2
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ Hint: simplify first before you plug in the values. You should get $\sin^2 20^\circ-\sin160^\circ\sin220^\circ+\sin^2 320^\circ=\frac34$. $\endgroup$ Aug 8, 2020 at 3:59
  • $\begingroup$ @user10354138 I came to $\sin^220^\circ(1+2\cos20^\circ+4\cos^220^\circ)$. Can you please hint more? $\endgroup$
    – Ken
    Aug 8, 2020 at 6:16
  • $\begingroup$ 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 $\cos40^\circ+\cos80^\circ=2\cos60^\circ\cos20^\circ=\cos20^\circ$. $\endgroup$ Aug 8, 2020 at 6:28
  • 1
    $\begingroup$ @user10354138 You should make that an answer. $\endgroup$
    – Toby Mak
    Aug 8, 2020 at 6:29
  • 2
    $\begingroup$ @TobyMak I was going to, then AsdrubalBeltran posted the exact simplication steps I used. $\endgroup$ Aug 8, 2020 at 6:34

2 Answers 2

3
$\begingroup$

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}

$\endgroup$
1
  • 1
    $\begingroup$ Excellent answer ! $\endgroup$
    – sai-kartik
    Aug 8, 2020 at 6:29
1
$\begingroup$

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 $$ \tan 6A=\tan 3(2A)=\frac{3\tan 2A-\tan^3 2A}{1-3\tan^2 2A}=\dots $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.