Question

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

Answer 1

Given :

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

$\Rightarrow \cos \theta ={\sin }^2\theta$ ..... $[sin²x+cos²x=1]$

Square on both sides ;

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

$1-{\sin }^2\theta ={\sin }^4\theta$ ...... $[sin²x+cos²x=1]$

${\sin }^4\theta +{\sin }^2\theta =1$ $\to$equation (1)

Now cube on both sides ;

$\Rightarrow {\sin }^{12}\theta +{\sin }^6\theta +3{\sin }^4\theta {\sin }^2\theta ({\sin }^4\theta +{\sin }^2\theta )=1$

$\Rightarrow {\sin }^{12}\theta +{\sin }^6\theta +3{\sin }^{10}\theta +3{\sin }^8\theta =1$

To obtain above result we add and subtract 2 on LHS side ;

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

From equation (1), $1={\sin }^4\theta +{\sin }^2\theta$

$\Rightarrow {\sin }^{12}\theta +{\sin }^6\theta +3{\sin }^{10}\theta +3{\sin }^8\theta +2({\sin }^4\theta +{\sin }^2\theta )-2=1$

$\Rightarrow {\sin }^{12}\theta +3{\sin }^{10}\theta +3{\sin }^8\theta +{\sin }^6\theta +2{\sin }^4\theta +2{\sin }^2\theta -2=1$

Hence proved