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Question
Options
- \[16 \cos^4 x - 12 \cos^2 x + 1\]
- \[16 \cos^4 x + 12 \cos^2 x + 1\]
- \[16 \cos^4 x - 12 \cos^2 x - 1\]
- \[16 \cos^4 x + 12 \cos^2 x - 1\]
Solution
\[\text{ To find } : \frac{\sin 5x}{\text{ sin } x}\]
\[\text{ Now} , \]
\[\sin5x = \sin\left( 3x + 2x \right)\]
\[ = \sin3x\cos2x + \cos3x\sin2x\]
\[ = \left( 3\text{ sin } x - 4 \sin^3 x \right)\left( 1 - 2 \sin^2 x \right) + \left( 4 \cos^3 x - 3\text{ cos } x \right)\left( 2\text{ sin } x\text{ cos } x \right)\]
\[ = \left( 3\sin x - 6 \sin^3 x - 4 \sin^3 x + 8 \sin^5 x \right) + 2\text{ sin } x \cos^2 x\left( 4 \cos^2 x - 3 \right)\]
\[ = \left( 3\sin x - 10 \sin^3 x + 8 \sin^5 x \right) + 2\text{ sin } x\left( 1 - \sin^2 x \right)\left[ 4\left( 1 - \sin^2 x \right) - 3 \right]\]
\[ = \left( 3\sin x - 10 \sin^3 x + 8 \sin^5 x \right) + \left( 2\text{ sin } x - 2 \sin^3 x \right)\left( 4 - 4 \sin^2 x - 3 \right)\]
\[ = \left( 3\sin x - 10 \sin^3 x + 8 \sin^5 x \right) + \left( 2\text{ sin } x - 8 \sin^3 x - 2 \sin^3 x + 8 \sin^5 x \right)\]
\[ = 5\text{ sin } x - 20 \sin^3 x + 16 \sin^5 x\]
\[ \therefore \frac{\sin 5x}{\text{ sin } x} = \frac{5\text{ sin } x - 20 \sin^3 x + 16 \sin^5 x}{\text{ sin } x} \]
\[ = 5 - 20 \sin^2 x + 16 \sin^4 x \]
\[ = 5 - 20\left( 1 - \cos^2 x \right) + 16 \left( 1 - \cos^2 x \right)^2 \]
\[ = 5 - 20 + 20 \cos^2 x + 16\left( 1 + \cos^4 x - 2 \cos^2 x \right)\]
\[ = 5 - 20 + 20 \cos^2 x + 16 + 16 \cos^4 x - 32 \cos^2 x\]
\[ = 16 \cos^4 x - 12 \cos^2 x + 1\]
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