Question

prove: 2cosx+2cos(square)x over sin2x = sinx over 1-cosx ???

Answer 1

can you clarify it more....

Answer 2

"confused"

Answer 3

\[\frac{2\cos x+2\cos^2x}{\sin(2x)}=\frac{\sin x}{1-\cos^2x}\]this or no?

Answer 4

YES!!!! :) thanks.

Answer 5

no wait, you have\[\frac{2\cos x+2\cos^2x}{\sin(2x)}=\frac{\sin x}{1-\cos x}\] which is it?
you need to be sure, or else we can't help.

Answer 6

(I took the square off the cosine on the right)

Answer 7

Oo! Second one. :)

Answer 8

oh I got it...

Answer 9

\[\frac{2\cos x (1 + \cos x)}{\sin (2x)} = \frac{\sin x}{1-\cos x}\]

Answer 10

divide 1+ cos x on both sides

Answer 11

1 - cos^2 x = sin^2 x

Answer 12

then you get sin (2x) = 2 cos x sin x

Answer 13

ohh yay thank you guys!!!! :)

Answer 14

I never know how to start trig identity questions.. is there any stragedy to solve it? :P

Answer 15

always convert tangent in terms of cosine and sine

Answer 16

things like that (2x) have to be on both sides or neither.
Definitely need to take care of that in my opinion.
That said I did it slightly differently than moneybird, so there are different ways. I can post it but it would take a sec..

Answer 17

I mean the (2x) in the sine in the denominator...

Answer 18

I would appreciate your solution! :)

Answer 19

I think I cut moneybird off, he wasn't done yet...
sorry moneybird

Answer 20

(1 - cos x) (1 + cos x) = 1 - cos^2 x

Answer 21

If this is an identity that is supposed to be proven, you cannot treat it like an equation where you do the same thing to both sides, moneybird.

Answer 22

moneybird, I really appreciate your help! I am still confused what has happened.. so [2cos ^2x becomes (1+cosx)]?

Answer 23

\[\frac{2\cos x+2\cos^2x}{\sin(2x)}=\frac{\sin x}{1-\cos x}\]like I said take care of the sin(2x) with the formula\[\sin(2x)=2\sin x\cos x\]so we get\[\frac{2\cos x+2\cos^2x}{2\sin x\cos x}=\frac{\sin x}{1-\cos x}\]cancel out a 2 and a cosine on the left\[\frac{1+\cos x}{\sin x}=\frac{\sin x}{1-\cos x}\]Now we want to see if we can get something familiar out of this, like an identity we know. It may take practice just to see the next moves coming:
multiply by the denominator of each side\[(1-\cos x)(1+\cos x)=\sin^2x\]\[1-\cos^2x=\sin^2x\]\[1=\sin^2x+\cos^2x\]which is always true.

Answer 24

cancel out a 2 and a cosine on the left?