Let $c(x)=\dfrac{3^x+3^{-x}}{2}$ and $s(x)=\dfrac{3^x-3^{-x}}{2}$.
Show that $(c(x))^2=\frac{1}{2}(c(2x)+1)$.
How does one go about solving this?
I have honesty tried substituting in logarithms but it doesn't seem to reduce to any simpler form.
My intuition is to equate the coefficients of c(x) and s(x) on both sides and find a common factor.
Let's just calculate, and see what happens: $$ (c(x))^2 = \left(\frac{3^x + 3^{-x}}{2}\right)^2 = \frac{3^{2x} + 2 + 3^{-2x}}{4}\\ = \frac{1}{2}\left(\frac{3^{2x} + 3^{-2x}+2}{2}\right) = \frac{1}{2}\left(\frac{3^{2x} + 3^{-2x}}{2} + 1\right) = \frac12(c(2x) + 1) $$