Question

Express cos6x in terms of cosx.

Answer 1

do you know complex numbers?

Answer 2

we can use the formula of \(De\;Moivre\), so we can write this:
\[\begin{gathered}
{\left( {\cos x + i\sin x} \right)^6} = {\left( {\cos x} \right)^6} + 6i{\left( {\cos x} \right)^5}\left( {\sin x} \right) - 15{\left( {\cos x} \right)^4}{\left( {\sin x} \right)^2} + \hfill \\
\hfill \\
- 20i{\left( {\cos x} \right)^3}{\left( {\sin x} \right)^3} + 15{\left( {\cos x} \right)^2}{\left( {\sin x} \right)^4} + 6i\left( {\cos x} \right){\left( {\sin x} \right)^5} + \hfill \\
\hfill \\
- {\left( {\sin x} \right)^6} \hfill \\
\end{gathered} \]

Answer 3

cos 6x= cos (2.3x)=2 (cos 3x )^2-1 =2[4cos^3x - 3cosx ]^2-1
=2[16 cos^6x+9cos^2x-24cos^4x]-1=32cos^6x+18cos^2x-48cos^4x-1

Answer 4

next, we have to separate the imaginary part from the real part at the right side

Answer 5

I have used this identity:
\[{\left( {\cos x + i\sin x} \right)^6} = \cos \left( {6x} \right) + i\sin \left( {6x} \right)\]

Answer 6

@vandanaS if you know about complex numbers, you must use @Michele_Laino 's method. Otherwise use mine

Answer 7

thanks @jango_IN_DTOWN and @Michele_Laino