How to prove this trigonometry equation?

[Helly123]

• Homework Statement

I opened my old math book and interested in this

Proof Cos (A-B) - tan(A-D) = (2cos A)/(sin 2A + sin 2B)

Homework Equations

Cos(A-B) = cosA cosB + sinA sinB
Tan(A-D) = (Tan A - tan B)/(1+Tan A Tan B)

The Attempt at a Solution

cosA cosB + sinA sinB - (tan A - tan B)/(1+Tan A Tan B)
Become
CosA cos B + sin A sin B - $$\frac {sin A cos D - sin D cos A} {cos A cos D + sin A sin D} $$

How to simplify it and proof it? It gets more complicated
Is there anyone interested too?

 

[Ray Vickson]

• Homework Statement

I opened my old math book and interested in this

Proof Cos (A-B) - tan(A-D) = (2cos A)/(sin 2A + sin 2B)

Homework Equations

Cos(A-B) = cosA cosB + sinA sinB
Tan(A-D) = (Tan A - tan B)/(1+Tan A Tan B)

The Attempt at a Solution

cosA cosB + sinA sinB - (tan A - tan B)/(1+Tan A Tan B)
Become
CosA cos B + sin A sin B - $$\frac {sin A cos D - sin D cos A} {cos A cos D + sin A sin D} $$

How to simplify it and proof it? It gets more complicated
Is there anyone interested too?

You can give a proof of something, or you can prove something, but you cannot proof something.

Anyway, I cannot see how the result could possibly be true, because it involves A,B,D on one side and only A,B on the other side. Perhaps if the "D" was really a "B" the result could be true. It looks nasty in that case.

Added note: I think the result is false. Let us write $F(A,B) =\cos(A-B) - \tan(A-B) - 2 \cos(A)/ (\sin 2A + \sin 2B)$ in fully expanded form as $F(A,B) = N(A,B)/D(A,B)$, where $D(A,B) = \cos^4 A -\cos^4 B -\cos^2 A + \cos^2 B$ and $N(A,B)$ is some nasty 5th degree polynomial in $\cos A, \cos B, \sin A, \sin B$. We can do a multi-dimensional plot of $N(A,B)$ to see if it is zero. When we do that we find it is definitely non-zero. See the plot below, where the axes are A and B, going from $-\pi$ to $+\pi$. (All computations and plots were done in the computer algebra package Maple.)

 

[Helly123]

You can give a proof of something, or you can prove something, but you cannot proof something.

Anyway, I cannot see how the result could possibly be true, because it involves A,B,D on one side and only A,B on the other side. Perhaps if the "D" was really a "B" the result could be true. It looks nasty in that case.

Added note: I think the result is false. Let us write $F(A,B) =\cos(A-B) - \tan(A-B) - 2 \cos(A)/ (\sin 2A + \sin 2B)$ in fully expanded form as $F(A,B) = N(A,B)/D(A,B)$, where $D(A,B) = \cos^4 A -\cos^4 B -\cos^2 A + \cos^2 B$ and $N(A,B)$ is some nasty 5th degree polynomial in $\cos A, \cos B, \sin A, \sin B$. We can do a multi-dimensional plot of $N(A,B)$ to see if it is zero. When we do that we find it is definitely non-zero. See the plot below, where the axes are A and B, going from $-\pi$ to $+\pi$. (All computations and plots were done in the computer algebra package Maple.)

View attachment 211525

i realized too that why it's D.. maybe the question wasn't right

 

[Mark44]

You can give a proof of something, or you can prove something, but you cannot proof something.

Fixed the thread title. However, you can proof something, but the correct verb form in the context of this thread is prove. See the definitions of the verb forms of proof here: http://www.dictionary.com/browse/proof.

 

[Helly123]

Fixed the thread title. However, you can proof something, but the correct verb form in the context of this thread is prove. See the definitions of the verb forms of proof here: http://www.dictionary.com/browse/proof.

Idk how to edit it

 

[qspeechc]

I'm fairly sure the D should be B. Generally you simplify the two sides separately, so

L.H.S. (left hand side) = ... (simplify)
R.H.S. = ...

And you end up showing the two sides simplify to the same thing.

General hints are get everything in terms of functions of A and B, rather than 2A, (A-B), etc., and convert everything into sines and cosines (get rid of the tan).

 

[qspeechc]

To edit hover your cursor over the post, and you should see the Edit button near the bottom left corner.

 

[Ray Vickson]

I'm fairly sure the D should be B. Generally you simplify the two sides separately, so

L.H.S. (left hand side) = ... (simplify)
R.H.S. = ...

And you end up showing the two sides simplify to the same thing.

General hints are get everything in terms of functions of A and B, rather than 2A, (A-B), etc., and convert everything into sines and cosines (get rid of the tan).

As already shown: the stated result is false. See #2.

 

[qspeechc]

As already shown: the stated result is false. See #2.

You're right. I just sat down to check this. OP you should check the question with specific values for A and B. The identity isn't correct. Looks like your textbook could be riddled with errors.