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: Interpret the given relation geometrically.
From $a\sin\theta=b\cos\theta$ we get $\tan\theta=\dfrac{b}{a}$. Picture a right triangle with opposite side $b$ and adjacent side $a$, so the hypotenuse is $\sqrt{a^2+b^2}$.
Step 2: Read off sine and cosine.
$\sin\theta=\dfrac{b}{\sqrt{a^2+b^2}}$ and $\cos\theta=\dfrac{a}{\sqrt{a^2+b^2}}$.
Step 3: Use double-angle definitions directly.
$\cos2\theta=\cos^2\theta-\sin^2\theta=\dfrac{a^2-b^2}{a^2+b^2}$ and $\sin2\theta=2\sin\theta\cos\theta=\dfrac{2ab}{a^2+b^2}$.
Step 4: Form the required expression.
\[ a\cos2\theta+b\sin2\theta=\frac{a(a^2-b^2)+b(2ab)}{a^2+b^2}. \]
Step 5: Simplify the numerator.
$a^3-ab^2+2ab^2=a^3+ab^2=a(a^2+b^2)$.
Step 6: Cancel and conclude.
\[ \frac{a(a^2+b^2)}{a^2+b^2}=a. \] So the expression equals $a$. \[ \boxed{a\cos2\theta+b\sin2\theta=a} \]