Goniometric inequality

[Function]

Hello everyone

 

I was, with another topic in the back of my head (the one about the integration constant), making the following reasoning:

 

In a right triangle, let's say that the two acute angles are [math]\alpha[/math] and [math]\beta[/math].

 

Then is true that [math]\alpha+\beta=\frac{\pi}{2}[/math]

 

[math]\Leftrightarrow \alpha=\frac{\pi}{2}-\beta[/math]

[math]\Leftrightarrow \tan{\alpha}=\tan{\left(\frac{\pi}{2}-\beta\right)}[/math]

[math]\Leftrightarrow \frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\sin{\left(\frac{\pi}{2}-\beta\right)}}{\cos{\left(\frac{\pi}{2}-\beta\right)}}[/math]

[math]\Leftrightarrow \frac{\sin{\alpha}}{\cos{\alpha}}=\frac{\cos{\beta}}{\sin{\beta}}[/math]

[math]\Leftrightarrow \cos{\alpha}\cos{\beta}=\sin{\alpha}\sin{\beta}[/math]

 

According to Wolfram Alpha, left part and right part together would be (probably one of Simspon's formulas):

 

[math]\Leftrightarrow \frac{1}{2}\left(\cos{\left(\alpha-\beta\right)}+\cos{\left(\alpha+\beta\right)}\right)=\frac{1}{2}\left(\cos{\left(\alpha-\beta\right)}-\cos{\left(\alpha+\beta\right)}\right)[/math]

[math]\Leftrightarrow \cos{\left(\alpha+\beta\right)}=-\cos{\left(\alpha+\beta\right)}[/math]

[math]\Leftrightarrow 1=-1[/math]

 

Now, I will most certainly have made a mistake, and seen the result of this 'equality', a serious mistake, resulting in a pretty stupid topicquestion. Could someone help me? Perhaps I made a mistake in Wolfram? (I entered: cos(x)*cos(y) and sin(x)*sin(y) to get the equality after "According to ..."

 

Thanks.

 

Function

Edited February 20, 2014 by Function

[studiot]

 

 

Consider alpha = 20 degrees and beta = 50 degrees.

 

Are you really offering that cos (20+50) = - cos (70) ?

 

Some of these statements depend upon which quadrant your sum ends up in, have you heard of CAST?

Edited February 20, 2014 by studiot

[Function]

 

Consider alpha = 20 degrees and beta = 50 degrees.

 

Are you really offering that cos (20+50) = - cos (70) ?

 

Some of these statements depend upon which quadrant your sum ends up in, have you heard of CAST?

 

Let's say both alpha and beta are in the first quadrant; I haven't heard of CAST yet. Feel free to inform me, should you feel like doing so.

[John]

[math]\cos(\alpha + \beta) = \cos\left(\frac{\pi}{2}\right) = 0[/math], so the final line in the original post should read [math]0 = 0[/math] and not [math]1 = -1[/math].

Edited February 20, 2014 by John

[Function]

[math]\cos(\alpha + \beta) = \cos\left(\frac{\pi}{2}\right) = 0[/math], so the final line in the original post should read [math]0 = 0[/math] and not [math]1 = -1[/math].

 

Ah yes.. You see it was a stupid topicquestion

How could I've ever stated that if 0*a = 0*b, which is always true, a = b

[studiot]

Of course, John is quite right, the only entity for which A = -A is zero so you are just working on keeping zero to itself.

 

I was thinking about the general trig sum of two angles.

 

CAST is a way of remembering which trig function is positive in which quadrant, here is a list.

 

Edited February 21, 2014 by studiot