Question

If \( \left(\frac{\sin 3\theta}{\sin \theta}\right)^2 - \left(\frac{\cos 3\theta}{\cos \theta}\right)^2 = a \cos b\theta \), then \( a : b = \)

• A. \( 4:1 \)
• B. \( 8:1 \)
• C. \( 3:2 \)
• D. \( 2:1 \) Correct Answer

Hint:

Utilize trigonometric identities for multiple angles (e.g., \( \sin 3\theta, \cos 3\theta \)) to simplify expressions. The difference of squares factorization \( A^2 - B^2 = (A-B)(A+B) \) is often useful. Convert expressions involving \( \cos^2\theta \) to \( \cos 2\theta \) using \( \cos 2\theta = 2\cos^2\theta - 1 \).

Answer 1

The Correct Option is A

Step 1: Use the triple angle identities.
\( \sin 3\theta = 3\sin\theta - 4\sin^3\theta = \sin\theta (3-4\sin^2\theta) = \sin\theta (3-4(1-\cos^2\theta)) = \sin\theta (4\cos^2\theta - 1) \).
\( \cos 3\theta = 4\cos^3\theta - 3\cos\theta = \cos\theta (4\cos^2\theta - 3) \).
So, \( \frac{\sin 3\theta}{\sin \theta} = 4\cos^2\theta - 1 \), provided \( \sin\theta \ne 0 \).
And \( \frac{\cos 3\theta}{\cos \theta} = 4\cos^2\theta - 3 \), provided \( \cos\theta \ne 0 \).

Step 2: Substitute these into the given expression.
LHS = \( (4\cos^2\theta - 1)^2 - (4\cos^2\theta - 3)^2 \) This is of the form \( A^2 - B^2 = (A-B)(A+B) \).
Let \( A = 4\cos^2\theta - 1 \) and \( B = 4\cos^2\theta - 3 \).
\( A-B = (4\cos^2\theta - 1) - (4\cos^2\theta - 3) = -1+3 = 2 \).
\( A+B = (4\cos^2\theta - 1) + (4\cos^2\theta - 3) = 8\cos^2\theta - 4 \).
So, LHS = \( (2)(8\cos^2\theta - 4) = 16\cos^2\theta - 8 \).

Step 3: Express the result in terms of \( \cos b\theta \).
We know \( \cos 2\theta = 2\cos^2\theta - 1 \), so \( 2\cos^2\theta = \cos 2\theta + 1 \).
Then \( 8\cos^2\theta = 4(2\cos^2\theta) = 4(\cos 2\theta + 1) = 4\cos 2\theta + 4 \).
LHS = \( (16\cos^2\theta - 8) = 8(2\cos^2\theta - 1) = 8\cos 2\theta \).
The given equation is \( \text{LHS} = a \cos b\theta \).
So, \( 8\cos 2\theta = a \cos b\theta \).
By comparing, we get \( a=8 \) and \( b=2 \).

Step 4: Find the ratio \( a:b \).
\( a:b = 8:2 = 4:1 \).
This matches option (1).