# Question of the basic properties of tan

• thunderjolt
In summary, the student is having trouble solving a problem in their engineering homework involving the gamma function and the inverse tangent of 3. They are unsure if they did the problem correctly, or if they need to redraw it.

## Homework Statement

I've solved a problem in my engineering homework to a point where I have the following data:
ψ = arctan(3) = 71.57 (degrees) , inverse tangent of 3
δ = 30 degrees
solve for gamma and chi (χ):

tan(2gamma) = tan(2ψ)cos(δ)
sin(2χ) = sin(2ψ)sin(δ)

## Homework Equations

tan(2gamma) = tan(2ψ)cos(δ)
sin(2χ) = sin(2ψ)sin(δ)

## The Attempt at a Solution

My problem rests with the gamma solution. When I draw it (using other data in the problem), it is an angle larger than 70 degrees at least. When I solve for gamma with a calculator I get gamma = -16.5°.

my process:
tan(2ψ) = -.75
(-.75)cos(30) = -0.6495
arctan(-0.6495) = -33°
gamma = -33/2 = -16.5°

I know that there are properties of tangent that you need to take into consideration, but I just cannot remember them, and I am sure that is why I get this negative angle.

Or, did I do this correctly, and I just need to redraw my problem?

EDIT: the answer in the back of the book is gamma = 73.5 degrees

Last edited:
Without knowing the context of the problem, it's hard to say what's going on, but I don't think it's a coincidence that 73.5+16.5 = 90.

I, too, noticed that they summed to 90. Well, the entire problem, minus the EE info, is:

I have an equation that makes an elipse:

E = x[10cos(wt + 30°)] + y[30cos(wt)]

We can make wt our variable. x and y are the vector directions, I believe. (they have hats ^ above them)

gamma is the angle that the major axis of the elipse makes with the x-axis. When I draw the elipse, I can clearly see that said gamma is near positive 70°)

ψ is the angle that the amplitudes of the x and y components make with the x-axis (arctan(30/10) = 71.57)

and chi, χ, is an angle that is hard to describe, but not pertinent to what I need help with.

The tangent function has a period of 180°, so when you take the arctan, you can add 180° to what the calculator gives you to get another solution.