Question

If \( \cos\theta - \sin\theta = \sqrt{5} \sin\theta \), then the value of \( \cos\theta + 4\sin\theta \) is:

• A. \( 5\cos\theta \)
• B. \( \sqrt{5}\sin\theta \)
• C. \( 5\sin\theta \)
• D. \( \sqrt{5}\cos\theta \)

Hint:

Try expressing one trigonometric function in terms of another, then use Pythagorean identities to simplify.

Answer 1

The Correct Option is D

We are given: \[ \cos\theta - \sin\theta = \sqrt{5} \sin\theta \Rightarrow \cos\theta = (1 + \sqrt{5}) \sin\theta \] Now compute: \[ \cos\theta + 4\sin\theta = (1 + \sqrt{5}) \sin\theta + 4\sin\theta = (5 + \sqrt{5})\sin\theta \] However, this contradicts the expected answer. Let us square both sides: \[ \cos\theta - \sin\theta = \sqrt{5}\sin\theta \Rightarrow \cos\theta = (1 + \sqrt{5})\sin\theta \] Divide both sides: \[ \begin{align} \tan\theta = \frac{1}{1 + \sqrt{5}} \Rightarrow \cos\theta = \frac{1}{\sqrt{1 + \tan^2\theta}} = \sqrt{\frac{1}{1 + \left( \frac{1}{(1 + \sqrt{5})^2} \right)}} \] Alternatively, use the identity: Let \( \cos\theta = a,\ \sin\theta = b \) Given: \( a - b = \sqrt{5}b \Rightarrow a = b(\sqrt{5} + 1) \) Also, \( a^2 + b^2 = 1 \Rightarrow b^2(\sqrt{5} + 1)^2 + b^2 = 1 \Rightarrow b^2[(\sqrt{5} + 1)^2 + 1] = 1 \) \[ (\sqrt{5} + 1)^2 = 6 + 2\sqrt{5}, \quad \text{So: } b^2(7 + 2\sqrt{5}) = 1 \Rightarrow b = \frac{1}{\sqrt{7 + 2\sqrt{5}}} \] Then, \( \cos\theta + 4\sin\theta = a + 4b = b(\sqrt{5} + 1) + 4b = b(\sqrt{5} + 5) \) Now check the expression: \[ \cos\theta + 4\sin\theta = b(\sqrt{5} + 5) = \cos\theta \cdot \frac{\sqrt{5} + 5}{\sqrt{5} + 1} = \sqrt{5} \cos\theta \]