Date: 01/18/2002 at 19:52:37 From: Patricia Henry Subject: Trigonometry I am taking a 1st year math course at university. I have been through the text, my notes, and math software (The Princeton Review Math Library) but I cannot figure this out. In the text, a sample problem includes: tan(theta) = -sqrt(3) theta = 2pi/3 I cannot figure out how the solution for theta was found. Can you tell me? I need to know in order to answer the questions at the end of the chapter. Thanks, Patricia
Date: 01/18/2002 at 22:54:32 From: Doctor Peterson Subject: Re: Trigonometry Hi, Patricia. This is one of the "special angles" whose trig functions should be familiar to you: 30-60-90 and 45-45-90 Triangles http://mathforum.org/dr.math/problems/kristina3.15.99.html Remembering Trig Functions http://mathforum.org/dr.math/problems/kim.09.27.01.html If you don't recall those offhand, but suspect (since they asked) that the answer must be simple, you can work backward and construct the triangle. First, ignore the signs and just make a right triangle where the ratio of the legs is sqrt(3): + /| / | / |sqrt(3) / | /A | +-----+ 1 How can we see if this is a familiar triangle? Use Pythagoras to find the hypotenuse: 1^2 + sqrt(3)^2 = 1 + 3 = 4 so the hypotenuse is 2. Hmmm ... that's twice the bottom leg, so if we double the right triangle, we get ... + /|\ / | \ 2/ | \2 / | \ /A | \ +-----+-----+ 1 1 ... an equilateral triangle! So angle A must be 60 degrees. Now you have to handle the sign. To do this, picture a unit circle. The tangent will be negative in the second and fourth quadrants, so we have + |\ | \ sqrt(3)| \ | \ | 60\ theta +-----o----------- -1 So theta is 180-60 = 120 degrees; or, in radians, pi - pi/3 = 2pi/3. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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