Question

If sin x + sin² x = 1, then write the value of cos⁸ x + 2 cos⁶ x + cos⁴ x.

 

Answer 1

We have: 
\[\sin x + \sin^2 x = 1 \left( 1 \right)\]
\[ \Rightarrow \sin x = 1 - \sin^2 x\]
\[ \Rightarrow \sin x = co s^2 x \left( 2 \right)\]
Now, taking square of ( 1 ) : 
\[ \Rightarrow \left( \sin x + \sin^2 x \right)^2 = \left( 1 \right)^2 \]
\[ \Rightarrow \left( \sin x \right)^2 + \left( \sin^2 x \right)^2 + 2\left( \sin x \right) \left( \sin^2 x \right) = 1\]
\[ \Rightarrow \left( \sin x \right)^2 + \left( \sin x \right)^4 + 2 \left( \sin x \right)^3 = 1\]
\[ \Rightarrow \left( \sin x \right)^2 + 2 \left( \sin x \right)^3 + \left( \sin x \right)^4 = 1\]
\[ \Rightarrow \left( \cos^2 x \right)^2 + 2 \left( \cos^2 x \right)^3 + \left( \cos^2 x \right)^4 = 1\]
\[ \Rightarrow \cos^4 x + 2 \cos^6 x + \cos^8 x = 1\]
\[ \therefore \cos^8 x + 2 \cos^6 x + \cos^4 x = 1\]