1. If $$cosx+cos^{2}x=1$$, the numerical value of $$(sin^{12}x+3sin^{10}x+3sin^{5}x+sin^{6}x-1)$$ is :
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By: anil on 05 May 2019 01.57 am
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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$$sin^2x = cos x $$ .........(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