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Trigonometry Practice Test 2

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Question : 13 of 22

Marks: +1, -0

The value of

\sin\left(67\frac{1}{2}^{\circ}\right)\sin\left(22\frac{1}{2}^{\circ}\right)

is:

$-2\sqrt{2}$
$2\sqrt{2}$
$\frac{1}{2\sqrt{2}}$
$-\frac{1}{2\sqrt{2}}$

Solution:

Concept:

Use the product-to-sum identity:

\sin A \sin B = \frac{1}{2}[\cos(A-B)-\cos(A+B)]

.

Explanation:

Let

A=67\frac{1}{2}^{\circ}

and

B=22\frac{1}{2}^{\circ}

.

Then

A-B=45^{\circ}

and

A+B=90^{\circ}

.

So,

\sin 67\frac{1}{2}^{\circ}\sin 22\frac{1}{2}^{\circ}=\frac{1}{2}[\cos45^{\circ}-\cos90^{\circ}]

.

Substitute

\cos45^{\circ}=\frac{1}{\sqrt{2}}

and

\cos90^{\circ}=0

.

Thus, value

=\frac{1}{2}\left(\frac{1}{\sqrt{2}}-0\right)=\frac{1}{2\sqrt{2}}

.

Answer:

Option C:

\frac{1}{2\sqrt{2}}

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