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Evaluating a trigs of trigs

Attachment 25364

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.

Re: Evaluating a trigs of trigs

Quote:

Originally Posted by

**kingsolomonsgrave** Attachment 25364
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.

$\displaystyle \tan(\theta)=\frac{1}{3}$ so $\displaystyle \sin(\theta)=\frac{1}{\sqrt{10}}~\&~\cos(\theta)$$\displaystyle =\frac{3}{\sqrt{10}}$

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

1 Attachment(s)

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

(Use-modus is degree, not radian)

Best regards,

Stefan