Test Index

WAEC Further Mathematics 2015 Paper

© africaexams.com

Question : 35

Total: 40

If

2\sin^{2}\theta = 1 + \cos\theta, 0^{\circ} \leq \theta \leq 90^{\circ}

, find the value of

\theta

.

90°
60°
45°
30°

Solution:

Concept:

Use the identity

\sin^{2}\theta = 1 - \cos^{2}\theta

to form a quadratic equation in

\cos\theta

.

Explanation:

Given:

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

.

Substitute

\sin^{2}\theta = 1 - \cos^{2}\theta

:

2(1 - \cos^{2}\theta) = 1 + \cos\theta

.

Expand:

2 - 2\cos^{2}\theta = 1 + \cos\theta

.

Rearrange:

2\cos^{2}\theta + \cos\theta - 1 = 0

.

Factorize:

(\cos\theta + 1)(2\cos\theta - 1) = 0

.

So

\cos\theta = -1

or

\cos\theta = \frac{1}{2}

.

Since

0^{\circ} \leq \theta \leq 90^{\circ}

,

\cos\theta

is positive, so

\cos\theta \neq -1

.

Thus

\cos\theta = \frac{1}{2}

, giving

\theta = 60^{\circ}

.

Answer:

\theta = 60^{\circ}

, which is option B.

© africaexams.com

Go to Question:

Prev Question

Next Question