Trigonometry & Inverse Trigonometry
Trigonometric Equations
Grade 11
Question:
<p>If \(2\cos A = \cos B + \cos 3B\) and \(2\sin A = \sin B - \sin 3B\), then the possible value of \(\sin(A - B)\) is/are:</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{1}{3}\)</p>
<p>(c) \(-\frac{1}{2}\)</p>
<p>(d) \(-\frac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: Square both equations and add them to find a relationship, then use the constraint to find \(\sin(A-B)\).
<p>Square and add the two given equations: \(4\cos^2 A + 4\sin^2 A = (\cos B + \cos 3B)^2 + (\sin B - \sin 3B)^2\). Expanding and simplifying: \(4 = 2 + 2\cos(B - 3B) = 2 + 2\cos 2B\), so \(\cos 2B = 1\). From the original equations, \(\sin(A-B)\) can be determined.</p>
Correct Answer: b, d