Trigonometric Functions

[mathgeek]

Hi guys, I was wondering if you could figure this problem out

(1+cotx)/(1-cotx)=(1+tanx)\tanx-1)

Its a proof :evil: . Please show me the steps because i have no clue how to do it.

 

[mmm4444bot]

You're supposed to apply trigonometric identities to either side (or both), until you get the same expression on both sides.

I would start by rewriting everything in terms of sine and cosine.

Have you worked with the basic trigonometric identities, yet?

 

[mathgeek]

I have worked with basic trig identities, but i don't really know what to do with this problem. Here is the work i have.

(1+cotx)/(1-cotx)=(1+tanx)\tanx-1)

(1+(cosx/sinx))/(1-cosx/sinx)=

After that I am not really sure what to do.

 

[mmm4444bot]

The next step is to express tangent in terms of sine and cosine, also.

 

[mathgeek]

Aren't you only supposed to work with one side of the equation. That's what my Precalc with Trig teacher said.

 

[mmm4444bot]

Not as far as I'm concerned.

As long as you end up with identical expressions on both sides, and all of the steps leading to those two sides are valid, it doesn't matter.

If you've received explicit instructions to the contrary, then let me know.

Can you simplify what you've typed for the lefthand side?

In other words, can you get a common denominator and add 1 + cos(x)/sin(x) ?

Can you do the same with 1 - cos(x)/sin(x) ?

After you've combined the ratios on top and bottom, do you know how to simplify the compound ratio?

If you do this on both sides, you'll end up with the following (unless I made a goof).

$\displaystyle \frac{sin(x) + cos(x)}{sin(x) - cos(x)} \;=\; \frac{sin(x) + cos(x)}{sin(x) - cos(x)}$

 

[mathgeek]

I got to the point where (sin(x)+cos(x))/(sin(x)-cos(x))=

Don't you have to take it all the way till it equals (1+tan(x))/(tan(x)-1)

After (sin(x)+cos(x))/(sin(x)-cos(x)),

(sin(x)+cos(x))=1

so,

1/(sin(x)-cos(x))=(1+tan(x))/(tan(x)-1)

At least that's what i'm getting from it. After that i'm not sure

 

[mmm4444bot]

I got to the point where (sin(x)+cos(x))/(sin(x)-cos(x)) = GOOD!

Work the other side the same way, and you'll end up with the same result.

That proves that the given equation is an identity.

 

[mmm4444bot]

If you feel that you must not touch the righthand side, then here's the last.

You got the lefthand side to the following.

(sin + cos)/(sin - cos)

Multiply that by 1, in the form of (1/cos)/(1/cos). :|

 

[Mrspi]

I got to the point where (sin(x)+cos(x))/(sin(x)-cos(x))=

Don't you have to take it all the way till it equals (1+tan(x))/(tan(x)-1)

After (sin(x)+cos(x))/(sin(x)-cos(x)),

(sin(x)+cos(x))=1<-----you have an error here. sin(x) + cos(x) is NOT equal to 1. The fundamental identities say that
sin[sup:9yxr53k7]2[/sup:9yxr53k7](x) + cos[sup:9yxr53k7]2[/sup:9yxr53k7](x) = 1 There IS a difference!

so,

1/(sin(x)-cos(x))=(1+tan(x))/(tan(x)-1)

At least that's what i'm getting from it. After that i'm not sure

 

[stapel]

Aren't you only supposed to work with one side of the equation. That's what my Precalc with Trig teacher said.

Yes, but that's only for the final result.

If you need to (and you may, for this identity), work on both sides until you've gotten the two sides to match. Then, for what you hand in, start with one side, work your way down to where they match, switch over to the other side, and work your way back up to the starting point for that side.

This is how the authors came up with their "magical" proofs, where they pulled something from behind their left ear to (apparently) effortlessly prove an identity. They did the same thing you're doing; they just don't show their work. :wink:

 

[mathgeek]

Many thanks to Mrspi for finding that problem. Today I asked my teacher about it and she confirmed that. Thanks again