trig problem: Tan^-1(1/3) + Tan^-1(1/2)

[duckers321]

The problem reads "The inverse tanget of 1/3 plus the inverse tangent of 1/2"

This is a non-calculator problem
I was able to get 45 degrees or pi/4 using a calculator, but i cannot seem to figure it out without a calculator.

 

[Loren]

Let tan x = 1/3 and tan y = 1/2.

Use the identity $\displaystyle \tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y}$

Plug in the fractions, do the arithmetic and discover that the tan(x+y) = 1. Therefore, x+y = pi/4.

 

[stapel]

The problem reads "The inverse tanget of 1/3 plus the inverse tangent of 1/2"

The inverse tangent's value A is an angle measure @ (which I'll use to stand for "theta"), where tan(@) = A. So you have been given:

. . . . .tan(x) = 1/3

. . . . .tan(y) = 1/2

...for some angle measures x and y. That is:

. . . . .arctan(1/3) = x

. . . . .arctan(1/2) = y

You are being asked to find the value of x + y.

Hint: Trying taking the tangent of the angle sum, and applying an angle-addition formula. :wink:

Eliz.

 

[soroban]

Hello, duckers321!

$\displaystyle \text{We'll use the identity: }\;\tan(A +B) \;=\;\frac{\tan(A) + \tan (B)}{1 - \tan(A)\tan(B)}$

$\displaystyle \text{Evaluate: }\;\arctan\!\left(\frac{1}{3}\right) + \arctan\!\left(\frac{1}{2}\right)$

$\displaystyle \text{Let: }\:\theta \;=\;\arctan\!\left(\frac{1}{3}\right) + \arctan\!\left(\frac{1}{2}\right)$

$\displaystyle \text{Then: }\;\tan\theta \;=\;\tan\left[\arctan\!\left(\frac{1}{3}\right) + \arctan\!\left(\frac{1}{2}\right)\right]$

. . . . . . . . . .$\displaystyle =\;\frac{\tan\!\left[\arctan\!\left(\frac{1}{3}\right)\right] + \tan\!\left[\arctan\!\left(\frac{1}{2}\right)\right]} {1 - \tan\!\left[\arctan\!\left(\frac{1}{3}\right)\right]\!\cdot\!\tan\!\left[\arctan\!\left(\frac{1}{2}\right)\right]}$

. . . . . . . . . .$\displaystyle = \;\;\frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3}\!\cdot\!\frac{1}{2}} \;\;=\;\;\frac{\frac{5}{6}}{\frac{5}{6}} \;\;=\;\;1$


\(\displaystyle \text{And we have: }\:\tan\theta \:=\:1 \quad \hdots \quad \text{ Therefore: }\:\boxed{\theta \:=\:\frac{\pi}{4}}\)