Question

$\sf\: Sin\theta + Sin^2 \theta = 1$

Then;

$\sf\: Cos^4 \theta + Cos^8 \theta + 2Cos^6 \theta$​

Answer 1

$\bf\huge\textbf {\underline {\underline {Solution :-}}} \\ \\ \\ \bf\huge {\implies { Sin\theta + Sin^2 \theta = 1}} \\ \\ \\ \bf\huge {\implies {Sin\theta = 1 - Sin^2 \theta}} \\ \\ \\ \bf\huge {\implies {Sin\theta = Cos^2 \theta }} \\ \\ \\ \bf\huge {\implies{Cos^4 \theta + Cos^8 \theta + 2Cos^6 \theta}} \\ \\ \\ \bf\huge {\implies { (Cos^2 \theta)^2 + (Cos^4 \theta)^2 + 2 Cos^2 \theta × Cos^4 \theta}} \\ \\ \\ \bf\huge {\implies{(Cos^2 \theta + Cos^4 \theta)^2}} \\ \\ \\ \bf\huge {\implies { (Cos^2 \theta + Sin^2 \theta )^2 = 1}} \\ \\ \\ \bf\huge\textbf {\underline {\underline {Identity \: Used :-}}} \\ \\ \\ \bf\huge{\implies { Sin^2 \theta + Cos^2 \theta = 1 }}$

Answer 2

Answer :-

→ 1 .

Step-by-step explanation :-

We have ,

$\because \sf\: \sin\theta + \sin^2 \theta = 1 . \\ \\ \sf \implies \sin \theta = 1 - { \sin}^{2} \theta. \\ \\ \large \sf \therefore \sin \theta = { \cos}^{2} \theta.$

Now,

To find :-

$\because\sf \cos^4 \theta + \cos^8 \theta + 2 \cos^6 \theta. \\ \\ = \sf {( { \cos}^{2} \theta) }^{2} + {( { \cos}^{4} \theta)}^{2} + 2 \times { \cos}^{2} \theta \times { \cos}^{4} \theta. \\ \\ \sf = {( { \cos}^{2} \theta + { \cos}^{4} \theta) }^{2} . \: \: \: \: \: \{ {a}^{2} + {b}^{2} - 2ab = {(a - b)}^{2} . \} \\ \\ = {( { \cos}^{2} \theta + { \cos}^{2} \theta \times { \cos}^{2} \theta ) }^{2}. \\ \\ = {( { \cos}^{2} \theta + \sin\theta \times \sin \theta) }^{2}. \: \: \: \: \: \: \: \{ { \cos}^{2} \theta = \sin \theta \} \\ \\ \sf = {( { \cos}^{2} \theta + { \sin}^{2} \theta) }^{2}. \\ \\ \sf = {(1)}^{2} . \\ \\ \huge \pink{ \boxed{ \boxed{ \it = 1.}}}$

Hence, it is solved.