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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.
=\frac{1}{3})
so =\frac{1}{\sqrt{10}}~\&~\cos(\theta))
Now you have =\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
