(a) Copy and complete the table of values for the relation \(y = 2 \sin

MATHEMATICS
WAEC 2014

(a) Copy and complete the table of values for the relation \(y = 2 \sin x + 1\)

x30°60°90°120°150°180°210°240°

270°

y1.02.70.0-0.7

(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 2 \sin x + 1, 0° \leq x \leq 270°\).

(c) Use the graph to find the values of x for which \(\sin x = \frac{1}{4}\).

(a) Copy and complete the table of values for the relation \(y = 2 \sin x + 1\)

x30°60°90°120°150°180°210°240°

270°

y1.02.70.0-0.7

(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 2 \sin x + 1, 0° \leq x \leq 270°\).

(c) Use the graph to find the values of x for which \(\sin x = \frac{1}{4}\).

Explanation

(a)

x30°60°90°120°150°180°210°240°

270°

y1.02.02.73.02.72.01.00.0-0.7-1.0

(b)graph

(c) \(\sin x = \frac{1}{4}\) (Given)

Multiply through by 2 ; \(2 \sin x = 2 \times \frac{1}{4} = \frac{1}{2}\)

\(2 \sin x = \frac{1}{2} \implies 2 \sin x - \frac{1}{2} = 0\)

Add \(1\frac{1}{2}\) to both sides ;

\(2 \sin x - \frac{1}{2} + 1\frac{1}{2} = 0 + 1\frac{1}{2}\)

\(\implies 2 \sin x + 1 = 1\frac{1}{2}\)

Draw the line \(y = 1\frac{1}{2}\) on the same axis as \(y = 2 \sin x + 1\). The line \(y = 1\frac{1}{2}\) cuts the curve at points P and Q where x = 15° and x = 168°. Hence, the values of x for which \(\sin x = \frac{1}{4}\)are 15° and 168°.



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