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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find the value of the expression :$3[\sin^4(\large\frac{3\pi}{2}-\normalsize\alpha)+\sin^4(3\pi+\alpha)]-2[\sin^6(\large\frac{\pi}{2}+\normalsize \alpha)+\sin^6(5\pi-\alpha)]$

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;2&(D)\;3\end{array} $

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1 Answer

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Toolbox:
  • $\sin (\large\frac{\pi}{2}$$-x)=\cos x$
  • $\sin (\large\frac{\pi}{2}$$+x)=\cos x$
  • $\sin(\pi-x)=\sin x$
  • $\sin(\pi+x)=-\sin x$
  • $\sin (\large\frac{3\pi}{2}$$-x)=-\cos x$
  • $\sin (\large\frac{3\pi}{2}$$+x)=-\cos x$
$3[\sin^4(\large\frac{3\pi}{2}-\normalsize\alpha)+\sin^4(3\pi+\alpha)]-2[\sin^6(\large\frac{\pi}{2}+\normalsize \alpha)+\sin^6(5\pi-\alpha)]$
$3[\cos^4\alpha+\sin^4\alpha]-2[\cos^6\alpha+\sin^6\alpha]$
$3[(\sin^2\alpha+\cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha]-2[(\cos^2\alpha+\sin^2\alpha)^3-3\sin^2\alpha\cos^2\alpha(\sin^2\alpha+\cos^2\alpha)]$
$a^2+b^2=(a+b)^2-2ab$
$a^3+b^3=(a+b)^3-3ab(a^2+b^2)$
$3[1-2\sin^2\alpha\cos^2\alpha]-2[1-3\sin^2\alpha\cos^2\alpha]$
$\sin^2\alpha+\cos^2\alpha=1$
$3-6\sin^2\alpha\cos^2\alpha-2+6\sin^2\alpha\cos^2\alpha$
$(\sin^2\alpha+\cos^2\alpha=1)$
$3-2=1$
Hence (B) is the correct answer.
answered Apr 20, 2014 by sreemathi.v
 

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