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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 36-37

Direction: For the next two (2) items that follow:

Consider a triangle ABC in which

cosA+cosB+cosC=\sqrt{3}sin\frac{\pi }{3}

Question : 36 of 120

Marks: +1, -0

What is the value of

sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2} ?

Solution:

Concept:

In a triangle,

A+B+C=180^\circ

is used along with sum-to-product formulas to simplify trigonometric expressions.

Explanation:

Assume the condition:

\cos A + \cos B + \cos C = \sqrt{3} \sin \frac{\pi}{3}

Using

A+B+C=180^\circ

, we have

\frac{A+B}{2}=90^\circ-\frac{C}{2}

, so

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

Start with the left-hand side:

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

Substitute

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

and

\cos C = 1-2\sin^2\frac{C}{2}

2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2} = \sqrt{3}\sin\frac{\pi}{3}

Factor:

1+2\sin\frac{C}{2}\left(\cos\frac{A-B}{2}-\sin\frac{C}{2}\right) = \sqrt{3}\sin\frac{\pi}{3}

Replace

\sin\frac{C}{2}

with

\cos\frac{A+B}{2}

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

, so

1+2\sin\frac{C}{2}\left(\cos\frac{A-B}{2}-\cos\frac{A+B}{2}\right) = \sqrt{3}\sin\frac{\pi}{3}

Use identity

\cos P - \cos Q = -2\sin\frac{P+Q}{2}\sin\frac{P-Q}{2}

\cos\frac{A-B}{2}-\cos\frac{A+B}{2} = -2\sin\frac{A}{2}\sin\left(-\frac{B}{2}\right) = 2\sin\frac{A}{2}\sin\frac{B}{2}

Thus

1+2\sin\frac{C}{2}\cdot 2\sin\frac{A}{2}\sin\frac{B}{2} = \sqrt{3}\sin\frac{\pi}{3}

Simplify:

1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \sqrt{3}\cdot \frac{\sqrt{3}}{2} = \frac{3}{2}

Subtract 1:

4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \frac{3}{2}-1 = \frac{1}{2}

Divide by 4:

\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \frac{1}{8}

Answer:

\frac{1}{8}

(Option C).

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