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Prove that: $\tan^6 20° - 33\tan^4 20° + 27\tan^2 20°=3$

My Attempt:

$$L.H.S=\tan^6 20 - 33\tan^4 20 + 27\tan^2 20°$$ $$=\tan^2 20°(\tan^4 20° - 33\tan^2 20°+27)$$ $$=(\sec^2 20° -1)(\tan^4 20° - 33\tan^2 20° + 27)$$

Please help me to continue from here..

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  • $\begingroup$ It is \tan not \Tan. $\endgroup$ – S.C.B. Feb 10 '17 at 15:28
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Let $\pm\sqrt3=\pm\tan60^\circ=\tan3x=\dfrac{3t-t^3}{1-3t^2}$ where $t=\tan x$

$$\implies t^3-3t=\pm\sqrt3(1-3t^2)$$

Square both sides

OR observe that $$\dfrac{t^6-33t^4+27t^2-3}{t^3-3t\mp\sqrt3(1-3t^2)}=t^3-3t\pm\sqrt3(1-3t^2)$$

So, $3x=180^\circ n\pm60^\circ$ where $n$ is any integer.

$\implies x=60^\circ n\pm20^\circ$

Here $n=0$

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  • $\begingroup$ You are so good at trig identities. $\endgroup$ – A---B Feb 10 '17 at 16:32
  • $\begingroup$ @ lab bhattacharjee, Divide what with what?? $\endgroup$ – pi-π Feb 10 '17 at 16:36
  • $\begingroup$ @S.Ramanujan, $$\dfrac{t^6-33t^4+27t^2-3}{t^3-3\sqrt3t^2-3t+\sqrt3}$$ $\endgroup$ – lab bhattacharjee Feb 10 '17 at 16:38
  • $\begingroup$ @A---B, Please find the updated answer. Thanks $\endgroup$ – lab bhattacharjee Feb 11 '17 at 14:06
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My idea: We know formulas for $\cos(20°)$; we can write it as a relation for $\sec(20°)$, and then use the identity $\sec^2(x)=1+\tan^2(x)$


First of all, we'll compute some formulas related to $\cos(20°)$ (well-known, but I don't renember them xD). We know that

$$ \cos(3x)=\cos(2x)\cos(x)-\sin(x)\sin(2x)=(2\cos^2(x)-1)\cos(x)-2\sin^2(x)\cos(x)$$

So, $\cos(3x)=2\cos^3(x)-\cos(x)-2(1-\cos^2(x))\cos(x)=4\cos^3(x)-3\cos(x)$. Plugging $x=20°$, and calling $a=\cos(20)$, we get

$\frac{1}{2}=\cos(60°)=4\cos^3(20^°)-3\cos(x)=4a^3-3a$

So, $1=8a^3-6a$. Squaring both sides, we get $1=64a^6-96a^4+36a^2$. Now, dividing by $a^6$ both sides, and calling $b=\frac{1}{a}=\frac{1}{\cos(20^°)}=\sec(20°)$, we have

$$ \frac{1}{a^6}=64-\frac{96}{a^2}+\frac{36}{a^4} \Rightarrow b^6=64-96b^2+36b^4 $$

The key step now is try to use the identity $\sec^2(x)=1+\tan^2(x)$. Calling $c=\tan(20^°)$, we can plug $b^2=1+c^2$ on our formula, to get

$$ \begin{align*} (1+c^2)^3 &= 64-96(1+c^2)+36(1+c^2)^2 \\ 1+3c^2+3c^4+c^6 &= 64-96(1+c^2)+36(1+2c^2+c^4) \\ 1+3c^3+3c^4+c^6 &= 64-96-96c^2+36+72c^2+36c^4 \\ 1+3c^3+3c^4+c^6 &= 4-24c^2+36c^4 \\ c^6-33c^4+27c^2 &= 3 \\ \tan(20°)^6-33\tan(20°)^4+27\tan(20°)^2 &= 3 \\ \end{align*}$$

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$$\tan60^{\circ}=\sqrt3$$ gives $$\frac{3\tan20^{\circ}-\tan^320^{\circ}}{1-3\tan^220^{\circ}}=\sqrt3,$$ which gives $$\left(\frac{3\tan20^{\circ}-\tan^320^{\circ}}{1-3\tan^220^{\circ}}\right)^2=3,$$ which gives your identity.

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