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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\sin x+\sin^2 x=1$ then the value of $\cos^2x+3\cos^{10}x+3\cos^8 x+\cos^6x$ is equal to

$(a)\;1\qquad(b)\;-1\qquad(c)\;2\qquad(d)\;-2$

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1 Answer

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$I=\cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x$
$\;\;=\cos^{12}x+3.\cos^8x(\cos^2x+1)+\cos^6x$
$\;\;=(\cos^4x)^3+3\cos^6x(\cos^4x+\cos^2x)+(\cos^2x)^3$
$\;\;=(\cos^4x+\cos^2x)^3$
$\;\;=(\cos^2x(\cos^2x+1))^3$
We have $\sin x=1-\sin^2x$
$\cos^2x+\sin^2x=1$
$\therefore \sin x=\cos^2x$
$\qquad\;\;\;=\large\frac{-1\pm \sqrt 5}{2}$
$I=\bigg(\large\frac{\sqrt 5-1}{2}\big(\large\frac{\sqrt 5+1}{2}\big)\bigg)^3$
$\;\;=1$
Hence (a) is the correct option.
answered Oct 7, 2013 by sreemathi.v
 

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