Question

prove the identity: tan(x)/(sin(x)-cos(x)) = (sin^2(x) + sin(x)cos(x))/(cos(x)-2cos^3(x))

Answer 1

i. hate. trig. identities.

Answer 2

that "2" in
\[\cos^2(x)-2\cos^3(x)\] is really messing things up
\[\frac{\frac{b}{a}}{b-a}=\frac{b^2+ab}{a-2a^3}\]
\[\frac{b}{a(b-a)}=\frac{b(b+a)}{a(1-2a^2)}\]
i can't figure out what to do with the 2!

Answer 3

multiply both sides by (sinx-cosx) to get

tanx = sinx/cosx = (sin^3x -cos^2xsinx)/(cosx-2cos^3x)

=sinx(Sin^2x-Cos^2x)/cosx(1-2cos^2x) and the bracketed terms are both equal to Cos2x

QED

Answer 4

Sorry, forgot the sign, both equal to - Cos2x

Answer 5

You happy with that?

Answer 6

I don't completely follow.

Answer 7

Which part?

Answer 8

Just a sec.

Answer 9

I shortened it a bit at the beginning because I didn't want have to type it all out.

Answer 10

I'm attaching a pdf with my attempt to use your solution. Can you point out where I've gone wrong?

Answer 11

I think I might know where I messed up. I'm trying again.

Answer 12

The first step, "multiply both sides by (sinx-cosx) to get"

this is just to get tan x on the left hand side by itself.

Then multiply out the top row.

Then factor sin out of the top and cos out of the bottom.

Answer 13

very nice.

Answer 14

I don't think the first step of your proof is legal. I'm supposed to take one side and transform it into the other side.

Answer 15

You're first step assumes that they're equal from the start. Then there's no reason to prove the identity.

Answer 16

My first step hasn't changed anything at all, it's the same thing you wrote rearranged.

Besides that, it is perfectly OK in a proof to make an assumption and then see if you get a contradiction.

If the two sides had not been equal, then I would not have been able to show equality as I did.

Answer 17

I think maybe you are assuming that that is what you have to do.

Probably there is a way, you can try now that you have seen the possibilities.

If all you were asked to do is prove equality, then my method is 100% satisfactory.

Answer 18

All I know is that if a properly trained teacher marked it wrong, he should be replaced.

Answer 19

How does 1-2cos^2x = -cos 2x? Haven't learned that identity yet.

Answer 20

Are you sure?

Cos 2x = cos^x - sin^x (sum formula).

Then use cos^2x + sin^2x =1 (Pythagorean identity)

Answer 21

Thanks. Sum and difference formulas are in the next chapter. There should be a way to do this proof without the -cos 2x cancellation. I'm still working on it.

Answer 22

OK, I did it your way (what a pain).

Start off the same way, get rid of sin/cos (for both of them) and you are left with

1/(sinx-cosx) = (sinx + cosx)/(1-2cos^2x) (I'm just using = for convenience)

Now multiply the top and bottom of the lhs by (sinx + cosx) to get rid of that (for both) leaving to show

1/((sin x-cosx)(sinx+cosx)) = 1/(1-2cos^2x) and I will leave it to you to show that both denominators are the same.

(no double angles, basic operations only, no intermingling, hope I haven't broken any more of your teacher's rules!)

Answer 23

Thanks for the help. I discovered a solution last night by factoring by grouping, working with the right hand side. It's attached. I'm still using your trick of multiplying top and bottom by sin x - cos x.