Here again is figure 73, so you can look at the plots of sin(x) and cos(x) while we discuss them. And here is a table of values of sin(x) and cos(x) for some common values of x. These results can all be derived from basic geometry.
x degrees sin(x) cos(x) 
I'd like you to think for a moment about the train on the circular track again. Suppose that train is at the station on Main Street and 10th Avenue East. What is different about when it backs up from when it moves forward? If it backs up 50 meters, doesn't it end up just as far east of Main Avenue as it does going forward 50 meters? Isn't the same true if it backs up any amount versus going forward by that same amount? Now recall that how far east or west of Main Avenue the train is is how we introduced cosine. So what this illustrates is a symmetry of the cosine functions: 
cos(x) = cos(x) eq. 7.1b1
In math lingo, we say that cosine is an even function. Any
f(x) that obeys the property,
Similarly, what happens differently to the train's northsouth position depending upon whether it goes backward 50 meters or forward 50 meters? Well this time, there is a difference. But there is also a symmetry here as well. The train will end up just as far north of Main Street by going 50 meters forward as it would end up south of Main Street by by going 50 meters backward. And again you could substitute any other amount for 50 meters, and the same would still be true. Now recall that how far north or south of Main Street the train is is how we introduced sine. So with sine, the relationship is
sin(x) = sin(x) eq. 7.1b2As you might have guessed, in math lingo we say that sine is an odd function. Any f(x) that obeys
(Food for thought:
Can you show that the only function that is simultaneously an even
function and an odd function is
Here is another property of sine and cosine that should be evident from the circular track model and the plot shown in figure 73. Both sine and cosine have a period of 2p. That is, wherever you are on the track, if you go 2p kilometers farther, you will end up at precisely the same place. Why? Because you will have gone full circle. And if you go 2p again, the same thing. Indeed if you go any multiple of 2p kilometers you will end up exactly where you started. And that means just as far north or south of Main Street, and just as far east or west of Main Avenue as you started. And what this means in terms of sine and cosine is that for any integer, n, it is always the case that
sin(x) = sin(x + 2np) eq. 7.1b3a cos(x) = cos(x + 2np) eq. 7.1b3b
We are ready now to review a whole raft of useful relationships among trig functions. If you can't memorize them, you should learn to derive them quickly. They will soon become tools you will need to do homework and exam problems.
Here again are the two that we developed in the main text:
sin^{2}(x) + cos^{2}(x) = 1 eq. 7.11and
cos(a+b) = cos(a)cos(b)  sin(a)sin(b) eq. 7.16From equation 7.11 you have immediately that
___________ cos(x) = Ö1  sin^{2}(x) eq. 7.11a ___________ sin(x) = Ö1  cos^{2}(x) eq. 7.11b
Now look at table 71. Notice that
cos(a  p/2) = cos(a)cos(p/2)  sin(a)sin(p/2) eq. 7.1b3a cos(a  p/2) = cos(a)×0  sin(a)×(1) eq. 7.1b3b cos(a  p/2) = sin(a) eq. 7.1b3cAnd because cosine is an even function, it follows as well that
cos(p/2  a) = sin(a) eq. 7.1b3dBoth of these are true for any real number, a. Now substitute
cos(u) = sin(p/2  u) eq. 7.1b3eWhich is true for any real number, u. But suppose instead you substituted
cos(u) = sin(u + p/2) eq. 7.1b3fwhich again is true for any real number, u.
Now lets take
cos(p/2  a  b) = cos(p/2  a)cos(b)  sin(p/2  a)sin(b) eq. 7.1b4aNow simply apply the identities we have so far:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b) eq. 7.1b4band we have a way of finding the sine of the sum of two angles. So now we have formulae for both sine and cosine for the sum of angles, but what about differences of angles? You can use the sum formulae together with the even and odd properties to, substituting b for b, to get:
cos(a  b) = cos(a)cos(b) + sin(a)sin(b) eq. 7.1b5a sin(a  b) = sin(a)cos(b)  cos(a)sin(b) eq. 7.1b5bYou can also use the sum formulae to derive expressions for sine and cosine of doubleangles. Simply observe that
cos(2x) = cos(x + x) = cos^{2}(x)  sin^{2}(x) eq. 7.1b6a sin(2x) = sin(x + x) = 2sin(x)cos(x) eq. 7.1b6bYou can do a sneaky trick on equation 7.1b6a to get a halfangle by substituting x/2 and combining it with
cos(x) + 1 = cos(x/2 + x/2) + 1 = cos^{2}(x/2)  sin^{2}(x/2) + 1 = cos^{2}(x/2)  sin^{2}(x/2) + sin^{2}(x/2) + cos^{2}(x/2) eq. 7.1b7aWhen you cancel and simplify, you get
cos(x) + 1 = 2cos^{2}(x/2) eq. 7.1b7b ______________ Öcos(x)/2 + 1/2 = cos(x/2) eq. 7.1b7cLikewise, you can change signs and get
1  cos(x) = 1  cos(x/2 + x/2) = 1  cos^{2}(x/2) + sin^{2}(x/2) = sin^{2}(x/2) + cos^{2}(x/2)  cos^{2}(x/2) + sin^{2}(x/2) eq. 7.1b8aWhen you cancel and simplify, you get
1  cos(x) = 2sin^{2}(x/2) eq. 7.1b8b ______________ Ö1/2  cos(x)/2 = sin(x/2) eq. 7.1b8cPutting in x instead of x/2, you can use the above to get formulae for sine squared and cosine squared:
sin^{2}(x) = 1/2  cos(2x)/2 eq. 7.1b9a cos^{2}(x) = 1/2 + cos(2x)/2 eq. 7.1b9b(Observe what happens when you add the righthand sides of those two equations)
Combining equations 7.16 and 7.1b5a, you get
cos(a+b) + cos(ab) = cos(a)cos(b)  sin(a)sin(b) + cos(a)cos(b) + sin(a)sin(b) = 2cos(a)cos(b) eq. 7.1b10aIf you divide out the 2, you can see that taking the product of the cosines of two numbers is the same as taking half the cosine of their sum plus half the cosine of their difference. Likewise
cos(a+b)  cos(ab) = cos(a)cos(b)  sin(a)sin(b) + cos(a)cos(b)  sin(a)sin(b) = 2sin(a)sin(b) eq. 7.110bAgain, if you divide out the 2, you can see that taking the product of the sines of two numbers is the same as taking half the cosine of their difference minus half the cosine of their sum. Similarly, you can combine equations 7.1b4b and 7.1b5b to get
sin(a+b) + sin(ab) = sin(a)cos(b) + cos(a)sin(b) + sin(a)cos(b)  cos(a)sin(b) = 2sin(a)cos(b) eq. 7.1b11a sin(a+b)  sin(ab) = sin(a)cos(b) + cos(a)sin(b) + sin(a)cos(b) + cos(a)sin(b) = 2cos(a)sin(b) eq. 7.1b11bHere's a cute one for you.
_ _ _ sin(x) + cos(x) = Ö2 (Ö2/2)sin(x) + (Ö2/2)cos(x) ) = _ Ö2 ( cos(p/4)sin(x) + sin(p/4)cos(x) ) = _ Ö2 sin(x + p/4) eq. 7.1b12It's also equal to
_ Ö2 cos(p/4  x)but I'm sure you can prove that using the identities we have so far.
Surely you recall from trig that sine and cosine were not the only trig functions you studied. They also introduced you to
sin(x) tan(x) =eq. 7.1b13a cos(x) cos(x) cot(x) =eq. 7.1b13b sin(x) 1 sec(x) =eq. 7.1b13c cos(x) 1 csc(x) =eq. 7.1b13d sin(x)
Figure 74 shows the geometric interpretation of these functions. Once
again, the radius of the circle is 1. The angle, x, is
still in radians. Observe that the point, L, has
coordinates of
Unlike the sine and cosine, these new functions are not continuous everywhere.
For example, according to
Similarly,
We shall be discussing more about continuity of trig functions in a later section. 

The functions, sec(x) and csc(x), follow a similar pattern. Observe in figure 76 that sec(x) (the blue trace) is discontinuous at odd multiples of p/2, and csc(x) (the green trace) is discontinuous at multiples of p. And the discontinuities occur for the very same reasons as they do in tan(x) and cot(x).
There are just a few identities that we'll go over concerning these
functions. First, you recall from
1 sec(x) =so it must also be true that 1 sec^{2}(x) = 
But we all know from eq. 7.11 that the 1 in the numerator of
the above is the same as
sin^{2}(x) + cos^{2}(x) sec^{2}(x) =With just a little algebra and a glance back ateq. 7.1b14a cos^{2}(x)
sec^{2}(x) = tan^{2}(x) + 1 eq. 7.1b14band using the identical approach you can show that
csc^{2}(x) = cot^{2}(x) + 1 eq. 7.1b14c
Developing a formula for tan(a+b) is a piece of cake when you use the formulae we already have for sin(a+b) and cos(a+b).
sin(a+b) tan(a+b) =If you divide numerator and denominator of eq. 7.1b15b by cos(a)cos(b), you get some cancellations.eq. 7.1b15a cos(a+b) sin(a)cos(b) + sin(b)cos(a) tan(a+b) =eq. 7.1b15b cos(a)cos(b)  sin(a)sin(b)
sin(a) sin(b)Finally, applying+cos(a) cos(b) tan(a+b) =eq. 7.1b15c sin(a)sin(b) 1 cos(a)cos(b)
tan(a) + tan(b) tan(a+b) =Using a very similar method you can come up one for the cotangent (which I will let you derive for yourself):eq. 7.1b15d 1  tan(a)tan(b)
cot(a)cot(b)  1 cot(a+b) =eq. 7.1b15e cot(a) + cot(b)
email me at hahn@netsrq.com