# Let $\frac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$

Let $$\dfrac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$$, find $$\tan6A$$

My attempt : \begin{align*} \dfrac{\tan2A}{2}=\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ\\ \tan2A=2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ) \end{align*} and \begin{align*} \tan6A&=\tan(2A-60^\circ)\tan2A\tan(2A+60^\circ)\\ &=(\dfrac{\tan2A-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(\tan2A)(\dfrac{\tan2A+\sqrt{3}}{1-\sqrt{3}\tan60^\circ}) \end{align*} give $$\tan6A=(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)-\sqrt{3}}{1+\sqrt{3}\tan60^\circ})(2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ))(\dfrac{2(\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ)+\sqrt{3}}{1-\sqrt{3}\tan60^\circ})$$ This method is incredibly long, there may be better way to deal with this problem.

• Hint: simplify first before you plug in the values. You should get $\sin^2 20^\circ-\sin160^\circ\sin220^\circ+\sin^2 320^\circ=\frac34$. – user10354138 Aug 8 '20 at 3:59
• @user10354138 I came to $\sin^220^\circ(1+2\cos20^\circ+4\cos^220^\circ)$. Can you please hint more? – Ken Aug 8 '20 at 6:16
• You have $\sin^2\theta(1+2\cos\theta+4\cos^2\theta)=\frac12(2+\cos\theta-\cos 2\theta-\cos 3\theta-\cos 4\theta)$. Now note that you have $\cos40^\circ+\cos80^\circ=2\cos60^\circ\cos20^\circ=\cos20^\circ$. – user10354138 Aug 8 '20 at 6:28
• @user10354138 You should make that an answer. – Toby Mak Aug 8 '20 at 6:29
• @TobyMak I was going to, then AsdrubalBeltran posted the exact simplication steps I used. – user10354138 Aug 8 '20 at 6:34

Note that: $$\begin{array}{} \dfrac{\tan2A}{2}&=&\sin^2160^\circ-\sin160^\circ\sin220^\circ+\sin^2220^\circ\\ \dfrac{\tan2A}{2}&=&(\sin160^\circ-\sin220^\circ)^2+\sin160^\circ\sin220^\circ\\ \dfrac{\tan2A}{2}&=&(\sin20^\circ+\sin40^\circ)^2-\sin20^\circ\sin40^\circ\end{array}$$ Apply product-sum and sum-product $$\begin{array}{}\dfrac{\tan2A}{2}&=&(2\sin30^\circ\cos10^\circ)^2-\dfrac{\cos20^\circ-\cos60^\circ}{2}\\ \dfrac{\tan2A}{2}&=&\cos^210^\circ-\dfrac{1-2\sin^210^\circ}{2}+\dfrac{1}{4}\\ \dfrac{\tan2A}{2}&=&\cos^210^\circ+\sin^210^\circ-\dfrac{1}{4}\\ \dfrac{\tan2A}{2}&=&\dfrac{3}{4}\\ \tan(2A)&=&\dfrac{3}{2} \end{array}$$

• Excellent answer ! – sai-kartik Aug 8 '20 at 6:29

Let me post a different route to the simplication, as OP has begun.

So we have $$\sin^2 20^\circ-\sin160^\circ\sin220^\circ+\sin^2 320^\circ= \sin^2 20^\circ(1+2\cos 20^\circ + 4\cos^2 20^\circ)$$ I claimed for all $$\theta$$, $$\sin^2\theta(1+2\cos\theta+4\cos^2\theta)=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta-\cos4\theta)$$ Proof of claim: \begin{align*} LHS&=\sin^2\theta(1+2\cos\theta+4\cos^2\theta)\\ &=\frac12(1-\cos 2\theta)(3+2\cos\theta+2\cos2\theta)\\ &=\frac12(3+2\cos\theta-\cos2\theta-2\cos\theta\cos2\theta- 2\cos^22\theta)\\ &=\frac12(2+2\cos\theta-\cos2\theta-2\cos\theta\cos2\theta- \cos4\theta)\\ &=\frac12(2+\cos\theta-\cos2\theta-(4\cos^3\theta-3\cos\theta)- \cos4\theta)\\ &=\frac12(2+\cos\theta-\cos2\theta-\cos3\theta- \cos4\theta)\\ &=RHS\quad\checkmark \end{align*}

So with $$\theta=20^\circ$$, this is \begin{align*} &\sin^2 20^\circ(1+2\cos 20^\circ + 4\cos 40^\circ)\\ &=\frac12(2+\cos 20^\circ-\cos40^\circ-\cos60^\circ- \cos80^\circ)\\ &=\frac12\left(\frac32+\cos 20^\circ-\cos40^\circ-\cos80^\circ\right)\\ &=\frac12\left(\frac32+\cos 20^\circ-2\cos60^\circ\cos20^\circ\right)\\ &=\frac34. \end{align*} So $$\tan 2A=\frac32$$ and $$\tan 6A=\tan 3(2A)=\frac{3\tan 2A-\tan^3 2A}{1-3\tan^2 2A}=\dots$$