Trigonometry & Inverse Trigonometry
Linear combinations of trigonometric functions
Grade 12
Question:
<p><strong>Ex. 65:</strong> Let $f(x) = ab \sin x + b\sqrt{1 - a^2} \cos x + c$, where $|a| < 1, b > 0$ then</p>
<p>(a) maximum value of $f(x)$ is $b + c$ when $c = 0$</p>
<p>(b) difference of maximum and minimum values of $f(x)$ is $2b$</p>
<p>(c) $f(x) = c$ if $x = -\cos^{-1} a$</p>
<p>(d) $f(x) = c$ if $x = \cos^{-1} a$</p>
Step-by-Step Solution
Key Concept: Express a linear combination of sine and cosine as a single sinusoidal function using the formula $A\sin x + B\cos x = \sqrt{A^2+B^2}\sin(x+\phi)$ to find extrema.
<p><strong>Step 1:</strong> Given $f(x) = ab \sin x + b\sqrt{1 - a^2} \cos x + c$ where $|a| < 1, b > 0$</p><p><strong>Step 2:</strong> The maximum value of $A\sin x + B\cos x$ is $\sqrt{A^2 + B^2}$</p><p><strong>Step 3:</strong> Here $A = ab$ and $B = b\sqrt{1-a^2}$</p><p><strong>Step 4:</strong> Maximum of sine-cosine combination: $\sqrt{(ab)^2 + (b\sqrt{1-a^2})^2} = \sqrt{a^2b^2 + b^2(1-a^2)} = \sqrt{b^2} = b$</p><p><strong>Step 5:</strong> Therefore, maximum value of $f(x)$ is $b + c$ and minimum value is $-b + c$</p><p><strong>Step 6:</strong> Difference of maximum and minimum = $(b+c) - (-b+c) = 2b$</p><p>∴ Answers (a), (b), (c) are correct.</p>
Correct Answer: A