Question

Sinx+sin2 x=1, then find the value of cos12+3cos10x+3cos8 x+cos6 x+2cos4 x+cos2 x-2

Answer 1

$\textbf{Given:}$

$sinx+sin^2x=1$

$\textbf{To find:}$

$\text{The value of cos^{12}x+3\,cos^{10}x+3\,cos^{8}x+cos^{6}x+2\,cos^{4}x+cos^{2}x-2 }$

$\textbf{Solution:}$

$\text{Consider,}$

$sinx+sin^2x=1$

$\implies\,sinx=1-sin^2x$

$\implies\,sinx=cos^2x$

$\text{Squaring on bothsides, we get}$

$sin^2x=cos^4x$

$1-cos^2x=cos^4x$

$1=cos^4x+cos^2x$........(1)

$\text{Raising bothsides to the power 3, we get}$

$1^3=(cos^4x+cos^2x)^3$

$\text{Using the identity,}$

$\boxed{(a+b)^3=a^3+b^3+3a^2b+3ab^2}$

$1=cos^{12}x+cos^{6}x+3\,cos^{10x}+3\,cos^{8}x$.......(2)

$\text{Now,}$

$cos^{12}x+3\,cos^{10}x+3\,cos^{8}x+cos^{6}x+2\,cos^{4}x+cos^{2}x-2$

$=(cos^{12}x+3\,cos^{10}x+3\,cos^{8}x+cos^{6}x)+2\,cos^{4}x+cos^{2}x-2$

$\text{Using (2), we get}$

$=(1)+2\,cos^{4}x+cos^{2}x-2$

$=1+2(cos^{4}x-1)+cos^{2}x$

$\text{Using (1), we get}$

$=1+2(-cos^{2}x)+cos^{2}x$

$=1-2cos^{2}x+cos^{2}x$

$=1-cos^{2}x$

$=sin^2x$

$\textbf{Answer:}$

$\boxed{\bf\,cos^{12}x+3\,cos^{10}x+3\,cos^{8}x+cos^{6}x+2\,cos^{4}x+cos^{2}x-2=sin^2x}$