MCQ
$\frac{\sin5\text{x}}{\sin\text{x}}$ is equal to:
- A$16\cos^4-12\cos^2\text{x}+1$
- B$16\cos^4\text{x}+12\cos^2\text{x}+1$
- C$16\cos^4\text{x}-12\cos^2\text{x}-1$
- D$16\cos^4\text{x}+12\cos^2\text{x}-1$
Solution:
To find: $\frac{\sin5\text{x}}{\sin\text{x}}$
Now,
$\sin5\text{x}=\sin(3\text{x}+2\text{x})$
$=(3\sin\text{x}-4\sin^3\text{x})(1-2\sin^2\text{x})+(4\cos^3\text{x}-3\cos\text{x})(2\sin\text{x}\cos\text{x})$
$=(3\sin\text{x}-4\sin^3\text{x}-4\sin^3\text{x}+8\sin^5\text{x})+2\sin\text{x}\cos^2\text{x}(4\cos^2\text{x}-3)$
$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+2\sin\text{x}(1-\sin^2\text{x})[2(1-\sin^2\text{x})-3]$
$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-2\sin^3\text{x})(4-4\sin^2\text{x}-3)]$
$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-8\sin^3\text{x}2\sin^3\text{x}+8\sin^5\text{x})]$
$=5\sin\text{x}-20\sin^3+16\sin^5\text{x}$
$\therefore\frac{\sin5\text{x}}{\sin\text{x}}=\frac{5\sin\text{x}-20\sin^3\text{x}+16^5\text{x}}{\sin\text{x}}$
$=5-20\sin^2\text{x}+16\sin^4\text{x}$
$=5-20(1-\cos^2\text{x})+16(1-\cos^2\text{x})^2$
$=5-20+20\cos^2\text{x}+16(1+\cos^4\text{x}-2\cos^4\text{x})$
$=5-20+20\cos^2\text{x}+16+16\cos^4\text{x}-32\cos^2\text{x}$
$=16\cos^4-12\cos^2\text{x}+1$
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