4
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

Below is a solution based on right triangles.

enter image description here

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, $\frac{b}{a} = \frac{7}{4}$.


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.

$\endgroup$
12
$\begingroup$

Hint

Use the formula $\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$

$\endgroup$
6
$\begingroup$

Let $A=\arctan\frac 23,B=\arctan\frac 12$.

Then, $$\tan x=\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac 23+\frac 12}{1-\frac 23\cdot \frac 12}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.