How do you simplify $\tan 10A$ in terms of $5A$? How do you simplify $\tan 10A$ in terms of $5A$?
I just need a few steps to get me going. All help is appreciated.
Thanks!

Using de Moivre's formula,
$$\cos 10x+i\sin 10x=(\cos x+i\sin x)^{10}$$
$$=\sum_{0\le r\le 10}\binom {10}r(\cos x)^{10-r}(i\sin x)^r$$
$$=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(i\sin x)^{2s}+\binom {10}{2s+1}(\cos x)^{2n-2s-1}(i\sin x)^{2s+1}$$
$$=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(\sin x)^{2s}(-1)^s+i\binom {10}{2s+1}(\cos x)^{2n-2s-1}(\sin x)^{2s+1}(-1)^s$$
Equating the real & the imaginary parts, 
$$\cos10x=\sum_{0\le 2s+1\le 10}\binom {10}{2s}(\cos x)^{10-2s}(\sin x)^{2s}(-1)^s$$
$$=(\cos x)^{10}-\binom {10}2(\cos x)^8(\sin x)^{2}+\binom {10}4(\cos x)^6(\sin x)^4-\binom {10}6(\cos x)^4(\sin x)^6+\binom {10}8(\cos x)^2(\sin x)^8-(\sin x)^{10}$$
$$=(\cos x)^{10}\{1-(\tan x)^{2}+\binom {10}4 (\tan x)^4-\binom {10}6 (\tan x)^6+\binom {10}8 (\tan x)^8-(\tan x)^{10}\}$$
Simailrly, 
$$\sin10x=(\cos x)^{10}\{\binom {10}1 (\tan x)-\binom {10}3 (\tan x)^3+\binom {10}5 (\tan x)^5-\binom {10}7 (\tan x)^7+\binom {10}9 (\tan x)^9\}$$
Divide to get $\tan10x$ in terms of $\tan x$
Validate here.
