Question

\[ \sin^25^\circ+\sin^210^\circ+\sin^215^\circ+\cdots+\sin^285^\circ+\sin^290^\circ \]

is equal to

(a) \(7\)
(b) \(8\)
(c) \(9.5\)
(d) \(10\)

Solution

Using identity

\[ \sin^2\theta+\sin^2(90^\circ-\theta)=1 \]

Pair the terms:

\[ \sin^25^\circ+\sin^285^\circ=1 \]

\[ \sin^210^\circ+\sin^280^\circ=1 \]

\[ \sin^215^\circ+\sin^275^\circ=1 \]

Similarly,

\[ 8 \text{ such pairs } =8 \]

Middle term:

\[ \sin^245^\circ=\frac12 \]

Last term:

\[ \sin^290^\circ=1 \]

Therefore,

\[ 8+\frac12+1 = \frac{19}{2} = 9.5 \]

Answer

\[ \boxed{9.5} \]

Correct Option: (c)