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How to prove that $$\tan^2 20^{\circ} + \frac{3}{16}\csc^2 40^ {\circ}\sec^2 20^{\circ}-\frac{\sqrt3}{4}\tan 20^{\circ} \sec^2 20^{\circ} = 4 \sin^2 20^{\circ}$$

Can I express them in terms of $\tan 20^{\circ}$ and prove it?

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  • $\begingroup$ prove the identity, that means prove that left hand side equals right hand side $\endgroup$ – Ray Cheng Jun 6 '17 at 5:01
  • $\begingroup$ Are these all degrees or radian? $\endgroup$ – Jaideep Khare Jun 6 '17 at 5:04
  • $\begingroup$ all degrees, I am sure $\endgroup$ – Ray Cheng Jun 6 '17 at 5:04
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    $\begingroup$ What is the source? $\endgroup$ – lab bhattacharjee Jun 6 '17 at 5:14
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    $\begingroup$ What have you tried. I find it easiest to put this in terms for sins and cosines. A lot should simplify. the only tricky part I see is $ \csc^2 40 = \frac 1{\sin (20 + 20)}$ which should be straight forward. $\endgroup$ – fleablood Jun 6 '17 at 5:21
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Let $x=20^\circ$. Then \begin{eqnarray} &&\tan^2 20^{\circ} + \frac{3}{16}\csc^2 40^ {\circ}\sec^2 20^{\circ}-\frac{\sqrt3}{4}\tan 20^{\circ} \sec^2 20^{\circ}-4\sin^2 20^\circ\\ &=&\frac{1}{\cos^2 x\sin^2(2x)}\bigg[\frac{3}{16}-\sqrt3\cos x\sin^3x+4\cos^2x\sin^4x-\sin^4(2x)\bigg].\\ \end{eqnarray} Noting $$ \sin 60^\circ=\sin(3x)=-4\sin^3x+3\sin x=\frac{\sqrt3}{2}, \cos 60^\circ=\cos(3x)=4\cos^3x-3\cos x=\frac{1}{2}$$ and $$ \sin 120^\circ=\sin(6x)=-4\sin^3(2x)+3\sin(2x)=\frac{\sqrt3}{2}$$ one has $$ 4\sin^3x=-\frac{\sqrt3}{2}+3\sin x,4\cos^3x=\frac12+3\cos x, 4\sin^3(2x)=-\frac{\sqrt3}{2}+3\sin(2x) $$ and hence \begin{eqnarray} &&\frac{3}{16}-\sqrt3\cos x\sin^3x+4\cos^2x\sin^4x-\sin^4(2x)\\ &=&\frac{3}{16}-\frac{\sqrt3}{4}\cos x\bigg(-\frac{\sqrt3}{2}+3\sin x\bigg)+\cos^2 x\sin x\bigg(-\frac{\sqrt3}{2}+3\sin x\bigg)\\ &&-\frac14\sin(2x)\bigg(-\frac{\sqrt3}{2}+3\sin(2x)\bigg)\\ &=&\frac{3}{16}+\frac{3}{8}\cos x-\frac{\sqrt{3}}{4}\sin (2x)-\frac{\sqrt{3}}{4}\cos x \sin(2x)\\ &=&\frac{3}{16}+\frac{3}{8}\cos x-\frac{\sqrt{3}}{4}\sin (2x)-\frac{\sqrt{3}}{8}(\sin 60°+\sin x)\\ &=&\frac18 \bigg(3\cos x-\sqrt3 \bigg(\sin x+2\sin(2x)\bigg)\bigg). \end{eqnarray} Noting $$ \sin x+2\sin(2x)=\sin x+2\sin(60^\circ-x)=\sqrt3\cos x$$ it is easy to check that $$ 3\cos x-\sqrt3(\sin x+2\sin(2x))=0 $$ and hence $$ \frac{3}{16}-\sqrt3\cos x\sin^3x+4\cos^2x\sin^4x-\sin^4(2x)=0.$$

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  • $\begingroup$ Could you show some steps after changing the trigonmetric function into imaginary number form? $\endgroup$ – Ray Cheng Jul 14 '17 at 3:16
  • $\begingroup$ @RayCheng, check the update of my answer. $\endgroup$ – xpaul Jul 14 '17 at 14:15
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https://www.quora.com/How-can-I-prove-that-sin-20%C2%B0-+-2-sin-40%C2%B0-sqrt-3-cos-20%C2%B0/answer/Peter-Tran-93?filter=all&nsrc=1&snid3=1261438578

After I asked a question at Quora, I am sure that in order to show that $3\cos x-\sqrt3(\sin x+2\sin(2x))=0$, where $x=20°$, this is actually equivalent to show that $\tan 60°\cos20°-\sin20°=2\sin 40°$. Using this approach, we don't need to use imaginary numbers (Euler's formula).

$\tan 60°\cos20°-\sin20°=\frac{1}{\cos 60°}(\sin 60° \cos 20°-\cos 60° \sin 20°)=2 \sin(60°-20°)=2 \sin 40°.$

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