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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


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 


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.


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)
     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
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Associated Topics:
High School Trigonometry

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