Calculating Cosine Values Graphically and Algebraically
Date: 05/01/98 at 14:31:10 From: Vern Lindsey Subject: Trigonometry I'm trying to upgrade myself and I need more math. Just looking for help any where I can find it. I took college algebra, and I now need trig. I am self teaching before I take the class. I'm studying out of College Algebra and Trigonometry by Linda L. Exley and Vincent K. Smith, Dekalb College. Chapter 9 Section 9.3, about sine and cosine functions, lists problem #55: Suppose cos(t) = -2/5. Find cos(-t), sin(pi/2 - t), cos(t + 2pi), and cos(t - 2pi) I see that cos(-t) = -2/5 by definition, since cos(-t) = cos(t). However, I'm not seeing what I'm supposed to see on the others. I've tried the fundamental identity cos^2(t) + sin^2(t) = 1, where cos^2(t) is actually read cosine squared t. Anyway I'm lost as to what the lesson wants me to see here. Can you help? Thanks.
Date: 05/01/98 at 17:08:09 From: Doctor Sam Subject: Re: Trigonometry Vern, I'm not sure where you are in your study of trigonometry, so I'll suggest two different ways to evaluate these trig functions. One method is algebraic and uses the subtraction identities for sines and cosines. The other method is graphical and will let you visualize the solutions. Graphical The graphs of y = sin(t) and y = cos(t) are very similar. Both have period 2pi. Physicists say that the two waves are "pi/2 radians out of phase." A mathematician might say that you can get the graph of y = cos(t) by shifting the graph of y = sin(t) by pi/2 radians to the left. In fact, if you look at the graphs of y = sin(t) and y = cos(t), you will see the effect of shifting y = sin(t) by multiples of pi/2: one shift: sin(t + pi/2) = cos(t) two shifts: sin(t + pi) = - sin(t) three shifts: sin(t + 3pi/2) = - cos(t) four shifts: sin(t + 2pi) = cos(t) That's all we need to compute the desired values. 1. sin(pi/2 - t) = sin(-[t - pi/2]) = - sin(t - pi/2) This shifts the sine graph pi/2 units to the right, which is the same as shifting it 3pi/2 units to the left. So - sin(t - pi/2) = - [- cos(t)] = cos(t) = -2/5 2. cos(t + 2pi) = cos(t) = -2/5 because 2pi is the period of cos(t). 3. cos(t - 2pi) = cos(t) = -2/5 for the same reason. Algebraic Somewhere in your book, you will find several useful trig identities called addition formulas. They are used to change things like sin(A + B) into expressions involving only sines and cosines of A and B separately. Here they are: sin(A + B) = sin(A)cos(B) + cos(A)sin(B) and cos(A + B) = cos(A)cos(B) - sin(A)sin(B) A and B can be positive or negative. 1. sin(pi/2 - t). Use A = pi/2, and B = -t: sin(pi/2 - t) = sin(pi/2)cos(t) + cos(pi/2)sin(t) Now sin(pi/2) = 1, and cos(pi/2) = 0, so sin(pi/2 - t) = cos(t) = -2/5 2. cos(t + 2pi). Use A = t, and B = 2pi cos(t + 2pi) = cos(t)cos(2pi) - sin(t)sin(2pi) Since sin(2pi) = 0 and cos(2pi) = 1, cos(t + 2pi) = cos(t) = -2/5 3. cos(t - 2pi). Use A = t, and B = -2pi. cos(t - 2pi) = cos(t)cos(-2pi) - sin(t)sin(-2pi) Since cos(-2pi) = 1 and sin(-2pi) = 0, cos(t - 2pi) = cos(t) = -2/5 I hope that helps! -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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