How to simplify Arctan(cot(49$\pi$/15)) to 7$\pi$/30

Question

I tried to write it as Arctan(1/tan) but I got stuck there, and the solution on symbolab was far more complicated than what was covered in class.

Thanks!

Answer 1

First, we recognize that this problem can be solved in terms of a reference triangle:

Which means we have the same reference triangle for $49\pi/15 \mod 2\pi = 19\pi/15$:

This reference triangle has these parameters:

width: $-\cos(19\pi/15)$

height: $-\sin(19\pi/15)$

By definition of the cotangent function:

$$ \cot(19\pi/15)=\frac{-\cos(19\pi/15)}{-\sin(19\pi/15)}=\frac{\cos(19\pi/15)}{\sin(19\pi/15)} $$

We can shift the cosine and the sine to express the cotangent as a tangent:

$$ \tan(x)=\frac{\cos(19\pi/15-\pi/2)}{\sin(19\pi/15+\pi/2)} $$

We get:

$$ \tan(x)=\frac{\sin(23\pi/30)}{\cos(53\pi/30)} $$

Here is the reference triangle for $23\pi/30$:

height: $\sin(7\pi/30)$

Here is the reference triangle for $53\pi/30$:

width: $\cos(7\pi/30)$

So we really get:

$$ \tan(x)=\frac{\sin(7\pi/30)}{\cos(7\pi/30)} $$

By definition of the tangent function, $x=7\pi/30$. This is important because we were looking for $x$ in $\arctan(\tan(x))=x$. That's why we re-expressed our cotangent as this tangent.

Answer 2

You can use:

$$\arctan(x)+\arctan\left({1\over x}\right)={\pi\over2}$$ $$\arctan\left(\cot{49\pi\over15}\right)+\arctan\left(\tan{49\pi\over15}\right)={\pi\over2}$$ $$\arctan\left(\cot{49\pi\over15}\right)=-{49\pi\over15}+{\pi\over2}=-{83\pi\over30}={7\pi\over30}-3\pi$$

and you take ${7\pi\over30}$ because the range of $\arctan x$ is $-{\pi\over2}\le \alpha\le{\pi\over2}$.

Answer 3

Write $x\equiv\arctan(\cot(a))$, where $a\equiv49\pi/2$. Then take $\tan$ of both sides. After that, you can just use trigonometric addition formulas to find $x$.

You will probably not find $x=7\pi/30$ by doing this. This is because $\arctan$ by definition produces a result in the range $-\tfrac{\pi}{2}\leq x \leq \tfrac{\pi}{2}$, and we lost this information when we took $\tan$ of both sides. Hence we need to add to our answer the appropriate multiple of $\pi$ that puts it in this range.