Question

If $\cos \theta +{\cos }^2\theta =1$, prove that
${\sin }^{12}\theta +3{\sin }^{10}\theta +3{\sin }^8\theta +{\sin }^6\theta +2{\sin }^4\theta +2{\sin }^2\theta -2=1$

Answer 1

Given $\cos {\sin }^{10}\theta +{\cos }^2{\sin }^{10}\theta =1---\left(1\right)$

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

$\Rightarrow \cos \theta ={\sin }^2\theta ---\left(2\right)$

Now, to prove

${\sin }^{12}\theta +3.{\sin }^{10}\theta +3.{\sin }^8\theta +2.{\sin }^4\theta +2.{\sin }^2\theta -2=1$

$\Rightarrow L.H.S={\sin }^{12}\theta +3.{\sin }^{10}\theta +3.{\sin }^8\theta +1{\sin }^4\theta +2{\sin }^2\theta -2$

$=\left[\right({\sin }^4\theta {)}^2+3.{\sin }^6\theta -({\sin }^4\theta +{\sin }^2\theta )+\left({\sin }^2\theta {)}^3\right]+2({\sin }^4\theta +{\sin }^2\theta -1)$

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

(Using $(a+b{)}^3=a^3+b^3+3ab(a+b)$ using $\left(2\right)$ from above)

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

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

(using $\left(2\right)$ from above)

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

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

(using ${\cos }^2\theta +{\sin }^2\theta =1$ from equation $\left(a\right)$)

$=1+2\left(0\right)$

$=1+0$

$=1$

$=R.H.S$