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

• Feb 13th 2010, 12:57 PM
damanwithdaplan
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! :)
• Feb 13th 2010, 09:09 PM
Soroban
Hello, damanwithdaplan!

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

Quote:

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