(a) Copy and complete the table of values for the relation \(y = 2 \sin x + 1\)
(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 \(\sin x = \frac{1}{4}\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° |
| y | 1.0 | 2.7 | 0.0 | -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 \(\sin x = \frac{1}{4}\).
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Correct Answer: Option n
Explanation:
(a)
(b)
(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°.
(a)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° |
| y | 1.0 | 2.0 | 2.7 | 3.0 | 2.7 | 2.0 | 1.0 | 0.0 | -0.7 | -1.0 |
(b)
(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°.