
This page contains sample problems  identities high school students have been asked to prove.
Suppose you are asked to prove
(sin(x)cos(x)+1) / (sin(x)+cos(x)1) = (sin(x)+1) / cos(x)
The quickest way to convince yourself it's true is to crossmultiply:
(sin  cos + 1) cos = (sin + cos  1) (sin + 1)
sin cos  cos^{2} + cos = sin^{2} + sin cos  sin + sin + cos  1
sin cos  cos^{2} + cos = sin cos + sin^{2}  1 + cos
sin cos  cos^{2} + cos = sin cos  cos^{2} + cos
But does that prove the identity? No. What we've done here is assumed the identity, and then we used it to derive another more obvious identity. So it's not a proof at all. Although the method of crossmultiplying is very convincing, and usually disproves a false "identity", it does not prove the identity here, because we assumed the thing to be proved.
So many high school math teachers will simply not allow you to cross multiply as a method of solving this kind of problem. Instead they insist that you start from the left (or right) side, and make algebraic and trigonometric manipulations until you arrive at the expression on the right (or left) side. That's not always very easy, because it's not at all obvious what manipulations are needed.
There is a way to rescue the "proof" above, and turn it into an actual proof. It works by taking every step in reverse order:
sin cos  cos^{2} + cos = sin cos  cos^{2} + cos (true because left = right)
sin cos  cos^{2} + cos = sin cos + sin^{2}  1 + cos (true because cos^{2} = sin^{2}  1)
sin cos  cos^{2} + cos = sin^{2} + sin cos  sin + sin + cos  1 (true because sinsin=0, and I just moved things around)
(sin  cos + 1) cos = (sin + cos  1) (sin + 1) (true because I factored out cos on the left and sin on the right)
(sin  cos + 1) / (sin + cos  1) = (sin + 1) / cos (true by dividing both sides by cos and by (sin + cos  1) )
This is a valid proof. However, your teacher may still disapprove. Maybe you're pulling a fast one! Here's a plan "B" for you to try.
The identity given here, like so many you will encounter, is of the form
A / B = C / D
We can obey the teacher's rule by starting with A/B, and multiplying both the numerator and denominator by CD, giving us
A / B = (ACD) / (BCD)
Now, by crossmultiplying, we have shown that AD = BC, so we can substitute BC in place of AD in the numerator, giving us
= (BC^{2}) / (BCD)
And then we can cancel common factors, giving
= C / D
That is, if we form a linear combination of the two numerators, and divide that by the same linear combination of the two denominators, you will get yet another fraction that is equal to A/B and also to C/D.
Consider again, the problem
(sin(x)cos(x)+1) / (sin(x)+cos(x)1) = (sin(x)+1) / cos(x)
If we let A/B be cos(x) / (1sin(x)) and C/D be (1+sin(x)) / cos(x)
You can check that A/B = C/D by crossmultiplying, or by whatever "approved" method your teacher provides! (See "geometric progression identities" below for a whole bunch of similar cases)
So now, our problem boils down to
(CA) / (DB), which is equal by this "linear combination" method to C/D, proving our identity.
First, since A/B = C/D, we multiply both sides by B/C to derive
A/C = B/D
Now, we let k=A/C, so Ck=A and Dk=B, which gives us
(rA+sC)/(rB+sD) = ((rk+s)C) / ((rk+s)D) = C/D
We briefly saw one of these geometric progression identities, above, in the section "Linear combination of fractions". It was:
(1+sin(x)) / cos(x) = cos(x) / (1sin(x))
I call this a "geometric progression identity" because the quantities 1+sin(x), cos(x), and 1sin(x) are in geometric progression. You can check that by multiplying the first by the last, and you'll get the square of the middle. (1+sin(x))(1sin(x)) = 1sin^{2}(x) = cos^{2}(x).
Others in this category include
(1cos(x)) / sin(x) = sin(x) / (1+cos(x)), and this in fact is equal to tan(x/2)
Proof: tan(x/2) = 2sin^{2}(x/2) / (2 sin(x/2) cos(x/2)) = (1cos^{2}(x/2)+sin^{2}(x/2))/sin(x) = (1cos(x))/sin(x)
(sec(x)1) / tan(x) = tan(x) / (sec(x)+1))
(sec(x)tan(x)) = 1 / (sec(x)+tan(x))
(csc(x)1) / cot(x) = cot(x) / (csc(x)+1))
(csc(x)cot(x)) = 1/ (csc(x)+cot(x))
Note, too, that these identities can be easily disguised by moving cot(x) from the denominator making it tan(x) in the numerator, etc.
Tan and cot can be expanded to sin/cos and cos/sin, respectively, which will help you prove some identities. For example,
tan(x)+cot(x) = 1/(sin(x) cos(x)) = 2/sin(2x)
I'll leave off the x's to make it easier to follow:
sin/cos + cos/sin = sin²/(sin cos) + cos²/(sin cos) = 1/(sin cos) = 1/(1/2 sin(2x) = 2/sin(2x)
example 2: tan(x)cot(x) = 2 cot(2x)
sin/cos  cos/sin = sin²/(sin cos)  cos²/(sin cos) = (12 cos²)/(sin cos) = cos(2x)/(1/2 sin(2x)) = 2 cot(2x)
The webmaster and author of this Math Help site is Graeme McRae.