If we start off with a right angle triangle with a hypotenuse of length 1, we can use the Pythagorean theorem to get
x2 + y2 = 1
But, since
cos a = x/1 = x
sin a = y/1 = y
we can substitute to get
cos2a + sin2a = 1
The other things to remember are that tan = sin/cos, cot = cos/sin, sec = 1/cos and csc = 1/sin.
On top of that, it helps to know
cos(2a) = cos2a - sin2a
sin(2a) = 2(cos a) (sin a)
stemming from the facts that
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
Knowing these basic facts will help you solve any trig identity problems you might encounter. For example, if we take the basic squared equation above and divide both sides by cos2a we get
1 + tan2a = sec2a
Similarly, if we divide the basic squared equation by sin2a we get
cot2a + 1 = csc2a
In solving trig identity problems, I usually find that converting everything to sine and cosine keeps my life simpler. It also helps to know how to multiply and divide fractions.
Example 1: Prove sin2a + cos2a = (sina)(csca)
On the left hand side, we know that it equals 1. All we have to do is show that the right hand side is 1 as well. We know that csc = 1/sin. So,
(sina)(csca)
= (sina) x 1/(sina)
= (sina)/(sina) = 1
Sometimes, a little bit of intuition comes in handy, especially when you're missing something on one side of the equation that you have on the other. This next example is a case in point.
Example 2: Prove 2cos2a = cos(2a) + 1
There's not much we can do with the left hand side. So, let's expand the right hand side.
cos(2a) + 1
= cos2a - sin2a + 1
Isn't that wonderful? We've got a sin2 on the right hand side but not on the left. And what do we do with that 1 on the right hand side? This is where knowing your basic trig identities comes to the rescue. We can use the fact that sin2 + cos2 = 1.
cos2a - sin2a + 1
= cos2a - sin2a + sin2a + cos2a
= cos2a + cos2a
= 2cos2a
which is the same as the left hand side. Notice that in this problem, we didn't touch the left hand side at all.
This last example illustrates the point that converting everything to sin and cos and knowing how to multiply and divide fractions can make your life easier.
Example 3: Prove (csc a)/ (cot2a) = (tan a)(sec a).
They don't get much uglier than this do they? Well, lets start with the left hand side.
(csc a)/ (cot2a)
= (1/sin a)/ [(cos2a)/ (sin2a)]
= = (1/sin a) x [(sin2a)/ (cos2a)]
= (sin2a)/ [(sin a) (cos2a)]
= (sin a)/ (cos2a)
which is as far as we can go with the left hand side. Now the right hand side.
(tan a)(sec a)
= (sin a)/(cos a) x 1/(cos a)
= (sin a)/ (cos2a)
and the two sides are the same.