# Question of the basic properties of tan

## 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:

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vela
Staff Emeritus
Homework Helper
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.

vela
Staff Emeritus