Question

If $a\sin\theta = b\cos\theta$, where $a, b \neq 0$, then $a\cos 2\theta + b\sin 2\theta = $

• A. $ab$
• B. $a$
• C. $b$
• D. $\frac{a}{b}$

Hint:

To solve symmetric identities like this instantly without algebraic expansions, pick simple numbers that satisfy the given condition. For instance, if $a=1$ and $b=1$, then $\tan\theta = 1 \implies \theta = 45^\circ$. Substituting these values into the target expression gives: $1\cdot\cos(90^\circ) + 1\cdot\sin(90^\circ) = 0 + 1 = 1$, which matches the value of $a$. This technique confirms option (B) immediately!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a linear trigonometric relationship $a\sin\theta = b\cos\theta$ and need to evaluate the value of the double-angle expression $a\cos 2\theta + b\sin 2\theta$.

Step 2: Key Formula or Approach:
From the given condition, we can isolate the tangent of the angle: $$\tan\theta = \frac{b}{a}$$ We can rewrite the double-angle terms $\cos 2\theta$ and $\sin 2\theta$ using their standard tangent parametric identities: $$\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \quad \text{and} \quad \sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$$

Step 3: Detailed Explanation:
Substitute $\tan\theta = \frac{b}{a}$ into the double-angle parametric formulas: $$\cos 2\theta = \frac{1 - \left(\frac{b}{a}\right)^2}{1 + \left(\frac{b}{a}\right)^2} = \frac{\frac{a^2-b^2}{a^2}}{\frac{a^2+b^2}{a^2}} = \frac{a^2-b^2}{a^2+b^2}$$ $$\sin 2\theta = \frac{2\left(\frac{b}{a}\right)}{1 + \left(\frac{b}{a}\right)^2} = \frac{\frac{2b}{a}}{\frac{a^2+b^2}{a^2}} = \frac{2ab}{a^2+b^2}$$ Now, substitute these expressions into our target equation $a\cos 2\theta + b\sin 2\theta$: $$E = a\left(\frac{a^2-b^2}{a^2+b^2}\right) + b\left(\frac{2ab}{a^2+b^2}\right)$$ $$E = \frac{a^3 - ab^2 + 2ab^2}{a^2+b^2}$$ Combine the like terms in the numerator: $$E = \frac{a^3 + ab^2}{a^2+b^2}$$ Factor out $a$ from the numerator expression: $$E = \frac{a(a^2+b^2)}{a^2+b^2} = a$$

Step 4: Final Answer:
The expression simplifies perfectly to $a$, which matches option (B).