Finding $\alpha$ such that $\;\tan\alpha=\frac{8\sqrt3\cos^240^\circ+8\sqrt3\cos20^\circ-2\sqrt3}{(8\cos^240^\circ-1)(8\cos20^\circ-1)-3}$

Asked Jul 10 '22 at 07:45

Active Jul 10 '22 at 08:55

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I have to find $\alpha$ (in degrees) so it is satisfied $$\tan{\alpha}=\frac{8\sqrt{3}\cos^{2}40^{\circ}+8\sqrt{3}\cos20^{\circ}-2\sqrt{3}}{\left(8\cos^{2}40^{\circ}-1\right)\left(8\cos20^{\circ}-1\right)-3}$$

I tried to write $\cos40^{\circ}=2\cos^{2}20^{\circ}-1$. ALso i tried $\cos^{2}40^{\circ}=\frac{1+\cos80^{\circ}}{2}=\frac{1+\sin10^{\circ}}{2}$ and to simplify the expression but unsuccessfully.

edited Jul 10 '22 at 08:55

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asked Jul 10 '22 at 07:45

JohnB

Where did you get this problem? – Feng Jul 10 '22 at 08:58
I found it in my notes, but I didn't write the source. :( – JohnB Jul 10 '22 at 12:39
1

I calculate it numerically and get $\alpha$ should be $40^\circ$. – JohnB Jul 13 '22 at 07:06

Finding $\alpha$ such that $\;\tan\alpha=\frac{8\sqrt3\cos^240^\circ+8\sqrt3\cos20^\circ-2\sqrt3}{(8\cos^240^\circ-1)(8\cos20^\circ-1)-3}$

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