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UPSC NDA II 2015 Mathematics Paper

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Question : 55 of 120

Marks: +1, -0

The value of

\sin^{2} 5^{\circ} + \sin^{2} 10^{\circ} + \sin^{2} 15^{\circ} + \sin^{2} 20^{\circ} + \dots + \sin^{2} 90^{\circ}

is

7
8
9
$\frac{19}{2}$

Solution:

Concept:

Use the identities

\sin(90^{\circ} - \theta) = \cos \theta

and

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

to pair complementary angles.

Explanation:

The sum includes

\sin^2 5^{\circ}, \sin^2 10^{\circ}, \sin^2 15^{\circ}, \dots, \sin^2 90^{\circ}

(step of

5^{\circ}

).

Pair each angle with its complement:

\sin^2 5^{\circ} + \sin^2 85^{\circ} = \sin^2 5^{\circ} + \cos^2 5^{\circ} = 1

.

Similarly,

\sin^2 10^{\circ} + \sin^2 80^{\circ} = 1

,

\sin^2 15^{\circ} + \sin^2 75^{\circ} = 1

,

\dots

,

\sin^2 40^{\circ} + \sin^2 50^{\circ} = 1

.

This gives

8

pairs, each summing to

1

.

The unpaired terms are

\sin^2 45^{\circ} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}

and

\sin^2 90^{\circ} = 1

.

Total sum

= 8 \times 1 + \frac{1}{2} + 1 = 9.5 = \frac{19}{2}

.

Answer:

\frac{19}{2}

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