Trigonometry! #36

First of all, $sin(nx) = \frac{{e}^{inx} - {e}^{-inx}}{2i}$

${sin(x)}^{n} = {\frac{{e}^{ix} - {e}^{-ix}}{2i}}^{n} = f({e}^{ix}, {e}^{-ix})$ of degree n.

And as we have seen $sin(nx)$ is also $g({e}^{ix}, {e}^{-ix})$ of degree n.

So, $sin(nx) = f(sin x)$ of degree n.

And similarly for cos(x).

Hence, on LHS, we have a function of sin(x) of degree 6.

Now, we know, ${sin(x)}^{2} = 1 - {cos(x)}^{2}$ .

${sin(x)}^{6} = {(1 - {cos(x)}^{2})}^{3}$

And so, it is also a function of cos(x) of degree 6.

Therefore, now we can gain use the above property and find that the value of n is $6$ .