Question

If $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1,$ then
${\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =.......$

Answer 1

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

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

Squaring both sides,

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

$\Rightarrow {\sin }^2\theta +{\sin }^6\theta +2{\sin }^4\theta =1+{\sin }^2\theta -2{\sin }^2\theta$ ....... $(a+b{)}^2=a^2+2ab+b^2$

$\Rightarrow (1-{\cos }^2\theta )+{(1-{\cos }^2\theta )}^3+2{(1-{\cos }^2\theta )}^2=1+{(1-{\cos }^2\theta )}^2-2(1-{\cos }^2\theta )$ ..... $[sin²\theta +cos²\theta =1]$

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

............ $(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$ and $(a-b{)}^2=a^2-2ab+b^2$

$\Rightarrow 1-{\cos }^2\theta +1-{\cos }^6\theta -3{\cos }^2\theta +3{\cos }^4\theta +2+2{\cos }^4\theta -4{\cos }^2\theta =1+1+{\cos }^4\theta -2{\cos }^2\theta -2+2{\cos }^2\theta$

$\Rightarrow -8{\cos }^2\theta -{\cos }^6\theta +5{\cos }^4\theta +4={\cos }^4\theta$

$\Rightarrow -{\cos }^6\theta +4{\cos }^4\theta -8{\cos }^2\theta +4=0$

$\Rightarrow -{\cos }^6\theta +4{\cos }^4\theta -8{\cos }^2\theta =-4$

$\Rightarrow {\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$

$\therefore {\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$