

\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}

Substitute:

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

Cancel

2 \sin\frac{x+y}{2}

(non‑zero because

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

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

Since

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

, we get

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

Divide:

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

→

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

→

x - y = 60^{\circ}

Given

x + y = 90^{\circ}

, solve:

x = 75^{\circ}

y = 15^{\circ}

Now

\cos^3 x + \cos^3 y = (\cos x + \cos y)(\cos^2 x + \cos^2 y - \cos x \cos y)

Rewrite as

(\cos x + \cos y)[(\cos x + \cos y)^2 - 3\cos x \cos y]

Compute

\cos 75^{\circ} = \frac{\sqrt{3}-1}{2\sqrt{2}}

\cos 15^{\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}

Sum:

\cos 75^{\circ} + \cos 15^{\circ} = \frac{\sqrt{3}-1+\sqrt{3}+1}{2\sqrt{2}} = \frac{2\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2}

Product:

\cos 75^{\circ} \cos 15^{\circ} = \frac{(\sqrt{3}-1)(\sqrt{3}+1)}{(2\sqrt{2})(2\sqrt{2})} = \frac{3-1}{8} = \frac{1}{4}

Then

(\cos x + \cos y)^2 = \left(\frac{\sqrt{6}}{2}\right)^2 = \frac{6}{4} = \frac{3}{2}

Substitute:

\cos^3 x + \cos^3 y = \frac{\sqrt{6}}{2} \left[ \frac{3}{2} - 3 \cdot \frac{1}{4} \right] = \frac{\sqrt{6}}{2} \left[ \frac{3}{2} - \frac{3}{4} \right] = \frac{\sqrt{6}}{2} \cdot \frac{3}{4} = \frac{3\sqrt{6}}{8}

Answer:

rac{3\sqrt{6}}{8}

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