The question is $x = \arctan\frac 23 + \arctan\frac 12$. What is $\tan(x)$?
I'm having trouble figuring out how to calculate the arctan values without a calculator, or do I not even need to find these values to calculate what $\tan(x)$ is? Any help is greatly appreciated (I would show some sort of work, but I am actually completely stuck).
I know that the range of arctan is restricted to $(–90^\circ, 90^\circ)$ or $(-\pi/2, \pi/2)$, but i'm not sure how this helps.
$\endgroup$3 Answers
$\begingroup$Hint
Use the formula $\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$
$\endgroup$ $\begingroup$Below is a solution based on right triangles.
Note that $y=\arctan\frac{2}{3}$, and $z=\arctan\frac{1}{2}$. It follows that $x = y + z$.
The question is then: Calculate $\frac{b}{a}$.
Since $a^2 + c^2 = 4$ and $\frac{a}{c} = \frac{2}{3}$, it follows that $a^{2} = \frac{16}{13}$.
Since $a^2 + b^2 = 5$, it follows that $b^2 = \frac{49}{13}$.
Thus, $\frac{b}{a} = \frac{7}{4}$.
$\endgroup$ $\begingroup$The take-away should be that sometimes going to a more basic line of thought can get you unstuck on a problem. Also, this gives some insight on how you can derive the tangent of sums formula.
Let $A=\arctan\frac 23,B=\arctan\frac 12$.
Then, $$\tan x=\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac 23+\frac 12}{1-\frac 23\cdot \frac 12}.$$
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