Question

Trig Identity Prove:

Answer 1

\[\frac{ cotx+coty }{ 1-cotxcoty }=\frac{ cosxsiny+sinxcosy }{ sinxsiny-cosxcosy }\]

Answer 2

What have you tried here so far? :)

Answer 3

When I look at the identity, I see some similarities between the two and some differences.
One side has only cotangent (a quotient of cosine and sine), and the other has sines and cosines. The structure of both (two binomials divided) suggests that they are closely related already; perhaps just by converting cotangent to sines and cosines the secret can be revealed.

Answer 4

This is what I tried...

Answer 5

\[sinxsiny(\frac{ cosx }{ sinx }+\frac{ cosy }{ siny }) / sinxsiny(1-\frac{ cosx }{ sinx }*\frac{ cosy }{ siny })\]

Answer 6

\[\frac{ cosxsiny+cosysinx }{ sinxsiny-cosxsinycosysinx }\]

Answer 7

Isn't this as simplified as it gets though? I mean I see I could factor out sinx and siny... but that would not get me what I want.

Answer 8

@AccessDenied now at this part... I am stuck because the only thing I see is factor... yet factoring won't get me what I want.

Answer 9

for your problem (ORIGINAL PROBLEM)

the top of the fraction on the right is cos(x-y)
and the bottom of that same fraction is sin(x-y)

Answer 10

http://www.ies-math.com/math/java/trig/kahote/kahote.html

Answer 11

So this is dealing with differences?

Answer 12

well, that was just my first thought.

you would get tan(x-y) for the right side. I don;t think that would get you anywhere though :(

Got to go, sorry and bye.

Answer 13

I understand this step
\[ sinxsiny(\frac{ cosx }{ sinx }+\frac{ cosy }{ siny }) / sinxsiny(1-\frac{ cosx }{ sinx }*\frac{ cosy }{ siny }) \]
but I get
\[ \frac{ cosx siny+cosy sinx }{ sinxsiny-cosxcosy } \]

Answer 14

which matches your right hand side

you can simplify that to sum/differences of angles, and then -tan(x+y)
but that is not necessary.

Answer 15

So what did I do wrong?

Answer 16

You may want to retry the step where you multiplied top and bottom by sin x sin y.
It seems you did something like missing the sin x sin y in the denominator of cot x cot y = cos x cos y / (sin x sin y) when multiplying. :)

Answer 17

Or perhaps treated the cot x cot y altogether as cosine rather than cotangent, unless there was some typo here.

Answer 18

When you multiply the denominators you should get...
(sinxsiny) on the left side which I completely understand... but on the right how in the world do you end up with two cosines?

Answer 19

\( \displaystyle \frac{ \cot x + \cot y }{1 - \cot x \cot y} \)

\( \displaystyle \frac{\sin x \sin y \left( \cot x + \cot y \right) } { \sin x \sin y \left( 1 - \cot x \cot y \right) } \)

Answer 20

\( \displaystyle \cot x \cot y = \frac{\cos x}{\sin x} \frac{\cos y}{\sin y} \)
You agree with this?

Answer 21

Yes

Answer 22

By distributing through, you would get this:
\( \displaystyle \frac{\sin x \sin y \cot x + \sin x \sin y \cot y}{\sin x \sin y - \sin x \sin y \cot x \cot y } \)

which if we switch out cotangents for cosine/sine...
\( \displaystyle \frac{\sin x \sin y \frac{\cos x}{\sin x} + \sin x \sin y \frac{\cos y}{\sin y}} { \sin x \sin y - \sin x \sin y \frac{\cos x \cos y}{\sin x \sin y}} \)

Answer 23

Earlier I said... "convert cotangents to cosine and sines" or something to that effect. Stuff like this usually happens as a result, we can see things that cancel out like the sin x sin y * (cos x cos y)/(sin x sin y) there. :)

Answer 24

So why do I get this?

Answer 25

You multiply numerator and denominator by sin x sin y to obtain that expression. The end result is cancellations:
\( \displaystyle \frac{\cancel{\sin x} \sin y \frac{\cos x}{\cancel{\sin x}} + \sin x \cancel{\sin y} \frac{\cos y}{\cancel{\sin y}}} { \sin x \sin y - \cancel{\sin x \sin y} \frac{\cos x \cos y}{\cancel{ \sin x \sin y }}} \)
Or did I understand your question correctly?

Answer 26

Right, so they in the numerator you should have...

sinycosx + sinxcosy right?

Answer 27

Yes.

Answer 28

I don't understand the denominator...

Answer 29

Shouldn't you get...

Answer 30

Originally the denominator is \[sinxsiny (1-\frac{ cosx }{ sinx }~*\frac{ cosy }{ siny })\]

Answer 31

RIght?

Answer 32

That is correct.

Answer 33

So now when we distribute we should get...

Answer 34

\[sinxsiny-cosxsiny~*cosysinx\]

Answer 35

Right?

Answer 36

That would not be correct...
If you had...
\( \displaystyle A \left( 1 - \frac{B}{A} \right) \)
How would you distribute here?

Answer 37

A * 1 = A

Then for the fraction... wouldn't the A's reduce...

A - B

Answer 38

Yes. That is the same framework as the first one we did, as A = sin x sin y and B = cos x cos y.
\( \displaystyle \color{red}A \left( 1 - \frac{\color{green}B}{\color{red}A} \right) \)
\( \displaystyle \color{red}{\sin x \sin y} \left( 1 - \frac{ \color{green}{\cos x \cos y}}{\color{red}{\sin x \sin y}} \right) \)
The first one, you say, is A - B.
The comparison would be... \( \sin x \sin y - \cos x \cos y \), right?

Answer 39

oh... I think I know why I am messing up... you combined both and I have them separated as two separate fractions...

Answer 40

I see it now!

Answer 41

Ah! Excellent! :D

Answer 42

Thank You guys @AccessDenied and @phi ... I died and came back to life.

Answer 43

You're welcome! And I should say thanks to @phi as well for coming by while I was in limbo outside the website. lol :P