Trigonometry & Inverse Trigonometry
Trigonometric inequalities
Grade 11
Question:
<p>In the inequality below, the value of the angle is expressed in radian measure. Which one of the inequalities below is true?</p><p>(a) \(\sin 1 < \sin 2 < \sin 3\)</p><p>(b) \(\sin 3 < \sin 2 < \sin 1\)</p><p>(c) \(\sin 2 < \sin 1 < \sin 3\)</p><p>(d) \(\sin 3 < \sin 1 < \sin 2\)</p>
<p>(a) \(\sin 1 < \sin 2 < \sin 3\)</p>
<p>(b) \(\sin 3 < \sin 2 < \sin 1\)</p>
<p>(c) \(\sin 2 < \sin 1 < \sin 3\)</p>
<p>(d) \(\sin 3 < \sin 1 < \sin 2\)</p>
Step-by-Step Solution
Key Concept: Use the sum-to-product formula $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$ to compare sine values at different radian measures and combine the inequalities.
<p><strong>Step 1:</strong> We have $\sin 1 - \sin 2 = -2\cos\frac{3}{2}\sin\frac{1}{2} < 0$</p><p>Therefore, $\sin 1 < \sin 2$ — (i)</p><p><strong>Step 2:</strong> Similarly, $\sin 1 - \sin 3 = -2\cos 2 \sin 1 > 0$</p><p>Therefore, $\sin 1 > \sin 3$ — (ii)</p><p><strong>Step 3:</strong> From equations (i) and (ii), we get $\sin 3 < \sin 1 < \sin 2$</p><p>∴ Answer is (d).</p>
Correct Answer: D