Question

Show that:
\[\cos^2 (nx) = \frac{1 + \cos (2nx)}2\]

Answer 1

Does 1+cos(2x) = 2cos^2(x) not work here?

Answer 2

I would start with \(\cos^2nx+\sin^2nx=1\)

Answer 3

\[LHS=1-\sin^2nx\]

Answer 4

@Astrophysics
how can i show:
\[1+\cos(2x) = 2\cos^2(x)\]

Answer 5

@UnkleRhaukus use the identity for \(\cos(A+B)\)

Answer 6

\[1+\cos(2x) = \mathcal{R}(1+e^{i(2x)}) = \mathcal{R}(1+e^{ix}e^{ix}) \\~\\=
\mathcal{R}(1+(\cos x+i\sin x)(\cos x+i\sin x))\\~\\
=\cdots
\]

Answer 7

You can use cos(A+B) = cosAsinB+sinAcosB

cos(2x) = cos(x+x) = .... you see :P

Answer 8

\(\mathcal R\) is \(\Re\), the real component ?

Answer 9

Yes