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WAEC Mathematics 2010 Paper

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Question : 16

Total: 49

If sin 3y = cos 2y and 0^o

\leq

90^o, find the value of y

18^o
36^o
54^o
90^o

Solution:

Concept:

Use the co-function identity

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

and then equate the angle measures.

Explanation:

Given:

\sin 3y = \cos 2y

.

Rewrite the left side using the identity:

\sin 3y = \cos(90^\circ - 3y)

.

So,

\cos(90^\circ - 3y) = \cos 2y

.

Since the cosine angles are equal,

90^\circ - 3y = 2y

.

Solve for

y

:

90^\circ = 5y

.

Therefore,

y = \frac{90^\circ}{5} = 18^\circ

.

This value lies within the given range

0^\circ \leq y \leq 90^\circ

.

Answer:

y = 18^\circ

, which is option A.

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