# Thread: Evaluating a trigs of trigs

1. ## Evaluating a trigs of trigs

How does one find the value of tan theta that =1/3

I tried to work it out using a triangle and inverse trig functions

If arctan(1/3)=theta then theta = tan(1/3)

that means we should be able to draw a right angle triangle where the side opposte theta is 1 and the side adjacent is 3

this would make the hypotenuse sqrt(10)

however when I use arcsin and arccos to find the angles I get both angles near 18 which would imply that the triangle is not a right triangle.

Am I on the right track at all?

perhaps my calculator should be in radian mode? is that the what im missing? I'm not sure.

2. ## Re: Evaluating a trigs of trigs

Originally Posted by kingsolomonsgrave

How does one find the value of tan theta that =1/3
I tried to work it out using a triangle and inverse trig functions
If arctan(1/3)=theta then theta = tan(1/3)
You have that backwards.
$\tan(\theta)=\frac{1}{3}$ so $\sin(\theta)=\frac{1}{\sqrt{10}}~\&~\cos(\theta)$ $=\frac{3}{\sqrt{10}}$

Now you have $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$

3. ## Re: Evaluating a trigs of trigs

Evaluate
Hi,
I attached a word-file with the solution.
I hope, does help!
You can verify with the calculator:
tan^-1(0.3333333) =18.434932°
2*tan^-1(0.3333333) =36.869863°
cos(36.869863°) =0.800000