# Prove that the Given Equation is an (Trig) Identity

1. Jan 31, 2008

### TbbZz

1. The problem statement, all variables and given/known data
Prove that the Given Equation is an Identity:

Code (Text):
sin2A
------       =       cotA
1 - cos2A

2. Relevant equations
sin(A+B) = sinAcosB + cosAsinB
cos(A+B) = cosAcosB - sinAsinB
tan(A+B) = (tanA + tanB) / (1 - tanAtanB)

sin2A = 2sinAcosA
cos2A = cos$$^{}2$$A - sin$$^{}2$$A
tan2A = 2tanA / 1 - tan$$^{}2$$A

3. The attempt at a solution

I tried changing sin2A to sin(A+A) and arrived at 2sinAcosA at the top.

I also tried changing 1 - cos2A to 1 - cos$$^{}2$$A - sin$$^{}2$$A, but then I arrived at having a 0 in the denominator.

I'm really not sure where to start in trying to prove the identity. I understand that I should not touch the cotA on the right hand side, but no matter what I do to rewrite the left side I can't seem to arrive at the cotA.

I would appreciate it if someone could point me in the right direction. Thank you in advance for the assistance.

2. Jan 31, 2008

### Rudipoo

Okay so you know the identity cos2A=(cosA)^2-(sinA)^2, as you've written above.
Remember (sinA)^2+(cosA)^2=1 ?

Try substituting (cosA)^2 from the second equation into the first, then rearrange to find the denominator. Use sin2A=2sinAcosA for the numerator, and you should be okay! Hope this helps; i'll give more help if needed

3. Jan 31, 2008

### TbbZz

Thanks for the help Rudipoo, your assistance is appreciated. I successfully solved this problem and another one.

However, I am having difficulty on the problem after that. It is similar, but I can't seem to figure it out. I'm attaching a picture of my work so far.

I'm not sure what to do next or whether I'm heading in the right direction in the first place.
Again, thanks in advance for the help.

4. Feb 1, 2008

### HallsofIvy

Staff Emeritus
Do you know that $$tan(x/2)= \sqrt{\frac{1- cos(x)}{1+ cos(x)}}$$

(If not, you can prove it by writing tan(x/2)= sin(x/2)/cos(x/2) and using the half angle formulas for sine and cosine.)

Once you have $$\sqrt{\frac{1- cos(x)}{1+ cos(x)}}$$
multiply both numerator and denominator of that fraction by 1- cos(x).

5. Feb 1, 2008

### Rudipoo

Remember, you're trying to PROVE that tan(x/2) = sinx/(1+cosx), so don't start the proof by writing this!

I would suggest by writing down the right hand side (i.e. sinx/(1+cosx)) because this looks intuitively like it can be simplified, and then manipulate this to find it equal to tan(x/2)

My method was to use half angle formulae on sinx and cosx - these are same as double angle formula, but swap x for x/2. For sinx, you have only one option as before, and use this in the numerator. For cosx, because you're looking for a tan function, use an identity such that it includes just cosine terms, so you end up with a sine over a cosine.

Hope this helps, and if you need more clues, give me a message.

6. Feb 1, 2008

### VietDao29

In fact, I don't really think that this is an identity. Since the LHS can take negative value (say, when x = -2), whereas the RHS is always non-negative.

It should be an identity once you square both sides. :)

7. Feb 1, 2008

### TbbZz

Thanks for the help HallsofIvy, Rudipoo, and VietDao29. I was able to solve the problem correctly using your advice.