trigonometry
Given $\cos30 = \frac{\sqrt3}{2}$ use trigonometric identities to find the exact value of $\tan\frac{\pi}{3}$
I understand that $\cos30 = \frac{\sqrt3}{2}$ from the standard trig values chart and I know that $\frac{\pi}{3}$ is 60 degrees and I know the value of it from the same chart. I'm not understanding how to use identities to find the value.
Let $a=\sin^{-1}\left(-\frac{1}{2}\right)$ and $b=\tan^{-1}\left(\frac{3}{4}\right)$. We have $$\sin^2a+\cos^2a=1 \Rightarrow \frac{1}{4}+\cos^2a=1 \Rightarrow \cos a=\frac{\sqrt 3}{2} \Rightarrow \tan a = -\frac{\sqrt 3}{3}$$ since $\sin^{-1}x$ has $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ as image.
Also,
$$\frac{\sin b}{\cos b}=\frac{3}{4} \Rightarrow \frac{9}{16}\cos^2 b+\cos^2b=1 \Rightarrow \cos b=\frac{4}{5} \Rightarrow \sin b=\frac{3}{5}$$ since $\tan^{-1}x$ has $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ as image.
We then have $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b} = \frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3} \frac{3}{4}}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}\frac{12+3\sqrt 3}{12+3\sqrt 3}=$$ $$-\frac{108 + 48\sqrt 3 + 27\sqrt 3 +36}{144-27}=-\frac{144+75\sqrt 3}{117}=-\frac{48+25\sqrt 3}{39}$$
This is a table that expresses sin, cos and tan in terms of each other.
You don't need to remember the table... just substitute values in the triangle, find the third side using Pythagoras and find the value of the required function.
Also, you just need to remember which trigonometric function is positive in which quadrant using this table.
This can be remembered using the simple mnemonic.. After School To College.
Best Answer
Upto the sign, you can calculate the tan-value as $$\tan(x)=\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$$
Also note $$\cos(2x)=2\cos^2(x)-1$$ which allows you to calculate $\cos(\frac{\pi}{3})$ from $\cos(\frac{\pi}{6})$