I understand how to use the compound angle formula when solving $\sin(\pi/12)$.
However I dont understand how I can use a compound angle formula to show
$$\arctan(3)-\arctan(1/2)=\pi/4$$
Thankyou
Any references would be highly appreciated, I looked online but couldn't find any that targeted the arc-tan/sin/cos.
Best Answer
Hint:
I assume you know the formula for $\tan(A\pm B)$ (otherwise, your teacher is evil :) ).
Plug $A=\tan^{-1}a$, $B=\tan^{-1}b$ into that formula and rearrange it to get a formula for $tan^{-1}a - tan^{-1}b = tan^{-1}$(some expression involving $a$ and $b$).
We start with the known $\tan$ compound angle formula:
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
Substituting $A=\tan^{-1}a$, $B=\tan^{-1}b$ : $a = \tan A$, $b = \tan B$
$$\tan(\tan^{-1}a + \tan^{-1}b) = \frac{a + b}{1 - ab}$$
$$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 - ab}\right)$$
Now let $a = 3$, $b = -\frac{1}{2}$
Note than $\tan(-x) \equiv -\tan x$ for all $x$
$$\tan^{-1}3 - \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{3 - \frac{1}{2}}{1 + 3\cdot\frac{1}{2}}\right)$$
$$\tan^{-1}3 - \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{6 - 1}{2 + 3}\right) = \tan^{-1}1 = \frac{\pi}{4}$$