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UPSC CDS 1 2020 Math Paper

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Question : 58 of 100

Marks: +1, -0

What is the value of

\sin^2 6^\circ + \sin^2 12^\circ + \sin^2 18^\circ + \dots + \sin^2 84^\circ + \sin^2 90^\circ

?

1
2
4
8

Solution:

Concept:

Use the identity

\sin (90^\circ - heta) = \cos heta

and

\sin^2 heta + \cos^2 heta = 1

to pair terms.

Explanation:

The series is

\sin^2 6^\circ + \sin^2 12^\circ + \dots + \sin^2 84^\circ + \sin^2 90^\circ

.

Notice

\sin 84^\circ = \cos 6^\circ

,

\sin 78^\circ = \cos 12^\circ

, and so on.

Pair the first and last terms:

\sin^2 6^\circ + \sin^2 84^\circ = \sin^2 6^\circ + \cos^2 6^\circ = 1

.

Similarly,

\sin^2 12^\circ + \sin^2 78^\circ = 1

,

\sin^2 18^\circ + \sin^2 72^\circ = 1

,

\sin^2 24^\circ + \sin^2 66^\circ = 1

,

\sin^2 30^\circ + \sin^2 60^\circ = 1

,

\sin^2 36^\circ + \sin^2 54^\circ = 1

,

\sin^2 42^\circ + \sin^2 48^\circ = 1

.

That gives 7 pairs, each equal to 1, so sum so far =

7

.

Add the last term

\sin^2 90^\circ = 1

.

Total sum =

7 + 1 = 8

.

Answer:

8

(Option D).

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