Question

If $\sin \theta +{\sin }^2\theta =1,$ find the value of :-
${\cos }^{12}\theta +3{\cos }^{10}\theta +3{\cos }^8\theta +{\cos }^6\theta +2{\cos }^4\theta +2{\cos }^2\theta -2$.

Answer 1

Consider the given data.

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

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

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

We have,

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

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

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

Since, $\sin \theta ={\cos }^2\theta$

Therefore,

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

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

$=1+0$

$=1$

Hence, the required value is $1$.