Question

If \(x = 2 \sin 60^\circ \cos 60^\circ\) and \(y = \sin 230^\circ - \cos 230^\circ\), and \(x^2 = ky^2\), the value of \(k\) is

• A. \(\sqrt{3}\)
• B. \(-\sqrt{3}\)
• C. \(-3\)
• D. 3

Hint:

Use trigonometric identities carefully and rationalize when necessary.

Answer 1

The Correct Option is D

We are given that \( x = 2\sin 60^\circ \cos 60^\circ \) and \( y = \sin^2 30^\circ - \cos^2 30^\circ \). We also have \( x^2 = ky^2 \).

First, let's find the value of \( x \).
\( x = 2\sin 60^\circ \cos 60^\circ = 2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) = 2\left(\frac{\sqrt{3}}{4}\right) = \frac{\sqrt{3}}{2} \).

Next, let's find the value of \( y \).
\( y = \sin^2 30^\circ - \cos^2 30^\circ = \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} - \frac{3}{4} = \frac{-2}{4} = -\frac{1}{2} \).

Now, we have \( x^2 = ky^2 \). Substituting the values of \( x \) and \( y \), we get:
\( \left(\frac{\sqrt{3}}{2}\right)^2 = k\left(-\frac{1}{2}\right)^2 \)
\( \frac{3}{4} = k\left(\frac{1}{4}\right) \)
\( 3 = k \)

Therefore, the value of \( k \) is 3.

Final Answer: \( \boxed{3} \)