Question

If sin^2 +sinx = 1
then cos^10 x+5cos^12 x+10cos^14 x+10cos^16 x+5cos^18 x+cos^20 x

Answer 1

@Compassionate @dan815 @mathslover @nincompoop @TheSmartOne @eliassaab @bibby @shrutipande9

Answer 2

@AlexandervonHumboldt2

Answer 3

\(\large \begin{align} \color{black}{\sin^2x+\sin x=1\hspace{.33em}\\~\\
\sin x=1-\sin^2x----\color{red}{(1)}\hspace{.33em}\\~\\
\cos^{10}x+5\cos^{12}x+10\cos^{14}x+10\cos^{16}x+5\cos^{18}x+\cos^{20}x\hspace{.33em}\\~\\
=\cos^{10}x(1+5\cos^{2}x+10\cos^{4}x+10\cos^{6}x+5\cos^{8}x+\cos^{10}x)\hspace{.33em}\\~\\
=\cos^{10}x(1+5\cos^{2}x+10\cos^{4}x+10\cos^{6}x+5\cos^{8}x+\cos^{10}x)\hspace{.33em}\\~\\
\normalsize \text{put} ~~\cos^{2}x=a \hspace{.33em}\\~\\
=a^{5}(1+5a+10a^{2}+10a^{3}+5a^{4}+a^{5})\hspace{.33em}\\~\\
\normalsize \text{this is the binomial expansion of } ~~(a+1)^5 \hspace{.33em}\\~\\
=a^{5}(a+1)^5 \hspace{.33em}\\~\\
=\left(\cos^{2}x(\cos^{2}x+1)\right )^5 \hspace{.33em}\\~\\
=\left(\cos^{4}x+\cos^{2}x\right )^5 \hspace{.33em}\\~\\
=\left((\cos^{2}x)^2+\cos^{2}x\right )^5 \hspace{.33em}\\~\\
=\left((1-\sin^2x)^2+\cos^{2}x\right )^5 \hspace{.33em}\\~\\
=\left(\sin x)^2+\cos^{2}x\right )^5 \hspace{.33em}\\~\\
=1^5
}\end{align}\)

Answer 4

@mathmath333 Thanks