Trigonometry Practice Test 2

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Question : 13 of 22

Marks: +1, -0

The value of

\sin ({67 1/2} °)\sin ({22 1/2} °)

is:

$-2√2$
$2√2$
$1/{2√2}$
$-1/{2√2}$

Solution:

\sin ({67 1/2} °)\sin ({22 1/2} °)

=\sin (45°+{22 1/2}°)\sin (45°-{22 1/2}°)

=\sin^2 45°-\sin^2 {22 1/2}°

=(1/√2)^2-{1-\cos 45°}/2

=1/2-1/2(1-1/√2)

=1/{2√2}

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