Simplifying Expressions with Trig FunctionsDate: 04/04/2001 at 10:56:21 From: Betsey Subject: Cosine and Cotangent 4 csc(3pi/4) - cot(-pi/4) I am home schooling my niece and it has been over 25 years since I worked with this kind of stuff. Can you work this out for me so I can get a grasp and then be able to explain this to her? Date: 04/04/2001 at 12:45:15 From: Doctor Peterson Subject: Re: Cosine and Cotangent Hi, Betsey. I find it helpful to draw a rough sketch of the angle and a triangle to define the trig functions, and then use my knowledge of the angle to determine the values of the relevant sides. Let's do the first part of this, the cosecant of 3pi/4. First, what does this angle look like? It will be halfway between pi/2 and pi, so placed in standard position in a unit circle, it is something like this: *********** ***** | ***** (x,y)**** | **** **| | ** * | \ | * ** | \ 1 | ** * y| \ | * * | \ | * * | \ | * * | T \ | * *-------+-------------+---------------------* * -x | * * | * * | * * | * ** | ** * | * ** | ** **** | **** ***** | ***** *********** The acute angle T is 45 degrees (pi/4 radians); that makes the triangle a right isosceles triangle, and I know that the positive lengths -x and y are equal. Using the Pythagorean theorem, they're equal to sqrt(2)/2. Now, the cosecant is the reciprocal of the sine; the sine is y/1 (opposite/hypotenuse), so the cosecant is 1/y. This makes the cosecant equal to the reciprocal of sqrt(2)/2, which is sqrt(2). With angles like this outside the first quadrant, you have to watch the signs; in this case the sine, y, is positive, but the cosine, x, would be negative since the point (x,y) is on the left side of the circle. You can do the same thing to find the cotangent of -pi/4: *********** ***** | ***** **** | **** ** | ** * | * ** | ** * | * * | * * | * * | x * *---------------------+-------------+-------* * | \T | * * | \ 1 | * * | \ | * * | \ |-y * ** | \ | ** * | \ | * ** | |** **** | ****(x,y) ***** | ***** *********** Here the cotangent is the reciprocal of the tangent, which is y/x (opposite/adjacent). Again, we have a right isoscecles triangle, so the absolute values of x and y are equal. But since x is positive and y is negative, the tangent is 1-1 = -1, and the cotangent is x/y = -1 as well. If you have trouble with any of the steps I took, please write back and let me know, so I can give more details where they are needed. Also, don't miss our archived answers on high school trigonometry: http://mathforum.org/dr.math/tocs/trig.high.html Here are a few good introductory pages on finding functions of familiar angles: Deriving Sines of 30, 45, 60 and 90 Degrees http://mathforum.org/dr.math/problems/michael.6.02.99.html Sine, Cosine, and Tangent on a Circle http://mathforum.org/dr.math/problems/jongejan.04.14.99.html Our FAQ on trig formulas might be overwhelming at first, but there's a lot of good stuff there: http://mathforum.org/dr.math/faq/formulas/faq.trig.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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