The Value of \( \frac{\sin 5x}{\sin x} \)
Question
Find the value of
\[ \frac{\sin 5x}{\sin x} \]
(a) \(16\cos^4x-12\cos^2x+1\)
(b) \(16\cos^4x+12\cos^2x+1\)
(c) \(16\cos^4x-12\cos^2x-1\)
(d) \(16\cos^4x+12\cos^2x-1\)
Solution
Use the identity
\[ \sin 5x = 16\sin^5x – 20\sin^3x + 5\sin x \]
Dividing by \(\sin x\),
\[ \frac{\sin 5x}{\sin x} = 16\sin^4x – 20\sin^2x + 5 \]
Using
\[ \sin^2x=1-\cos^2x \]
\[ = 16(1-\cos^2x)^2 – 20(1-\cos^2x) + 5 \]
\[ = 16(1-2\cos^2x+\cos^4x) -20+20\cos^2x+5 \]
\[ = 16-32\cos^2x+16\cos^4x -20+20\cos^2x+5 \]
\[ = 16\cos^4x -12\cos^2x +1 \]
Final Answer
\[ \boxed{16\cos^4x-12\cos^2x+1} \]
Hence, the correct option is (a) \(16\cos^4x-12\cos^2x+1\).