Question

If sin x + sin² x = 1, then write the value of cos¹² x + 3 cos¹⁰ x + 3 cos⁸ x + cos⁶ 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 cube of  (1) :
\[\sin x + \sin^2 x = 1\]
\[ \Rightarrow \left( \sin x + \sin^2 x \right)^3 = \left( 1 \right)^3 \]
\[ \Rightarrow \left( \sin x \right)^3 + \left( \sin^2 x \right)^3 + 3 \left( \sin x \right)^2 \left( \sin^2 x \right) + 3\left( \sin x \right) \left( \sin^2 x \right)^2 = 1\]
\[ \Rightarrow \left( \sin x \right)^3 + \left( \sin x \right)^6 + 3 \left( \sin x \right)^4 + 3 \left( \sin x \right)^5 = 1\]
\[ \Rightarrow \left( \sin x \right)^6 + 3 \left( \sin x \right)^5 + 3 \left( \sin x \right)^4 + \left( \sin x \right)^3 = 1\]
\[ \Rightarrow \left( \cos^2 x \right)^6 + 3 \left( \cos^2 x \right)^5 + 3 \left( \cos^2 x \right)^4 + \left( \cos^2 x \right)^3 = 1\]
\[ \Rightarrow \cos^{12} x + 3 \cos^{10} x + 3 \cos^8 x + \cos^6 x = 1\]