Question 635089: Hello, I'm a little stuck...
The question is:
prove that (sin4x(1-cos2x))/(cos2x(1-cos4x)) = tanx , for 0
Thank you for your help!
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 
The keys to this are:- Knowing the Trig identities/properties. (For this problem we will use using sin(2x), cos(2x) and tan((1/2)x)); and
- Knowing how to use the identities/properties.
Knowing the identities is primarily a matter of memorization. Knowing how to use them comes when you realize that the x's, A's and B's in the formulas are just place holders. You can replace them with anything. When you look at:
sin(2x) = 2sin(x)cos(x)
you can't just see "This is how I can change sin(2x) into an expression with arguments of x" (and vice versa). You need to learn that this formula shows you how can change the sin of anything into an expression with arguments of half that much (and vice versa). Some examples:
sin(4x) = 2sin(2x)cos(2x)
sin(6x) = 2sin(3x)cos(3x)
sin(500q) = 2sin(250q)(cos(250q)
sin(x) = 2sin((1/2)x)cos((1/2)x)
sin(3p+10q) = 2sin(1.5p+5q)cos(1.5p+5q)
etc.
Our goal is to use identities and algebra to transform the left side until it matches the right side. So we will, at some point, have to change the arguments of 4x and 2x to arguments of x (like the argument on the right side). We will start by using the first example above to replace sin(4x):
The factors of cos(2x) cancel:
leaving:
One 4x is gone. Now we'll get rid of the other one, using cos(2x). Fortunately (or unfortunately, depending on your point of view) there are three variations of this formula:They all will work. Any of them can be used. But choosing a particular one will often make a problem easier. In this case, looking at where the cos(4x) is, 1-cos(4x), I will choose the third variation above. This is because I can see that the 1 in the formula will cancel with the 1 in 1-cos(4x) (because of the subtraction involved). This will make the denominator a simpler expression. Using replacing the cos(4x) with the expression in the third variation of cos(2x) we get:
The 1's and the minuses cancel in the denoinator:
And the factors of 2 and sin(2x) in the numerator cancels with the factors of 2 and one of the two factors of sin(2x) in the denominator:
Not only are the 4x's gone, but the expression is much simpler. And, believe it or not, we can actually finish this with one more (power-user) step!
Some explanation first: We have been using the double-angle formulas to divide an argument in half. (And if you don't understand what follows you can finish the problem by using the double angle formulas to change the 2x's to x's.) But as it turns out there are other formulas that can be used to divide an argument in half: The half-angle formulas.
You may recall that when I was describing how to use the formulas I kept adding "and vice versa". This meant that not only could you use the sin(2x) formula to change sin(40x) into 2sin(20x)cos(20x), you could use it in reverse. You can use it to change 2sin(14x)cos(14x) into sin(28x). As long as we fit the pattern of 2sin(something)cos(something) [The arguments must be the same!], we can use the sin(2x) formula to turn this into sin(2*something).
Using the double-angle formulas in one direction divides the argument in half. Using it in the other direction doubles the argument.
Now let's look at one of the half-angle formulas:
We can see that the argument on the left is half of the argument on the right. Going from left to right this formula will double an argument. For example:
Going in the other direction this formula will divide an argument in half.
So there are two ways to divide an argument in half:- Using the double-angle formulas from left to right; or
- Using the half-angle formulas from right to left.
So to change the arguments of the left side of our equation:
from 2x to x we can either use cos(2x) and sin(2x) or we can use a half-angle formula if we can match the right side of one of them.
There is a reason I've been using
as an example. Doesn't the left side of our equation match the pattern of right side of the formula? Answer: It does! So by virtue of this formula the left side of our equation must be equal to the tan of half the argument. So by using the tan((1/2)x) formula we can go from
to
in one step!.
As I said before, if this last step is too much for you to understand, then just us cos(2x) and sin(2x) as we did earlier on:
You'll eventually get to tan(x).
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