Question

If sin theta + sin2 theta = 1, find the value of cos12 theta+ 3cos10 theta + 3cos8 theta + cos6 theta + 2cos4 theta + 2cos2 theta - 2

Answer 1

Solution:

We have ,

$sin\theta+sin^{2}\theta=1$

$\implies sin\theta = 1-sin^{2}\theta$

$\implies sin\theta = cos^{2}\theta$

Now ,

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

=$\left(cos^{12}\theta+3cos^{10}\theta+3cos^{8}\\\theta+cos^{6}\theta\right)+2(cos^{4}\theta+cos^{2}\theta-1)$

= $[(cos^{4}\theta)^3+3\times(cos^{4}\theta)^{2}\times(cos^{2}\theta)+\\ 3\times(cos^{4}\theta)\times(cos^{2}\theta)^{2}+(cos^{2}\theta)^{3}]\\+2(cos^{4}\theta+cos^{2}\theta-1)$

=$\left(cos^{4}\theta+cos^{2}\theta\right)^{3}\\+2(cos^{4}\theta+cos^{2}\theta-1)$

_____________________

Here , we are using the following:

i)$cos^{2}\theta = sin\theta$

ii) $cos^{4}\theta = sin^{2}\theta$

_____________________

= $\left(sin^{2}\theta+cos^{2}\theta\right)^{3}\\+2(sin^{2}\theta+cos^{2}\theta-1)$

= $1+2(1-1)$

=$1$

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Answer 2

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