Prove the following identies? Cos2X

Pingu

Junior Member
Joined
Dec 22, 2005
Messages
56
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]
 
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).
 
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?
 
\(\displaystyle \L \frac{\cot{x} \, - \, \sin{(2x)}}{\cot{x}} \, = \, \frac{\cot{x}}{\cot{x}} - \frac{\sin{(2x)}}{\cot{x}} \, = \, 1 \, - \, \frac{\sin{(2x)}}{\cot{x}}\)
 
If I have (CotX-Sin2X)/CotX, can I simplify it to 1-Sin2X by cancelling the CotX on the top and bottom?
 
Nope, you have to divide both terms of the numerator by cot(x) as shown by Unco
 
I changed the (1-Sin2X)/CotX to (1-2SinXCosX)/CotX
What do I do now?
 
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
\(\displaystyle \L 1 \, - \, 2\sin^2{x}\)
 
why is the "1-" outside of the dividing place?

How did it get there?
 
Look back at his 5:30 post. That's where it became separate.
Gene
 
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?
 
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.
 
R.H.S = cotx - sin2x

=(cosx/sinx) - 2sinxcosx

=cosx - 2sinsquaredxcosx

= 1 - 2sinsquaredx

= cos2x

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