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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions

If $\sin x+\cos x=a$ then $\sin ^6x+\cos ^6x$=_______

$\begin{array}{1 1}(A)\;\large\frac{1}{4}\normalsize [4-3(a^2-1)^2]&(B)\;\large\frac{1}{2}\normalsize [2-3(a^2-1)^2]\\(C)\;a^2&(D)\;2a^6\end{array} $

1 Answer

Toolbox:
  • $a^3+b^3=(a+b)^3-3ab(a+b)$
  • $ab=\large\frac{(a+b)^2-(a^2+b^2)}{2}$
$\sin ^6x+\cos ^6x$
$(\sin ^2x)^3+(\cos ^2x)^3$
$\Rightarrow (\sin^2x+\cos ^2x)^3-3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)$
$\Rightarrow 1-3\sin ^2x\cos ^2x$
$\Rightarrow 1-3\big[\large\frac{(\sin x+\cos x)^2-(\sin ^2x+\cos ^2x)}{2}\big]^2$
$\Rightarrow 1-\large\frac{3(a^2-1)^2}{4}$
$\Rightarrow \large\frac{1}{4}$$[4-3(a^2-1)^2]$
Hence (A) is the correct answer.
answered Apr 22, 2014 by sreemathi.v
 
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