# Reference Triangles for Inverse Trig Functions

• Apr 7th 2010, 07:53 AM
KarlosK
Reference Triangles for Inverse Trig Functions
We are given basic inverse trig functions and asked to find the angles by using a reference triangle. I have gone to some websites and read up on this and am just not understanding how they are getting the angle.

EXAMPLE 1: Determine the value of tan -1(-1).

Let us look at the reference triangle for the angle a = tan -1 (-1) or tan a = -1.

What angle, a, has a triangle that has side lengths of 1 and -1, and the length of the hypotenuse is the http://faculty.eicc.edu/bwood/ma155s...l/Image335.gif
The angle, a, is - p /4.

I am following everything they do up until the last part where they give the angle. How is that number determined? Thanks for the help!
• Apr 7th 2010, 08:07 AM
mathemagister
Quote:

Originally Posted by KarlosK
We are given basic inverse trig functions and asked to find the angles by using a reference triangle. I have gone to some websites and read up on this and am just not understanding how they are getting the angle.

EXAMPLE 1: Determine the value of tan -1(-1).

Let us look at the reference triangle for the angle a = tan -1 (-1) or tan a = -1.

What angle, a, has a triangle that has side lengths of 1 and -1, and the length of the hypotenuse is the http://faculty.eicc.edu/bwood/ma155s...l/Image335.gif
The angle, a, is - p /4.

I am following everything they do up until the last part where they give the angle. How is that number determined? Thanks for the help!

Well, you know SOHCAHTOA, right? The tangent of an angle in a triangle is the length of the opposite side divided by the length of the adjacent side.

Just using that and these two triangles (and radian conversions), you can figure out all the values in the table below. You don't need to memorize the table. Just know SOHCAHTOA and the triangles well.

[IMG]http://uk.wrs.yahoo.com/_ylt=A0WTf2xErLxLXm8AU5FWBQx./SIG=12n8ilud8/EXP=1270742468/**http%3a//www.ltu.edu/courses/lowry/techmath/class/images/u4obj313.gif[/IMG]

Now, do you see how they got that angle? $\tan^{-1}(-1)=\frac{\pi}{4}$. Using the unit circle, you can figure out that $tan^{-1}(-1)=-\frac{\pi}{4}.$

Hope that helps :)

Feel free to ask for more help

Mathemagister
• Apr 7th 2010, 08:11 AM
KarlosK
Quote:

Originally Posted by mathemagister
Well, you know SOHCAHTOA, right? The tangent of an angle in a triangle is the length of the opposite side divided by the length of the adjacent side.

Just using that and these two triangles (and radian conversions), you can figure out all the values in the table below. You don't need to memorize the table. Just know SOHCAHTOA and the triangles well.

[IMG]http://uk.wrs.yahoo.com/_ylt=A0WTf2xErLxLXm8AU5FWBQx./SIG=12n8ilud8/EXP=1270742468/**http%3a//www.ltu.edu/courses/lowry/techmath/class/images/u4obj313.gif[/IMG]

Now, do you see how they got that angle? $\tan^{-1}(-1)=\frac{\pi}{4}$. Using the unit circle, you can figure out that [tex]tan^{-1}(-1)=-\frac{\pi}{4}.

Hope that helps :)

Feel free to ask for more help

Mathemagister

That was a really great explanation, and very helpful. It has been 5 years since pre-calc so this trig they teach in calculus is not easy for me. Thanks so much for the refresher
• Apr 7th 2010, 08:15 AM
mathemagister
Quote:

Originally Posted by KarlosK
That was a really great explanation, and very helpful. It has been 5 years since pre-calc so this trig they teach in calculus is not easy for me. Thanks so much for the refresher

No problem (Happy)

If you are taking strictly-timed exams where you will need to use this frequently, maybe, just to save time, you can memorize the table instead of the triangles.

However, unless you really feel this would make a significantly useful difference in your speed, you should just use the logic instead of simply using rote memory.