Question

If sin x + sin² x = 1, then write the value of cos¹² x + 3 cos¹⁰ x + 3 cos⁸ x + cos⁶ x.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \sin x+{\sin }^{2}x=1\left(1\right) \\ \Rightarrow \sin x=1-{\sin }^{2}x \\ \Rightarrow \sin x=co{s}^{2}x\left(2\right) \\ \mathrm{Now},\mathrm{taking}\mathrm{cube}\mathrm{of}\left(1\right): \\ \sin x+{\sin }^{2}x=1 \\ \Rightarrow {\left(\sin x+{\sin }^{2}x\right)}^3={\left(1\right)}^3 \\ \Rightarrow {\left(\sin x\right)}^3+{\left({\sin }^{2}x\right)}^3+3{\left(\sin x\right)}^2\left({\sin }^{2}x\right)+3\left(\sin x\right){\left({\sin }^{2}x\right)}^2=1 \\ \Rightarrow {\left(\sin x\right)}^3+{\left(\sin x\right)}^6+3{\left(\sin x\right)}^4+3{\left(\sin x\right)}^5=1 \\ \Rightarrow {\left(\sin x\right)}^6+3{\left(\sin x\right)}^5+3{\left(\sin x\right)}^4+{\left(\sin x\right)}^3=1 \\ \Rightarrow {\left({\cos }^2\mathrm{x}\right)}^6+3{\left({\cos }^2\mathrm{x}\right)}^5+3{\left({\cos }^2\mathrm{x}\right)}^4+{\left({\cos }^2\mathrm{x}\right)}^3=1 \\ \Rightarrow {\cos }^{12}\mathrm{x}+3{\cos }^{10}\mathrm{x}+3{\cos }^8\mathrm{x}+{\cos }^6\mathrm{x}=1 \end{gathered}$$