[MOVED] Trig Identity: [sin(x)cos(x)] / [1 - cos(x)] = csc(x

[anisapatel]

I have been stuck on this problem forever. I can't seem to get it started the correct way. Could someone give me a nudge on how to get started? (I am not asking for the answer; just a push in the right direction.)

. . .Prove the identity:

. . .[sin(x)cos(x)] / [1 - cos(x)] = csc(x) - sin(x) + cot(x)

Thanks in advance!

 

[stapel]

This is one where you sort of have to "cheat".

On the left-hand side, multiply top and bottom by the conjugate, 1 + cos(x). Convert the resulting denominator, using the Pythagorean Identity, into sin²(x); cancel off the duplicate factor.

On the right-hand side, convert everything to sines and cosines, and get a common denominator. Convert the 1 - sin²(x) in the numerator to cos²(x).

Compare. You should have (or be able to see how to get) the same thing on both sides. This is NOT the hand-in solution.

To get the hand-in solution, pick one of the sides to start on. Copy your work on that side, down to the line where both sides are the same expression. Then work up the other side, copying your steps backwards, until you get at the initial expression on the other side of the identity.

Eliz.

 

[soroban]

Re: [MOVED] Trig Identity: [sin(x)cos(x)] / [1 - cos(x)] = c

Hello, anisapatel!

Eliz. had the best approach ... and no "cheating" is necessary.

Prove the identity: \(\displaystyle \L\:\frac{\sin x\cos x}{1\,-\,\cos x}\:=\:\csc x\,-\,\sin x\,+\,\cot x\)

On the left side, multiply top and bottom by \(\displaystyle (1\,+\,\cos x):\)

. . \(\displaystyle \L\frac{\sin x\cos x}{1\,-\,\cos x}\,\cdot\,\frac{1\,+\,\cos x}{1\,+\,\cos x}\:=\:\frac{\sin x\cos x(1\,+\,\cos x)}{1\,-\,\cos^2x}\:=\:\frac{\sin x\cos x(1\,+\,\cos x)}{\sin^2x}\)

. . \(\displaystyle \L=\:\frac{\cos x(1\,+\,\cos x)}{\sin x} \:=\:\frac{\cos x\,+\,\cos^2x}{\sin x}\:=\:\frac{\cos x\,+\,1\,-\,\sin^2x}{\sin x}\)

. . \(\displaystyle \L=\:\frac{\cos x}{\sin x}\,+\,\frac{1}{\sin x}\,-\,\frac{\sin^2x}{\sin x} \:=\:\cot x\,+\,\csc x\,-\,\sin x\)