# Homework Help: Trigonometric question 2

1. Dec 2, 2011

### beborche

1. The problem statement, all variables and given/known data
Write arctan8 + arctan11 as an expression containing max one term with arctan.

2. Relevant equations
tan(u+v) = $\frac{tanu+tanv}{1-tanutanv}$
arctan(tanx) = x

3. The attempt at a solution
u = arctan8, v = arctan11
tan(u+v) = $\frac{tanu+tanv}{1-tanutanv}$ = $\frac{8+11}{1-8*11}$ = -(19/87) = tan(arctan8 + arctan11)

arctan(tan(arctan8 + arctan11)) = arctan(-(19/87)) = arctan8 + arctan11

arctan8 + arctan11 = arctan(-/19/87)) = -arctan(19/87)

But (arctan8 + arctan11) > 0 and -arctan(19/87) < 0

Where have I gone wrong?

Last edited: Dec 2, 2011
2. Dec 2, 2011

### ehild

Remember, the tangent function is periodic with pi: tanθ=tan(θ±π)
The range of the inverse function arctan is (-π/2,π/2). When you type in a number and hit tan-1, the calculator gives the principal value, an angle in the interval (-π/2,π/2).

You got it right, tan(arctan8 + arctan11)= -(19/87).

arctan8 + arctan11 is an angle in the second quadrant, and its tangent is the same as that of (arctan8 + arctan11-pi), an angle between -pi/2 and pi/2: That is what you get as result. Add pi to have the the real sum: arctan8 + arctan11=arctan(-19/87)+pi

ehild

3. Dec 2, 2011

### beborche

@ehild

Alright. I get it now. Tanv will produce the same result for all angles v + n*pi, where n=1, 2, 3... Thank you!