Common Trig Identities and their Derivations© 1997, 2003 by Karl Hahn |
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Here again is figure 7-3, 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)
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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: |
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cos(-x) = cos(x) eq. 7.1b-1
In math lingo, we say that cosine is an even function. Any f(x) that obeys the property, f(-x) = f(x) is said to be an even function. Note, for example, that x raised to any even power is an even function. That, by the way, is the origin of the calling such functions even.
Similarly, what happens differently to the train's north-south 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. 7.1b-2As you might have guessed, in math lingo we say that sine is an odd function. Any f(x) that obeys f(-x) = -f(x) is said to be an odd function. And as you'd expect, we get that terminology because x raised to any odd power is an odd function.
(Food for thought: Can you show that the only function that is simultaneously an even function and an odd function is f(x) = 0? Can you also show that every real function of a real variable is the sum of an even function and an odd function? Think about the sum and the difference of f(x) and f(-x).)
Here is another property of sine and cosine that should be evident from the circular track model and the plot shown in figure 7-3. Both sine and cosine have a period of 2π. That is, wherever you are on the track, if you go 2π kilometers farther, you will end up at precisely the same place. Why? Because you will have gone full circle. And if you go 2π again, the same thing. Indeed if you go any multiple of 2π 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 + 2nπ) eq. 7.1b-3a cos(x) = cos(x + 2nπ) eq. 7.1b-3b
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:
sin2(x) + cos2(x) = 1 eq. 7.1-2
and
cos(a+b) = cos(a)cos(b) - sin(a)sin(b) eq. 7.1-6
From equation 7.1-2 you have immediately that
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|cos(x)| = √1 - sin2(x) eq. 7.1-2a
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|sin(x)| = √1 - cos2(x) eq. 7.1-2b
Now look at table 7-1. Notice that
sin(π/2) = 1 and
cos(π/2) = 0.
According to the odd and even properties of sine and cosine, we also know
that sin(-π/2) = -1 and
cos(-π/2) = 0.
Suppose you let b = -π/2 and stick it into
equation 7.1-6.
cos(a - π/2) = cos(a)cos(-π/2) - sin(a)sin(-π/2) eq. 7.1b-4a
cos(a - π/2) = cos(a)×0 - sin(a)×(-1) eq. 7.1b-4b
cos(a - π/2) = sin(a) eq. 7.1b-4c
And because cosine is an even function, it follows as well that
cos(π/2 - a) = sin(a) eq. 7.1b-4d
Both of these are true for any real number, a.
Now substitute u = π/2 - a into equation 7.1b-4d and
you have
cos(u) = sin(π/2 - u) eq. 7.1b-4e
Which is true for any real number, u. But suppose instead you
substituted u = a - π/2 into
equation 7.1b-4c.
You'd get
cos(u) = sin(u + π/2) eq. 7.1b-4fwhich again is true for any real number, u.
Now lets take equation 7.1-6 and everywhere you see an a, replace it with π/2 - a and everywhere you see a b, replace it with -b
cos(π/2 - a - b) = cos(π/2 - a)cos(-b) - sin(π/2 - a)sin(-b)
eq. 7.1b-5a
Now simply apply the identities we have so far:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b) eq. 7.1b-5band we have a way of finding the sine of the sum of two angles. So now we have formulas for both sine and cosine for the sum of angles, but what about differences of angles? You can use the sum formulas 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. 7.1b-6a sin(a - b) = sin(a)cos(b) - cos(a)sin(b) eq. 7.1b-6bYou can also use the sum formulas to derive expressions for sine and cosine of double-angles. Simply observe that 2x = x + x, and substitute into the sum expressions.
cos(2x) = cos(x + x) = cos2(x) - sin2(x) eq. 7.1b-7a sin(2x) = sin(x + x) = 2sin(x)cos(x) eq. 7.1b-7bYou can do a sneaky trick on equation 7.1b-7a to get a half-angle by substituting x/2 and combining it with eq. 7.1-2.
cos(x) + 1 = cos(x/2 + x/2) + 1 = cos2(x/2) - sin2(x/2) + 1 =
cos2(x/2) - sin2(x/2) + sin2(x/2) + cos2(x/2) eq. 7.1b-8a
When you cancel and simplify, you get
cos(x) + 1 = 2cos2(x/2) eq. 7.1b-8b
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√cos(x)/2 + 1/2 = |cos(x/2)| eq. 7.1b-8c
Likewise, you can change signs and get
1 - cos(x) = 1 - cos(x/2 + x/2) = 1 - cos2(x/2) + sin2(x/2) =
sin2(x/2) + cos2(x/2) - cos2(x/2) + sin2(x/2) eq. 7.1b-9a
When you cancel and simplify, you get
1 - cos(x) = 2sin2(x/2) eq. 7.1b-9b
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√1/2 - cos(x)/2 = |sin(x/2)| eq. 7.1b-9c
Putting in x instead of x/2, you can use the above to
get formulas for sine squared and cosine squared:
sin2(x) = 1/2 - cos(2x)/2 eq. 7.1b-10a cos2(x) = 1/2 + cos(2x)/2 eq. 7.1b-10b(Observe what happens when you add the right-hand sides of those two equations)
Combining equations 7.1-6 and 7.1b-6a, you get
cos(a+b) + cos(a-b) = cos(a)cos(b) - sin(a)sin(b) +
cos(a)cos(b) + sin(a)sin(b) =
2cos(a)cos(b) eq. 7.1b-11a
If 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(a-b) = cos(a)cos(b) - sin(a)sin(b) +
-cos(a)cos(b) - sin(a)sin(b) =
-2sin(a)sin(b) eq. 7.1-11b
Again, 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 7.1b-5b and 7.1b-6b to get
sin(a+b) + sin(a-b) = sin(a)cos(b) + cos(a)sin(b) +
sin(a)cos(b) - cos(a)sin(b) =
2sin(a)cos(b) eq. 7.1b-12a
sin(a+b) - sin(a-b) = sin(a)cos(b) + cos(a)sin(b) +
-sin(a)cos(b) + cos(a)sin(b) =
2cos(a)sin(b) eq. 7.1b-12b
On any of the previous four identities you can substitute:
u = a + b and
v = a - b, then you also have
a = |
u + v |
and | b = |
u - v |
sin(u) - sin(v) = 2 cos |
u + v |
sin |
u - v |
eq. 7.1b-12c
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Here's a cute one for you.
_ _ _
sin(x) + cos(x) = √2 (√2/2)sin(x) + (√2/2)cos(x) ) =
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√2 ( cos(π/4)sin(x) + sin(π/4)cos(x) ) =
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√2 sin(x + π/4) eq. 7.1b-13
It's also equal to
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√2 cos(π/4 - x)
but I'm
sure you can prove that using the identities we have so far.
Tangent, Cotangent, Secant, and Cosecant
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. 7.1b-14a
cos(x)
cos(x)
cot(x) = eq. 7.1b-14b
sin(x)
1
sec(x) = eq. 7.1b-14c
cos(x)
1
csc(x) = eq. 7.1b-14d
sin(x)
Figure 7-4 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 (cos(x), sin(x) ). We have
- length of NK equal to tan(x).
- length of MJ equal to cot(x).
- length of OK equal to sec(x).
- length of OJ equal to csc(x).
Unlike the sine and cosine, these new functions are not continuous everywhere.
For example, according to
eq. 7.1b-14a, tan(x) has a
cos(x) in its denominator. So everywhere that cos(x) is
zero, tan(x) is discontinuous and undefined. This happens at
odd multiples of π/2. Look at the blue trace in figure 7-5 to
see what's going on here.
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Similarly, equation 7.1b-14b shows that cot(x) has a sin(x) in its denominator, so everywhere sin(x) is zero, cot(x) is undefined and discontinuous. This happens at multiples of π. Look at the green trace in figure 7-5 to see what's going on in this case. We shall be discussing more about continuity of trig functions in a later section. |
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The functions, sec(x) and csc(x), follow a similar pattern. Observe in figure 7-6 that sec(x) (the blue trace) is discontinuous at odd multiples of π/2, and csc(x) (the green trace) is discontinuous at multiples of π. 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 eq. 7.1b-14c that
1
sec(x) =
so it must also be true that
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But we all know from eq. 7.1-2 that the 1 in the numerator of the above is the same as sin2(x) + cos2(x). So by replacing the numerator with that, we get
sin2(x) + cos2(x)
sec2(x) = eq. 7.1b-15a
cos2(x)
With just a little algebra and a glance back at
eq. 7.1b-14a, you can see
that
sec2(x) = tan2(x) + 1 eq. 7.1b-15band using the identical approach you can show that
csc2(x) = cot2(x) + 1 eq. 7.1b-15c
Developing a formula for tan(a+b) is a piece of cake when you use the formulas we already have for sin(a+b) and cos(a+b).
sin(a+b)
tan(a+b) = eq. 7.1b-16a
cos(a+b)
sin(a)cos(b) + sin(b)cos(a)
tan(a+b) = eq. 7.1b-16b
cos(a)cos(b) - sin(a)sin(b)
If you divide numerator and denominator of eq. 7.1b-16b by
cos(a)cos(b), you get some cancellations.
sin(a) sin(b)
+
cos(a) cos(b)
tan(a+b) = eq. 7.1b-16c
sin(a)sin(b)
1 -
cos(a)cos(b)
Finally, applying
eq. 7.1b-14a, you get the identity
tan(a) + tan(b)
tan(a+b) = eq. 7.1b-16d
1 - tan(a)tan(b)
Using a very similar method you can come up one for the cotangent
(which I will let you derive for yourself):
cot(a)cot(b) - 1
cot(a+b) = eq. 7.1b-16e
cot(a) + cot(b)
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