Question

If \(3\cos\theta+4\sin\theta=5\), then \(4\sin\theta-3\cos\theta=\)

• A. 5
• B. 4
• C. 2
• D. 0

Hint:

Remember: \[ (a\cos\theta+b\sin\theta)^2+ (b\cos\theta-a\sin\theta)^2 =(a^2+b^2). \] This identity is very useful in trigonometric simplification.

Answer 1

The Correct Option is D

Concept: Use the identity \[ \boxed{\sin^2\theta+\cos^2\theta=1.} \] Also, \[ (3\cos\theta+4\sin\theta)^2+(4\sin\theta-3\cos\theta)^2 =(3^2+4^2)(\sin^2\theta+\cos^2\theta). \]

Step 1: Square the given expression.
Given, \[ 3\cos\theta+4\sin\theta=5. \] Therefore, \[ (3\cos\theta+4\sin\theta)^2=25. \]

Step 2: Apply the identity.
Using \[ (3\cos\theta+4\sin\theta)^2+(4\sin\theta-3\cos\theta)^2 =25, \] we get \[ 25+(4\sin\theta-3\cos\theta)^2=25. \] Hence, \[ (4\sin\theta-3\cos\theta)^2=0. \] Therefore, \[ 4\sin\theta-3\cos\theta=0. \]

Step 3: Final conclusion.
Thus, \[ \boxed{4\sin\theta-3\cos\theta=0.} \]