Prove the following identies? Cos2X

[Pingu]

I cannot figure out this question

It says
Prove the following identities

Cos2X=


CotX-Sin2X
__________
CotX


I started at the left side and changed Cos2X to CossquaredX-SinsquaredX
I don't know what to do next.[/img]

 

[Unco]

G'day, Pingu.

Rule of thumb: start with the more complicated-looking side, and simplify down to the less complicated-looking side.

So start with the right-hand side, divide the numerator through by cot(x) (as it is the denominator) and see what you can come up with.

Remember that cot(x) = cos(x)/sin(x) and sin(2x) = 2sin(x)cos(x).

 

[Pingu]

What do you mean by "divide the numerator through by cot(x)"?
Do you mean to cancel out the cotX on the top and the bottom and be left with 1-sin2X?

 

[Gene]

(cot(x)-sin(2x))/cot(x) =
1-sin(2x)/cot(x)

 

[Pingu]

(cot(x)-sin(2x))/cot(x) =
1-sin(2x)/cot(x)

So does that mean that CotX=1?

 

[Unco]

\(\displaystyle \L \frac{\cot{x} \, - \, \sin{(2x)}}{\cot{x}} \, = \, \frac{\cot{x}}{\cot{x}} - \frac{\sin{(2x)}}{\cot{x}} \, = \, 1 \, - \, \frac{\sin{(2x)}}{\cot{x}}\)

 

[Pingu]

If I have (CotX-Sin2X)/CotX, can I simplify it to 1-Sin2X by cancelling the CotX on the top and bottom?

 

[Gene]

Nope, you have to divide both terms of the numerator by cot(x) as shown by Unco

 

[Gene]

PS. Don't forget his first post.
G.

 

[Pingu]

I changed the (1-Sin2X)/CotX to (1-2SinXCosX)/CotX
What do I do now?

 

[Gene]

Change cot(x) to cos(x)/sin(x) and keep simplifying.

 

[Pingu]

So now I have (1-2SinXCosX)/(CosX/SinX)
What do I do now?

 

[Unco]

That should be \(\displaystyle \L 1 \, - \, \frac{2\sin{x}\cos{x}}{\left(\frac{\cos{x}}{\sin{x}}\right)}\)

Simplify the right-hand term and we have
\(\displaystyle \L 1 \, - \, 2\sin^2{x}\)

 

[Pingu]

why is the "1-" outside of the dividing place?

How did it get there?

 

[Gene]

Look back at his 5:30 post. That's where it became separate.
Gene

 

[Pingu]

I don't understand. When I did it, the "1-" was on top of the division line. How do you move it to the left, off of the division line, without messing up the equation?

 

[Gene]

Its just like

a+b    a   b          b
--- =  - + -  =  1 +  -
 a     a   a          a

 

[Gene]

I didn't notice your mastake.

I changed the (1-Sin2X)/CotX to (1-2SinXCosX)/CotX
What do I do now?

should have been

I changed the 1-(Sin2X)/CotX to 1-(2SinXCosX)/CotX
What do I do now?

Look at any of the places Unco and I wrote it.

 

[mathfun]

R.H.S = cotx - sin2x

=(cosx/sinx) - 2sinxcosx

=cosx - 2sinsquaredxcosx

= 1 - 2sinsquaredx

= cos2x

=L.H.S.

 

[Gene]

I'm not sure what Mathisfun is doing. Stick with what Unco and I have been saying.