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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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$3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin ^6x+\cos ^6x)$=_______

$\begin{array}{1 1}(A)\;11&(B)\;13\\(C)\;12&(D)\;14\end{array} $

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1 Answer

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  • $\sin 2x=2\sin x\cos x$
  • $(a\pm b)^2=a^2+b^2\pm 2ab$
  • $(a+ b)^3=a^3+b^3+ 3ab(a+b)$
$3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4(\sin ^6x+\cos ^6x)$-----(i)
$(\sin x-\cos x)^4=(\sin x-\cos x)^2(\sin x-\cos x)^2$
$\Rightarrow (\sin^2x+\cos ^2x-2\sin x\cos x)(\sin^2x+\cos ^2x-2\sin x\cos x)$
$\Rightarrow (1-2\sin x\cos x)(1-2\sin x\cos x)$
$\Rightarrow (1-\sin 2x)^2$
$\Rightarrow (1+\sin ^22x-2\sin 2x)$-----(1)
$(\sin x+\cos x)^2=\sin^2x+\cos ^2x+2\sin x\cos x$
$\Rightarrow 1+2\sin x\cos x$
$\Rightarrow 1+\sin 2x$------(2)
$\Rightarrow (\sin^2x+\cos^2x)^3-3\sin ^2x\cos ^2x(\sin^2x+\cos ^2x)$
$\Rightarrow (1-\large\frac{3\sin^22x}{4})$------(3)
Put (1),(2) & (3) in (i)
$3(1+\sin^22x-2\sin 2x)+6(1+\sin 2x)+4(1-\large\frac{3\sin^22x}{4})$
$3+3\sin^22x-6\sin 2x+6+6\sin 2x+4-3\sin ^22x$
$\Rightarrow 13$
Hence (B) is the correct answer.
answered May 20, 2014 by sreemathi.v

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