# Thread: Inverse Trig functions- Find sin(arctanp+arctan1/p) if p>0

1. ## Inverse Trig functions- Find sin(arctanp+arctan1/p) if p>0

Hi, I was given this question for homework but I'm stuck. Determine sin(arctanp+arctan1/p) if p>0
After using the compound angle formula and substituting functions, I have no idea what to do next:
I started off by using the compound angle formula, giving me: [sin(arctan(p))]*[cos(arctan(1/p)] + [cos(arctan(p)]*[sin(arctan(1/p)]. From there I Let arctan(p)=a. Through manipulation I therefore got tan(arctan(p))=tan(a), and so p=tan(a)...I did the same for arctan(1/p) but with b, giving me (1/p)=tan(b)...
I realize that the answer to this question should be 1 but have no idea how to get it. Any help would be greatly appreciated! Thanks!

2. Hello, damanwithdaplan!

If we look at it the right way, we don't need fancy formulas.

Determine: . $\sin\left[\arctan p+\arctan\frac{1}{p}\right]\;\text{ if }p>0$

Let: . $\begin{Bmatrix} \alpha &=& \arctan p \\ \beta &=& \arctan\frac{1}{p} \end{Bmatrix}$ .[1]

The problem becomes: . $\sin(\alpha + \beta)$

From [1], we have: . $\begin{Bmatrix}\tan\alpha &=& p \\ \tan\beta &=&\frac{1}{p}\end{Bmatrix}$

And we have this right triangle:
Code:
                        *
* β|
*     |
*        | p
*           |
* α            |
* - - - - - - - - *
1

Get it?
The two angles are from the same right triangle!
. . That is, $\alpha + \beta \:=\:90^o$

Therefore: . $\sin(\alpha + \beta) \;=\;\sin(90^o) \;=\;1$