Question

if sintheta + sin²theta=1
Prove that cos¹²theta+3cos^10theta +3cos^8theta+cos^6theta-1=0


Pls ans step by step asap
It's urgent
I will mark u as brainliest
Don't spam please ​

Answer 1

Answer:

Solution:

We have ,

sin\theta+sin^{2}\theta=1sinθ+sin

2

θ=1

\implies sin\theta = 1-sin^{2}\theta⟹sinθ=1−sin

2

θ

\implies sin\theta = cos^{2}\theta⟹sinθ=cos

2

θ

Now ,

\begin{gathered}cos^{12}\theta+3cos^{10}\theta+3cos^{8}\theta\\+cos^{6}\theta+2cos^{4}\theta+2cos^{2}\theta-2\end{gathered}

cos

12

θ+3cos

10

θ+3cos

8

θ

+cos

6

θ+2cos

4

θ+2cos

2

θ−2

=\begin{gathered}(cos^{12}\theta+3cos^{10}\theta+3cos^{8}\\\theta+cos^{6}\theta)+2(cos^{4}\theta+cos^{2}\theta-1)\end{gathered}

(cos

12

θ+3cos

10

θ+3cos

8

θ+cos

6

θ)+2(cos

4

θ+cos

2

θ−1)

= \begin{gathered}[(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)\end{gathered}

[(cos

4

θ)

3

+3×(cos

4

θ)

2

×(cos

2

θ)+

3×(cos

4

θ)×(cos

2

θ)

2

+(cos

2

θ)

3

]

+2(cos

4

θ+cos

2

θ−1)

=\begin{gathered}(cos^{4}\theta+cos^{2}\theta)^{3}\\+2(cos^{4}\theta+cos^{2}\theta-1)\end{gathered}

(cos

4

θ+cos

2

θ)

3

+2(cos

4

θ+cos

2

θ−1)

_____________________

Here , we are using the following:

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

2

θ=sinθ

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

4

θ=sin

2

θ

_____________________

= \begin{gathered}(sin^{2}\theta+cos^{2}\theta)^{3}\\+2(sin^{2}\theta+cos^{2}\theta-1)\end{gathered}

(sin

2

θ+cos

2

θ)

3

+2(sin

2

θ+cos

2

θ−1)

= 1+2(1-1)1+2(1−1)

=11