Question 142

If $$cosx+cos^{2}x=1$$, the numerical value of $$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$ is :

Solution

given that : $$cosx+cos^{2}x=1$$

$$cosx+cos^{2}x= sin^2 x +cos^2 x$$

$$sin^2x = cos x $$ .........(1)

Now we need to find value of :

$$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$

Usng equation 1 , $$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$ = $$(cos^{6}x+3cos^{5}x+3sin^{5}x+sin^{6}x-1)$$

= 1 + 3 -1

= 3

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