Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions
Grade 12
Question:
<p>Suppose <math>3\sin^{-1}(\log_2 x) + \cos^{-1}(\log_2 y) = \frac{\pi}{2}</math> and <math>\sin^{-1}(\log_2 x) + 2\cos^{-1}(\log_2 y) = \frac{11\pi}{6}</math> then the value of <math>x^2 + y^2</math> equals</p>
<p>(a) 6</p>
<p>(b) 7</p>
<p>(c) 5</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: Solve the system of inverse trigonometric equations by substitution to find the logarithmic values, then convert back to original variables.
<p><strong>Solution:</strong></p><p>Let <math>\sin^{-1}(\log_2 x) = a</math> and <math>\cos^{-1}(\log_2 y) = b</math></p><p>Then: <math>3a + b = \frac{\pi}{2}</math> and <math>a + 2b = \frac{11\pi}{6}</math></p><p>Solving these equations: <math>a = \frac{\pi}{6}</math> and <math>b = \frac{\pi}{6}</math></p><p>Hence, <math>\log_2 x = \sin\frac{\pi}{6} = \frac{1}{2}</math> and <math>\log_2 y = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}</math></p><p>Therefore, <math>x^2 + y^2 = 6</math></p>
Correct Answer: A