If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1

Solution:

Given: $\cos \theta+\cos ^{2} \theta=1$

We have to prove $\sin ^{12} \theta+3 \sin ^{10} \theta+3 \sin ^{8} \theta+\sin ^{6} \theta+2 \sin ^{4} \theta+2 \sin ^{2} \theta-2=1$

From the given equation, we have

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

$\Rightarrow \quad \cos \theta=1-\cos ^{2} \theta$

$\Rightarrow \quad \cos \theta=\sin ^{2} \theta$

 

$\Rightarrow \quad \sin ^{2} \theta=\cos \theta$

Therefore, we have

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

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

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

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

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

$=(1)^{3}+2(1)-2$

 

$=1$

Hence proved.