nice little problem

[Asimov Pupil]

what does

 

[math]\frac{\tan(\frac{\pi}{4} + a) - \tan(\frac{\pi}{4} - a)}{\tan(\frac{\pi}{4} + a) + \tan(\frac{\pi}{4} - a)}[/math]

 

equal to?

[Callipygous]

[hide]tan(pi/4)=1

after you know that much its not very hard to boil it down to 2a/2 which equals a

 

 

the above contains total neglect of parentheses... sorry [/hide]

[Asimov Pupil]

that's a good idea but i cannot prove that right and that is not the answer i am looking for

[Callipygous]

im confused about what kind of answer your looking for then. DOH

 

i take it back, thats not right : P

[P_Rog]

if tan(pi/4 + a) = tan(pi/4) + tan(a) then it's just tan(pi/4 + a)

 

I don't know if that is true, just a guess.

[Callipygous]

unfortunately, no. that doesnt work : P

[Algebracus]

We are trying to find

[MATH]\frac{\tan(\frac{\pi}{4} + a) - \tan(\frac{\pi}{4} - a)}{\tan(\frac{\pi}{4} + a) + \tan(\frac{\pi}{4} - a)}[/MATH]

which equals

[MATH]\frac{\frac{\sin {x}}{\cos {x}} - \frac{\sin {y}}{\cos {y}}}{\frac{\sin {x}}{\cos {x}} + \frac{\sin {y}}{\cos {y}}}[/MATH],

where [MATH]x = \frac{\pi}{4} + a[/MATH] and [MATH]y = \frac{\pi}{4} - a[/MATH].

Here is what follows:

[MATH]\frac{\frac{\sin {x}}{\cos {x}} - \frac{\sin {y}}{\cos {y}}}{\frac{\sin {x}}{\cos {x}} + \frac{\sin {y}}{\cos {y}}}[/MATH][MATH]= \frac{\sin {x}\cos{y} - \sin {y}\cos {x}}{\sin{x}\cos{y} + \sin {y}\cos {x}}[/MATH][MATH]=\frac{\sin (x - y)}{\sin (x + y)} = \sin {2a}[/MATH].

[Asimov Pupil]

Very good and if your teacher's are stingy and don't want you to leave it in double angles or if you solve it in three pages of work like i did then it is 2sin(a)cos(a)

 

by the way where did you learn to do it like that

it took me three pages doing addittion/subtraction formulas

[Algebracus]

You possibly used the addition- and subtraction-formulas for tan(x + y) and tan(x - y), and you have possibly learned those formulas at some instance. This would not be all that bad, but when it comes to those formulas, it is more than enough to learn the formulas for sinus and cosinus. To find the formulas for the tangent, just substitute tan x = sin x/cos x, and multiply with cos x cos y/cos x cos y, that is, 1.

 

So, what I did was just to use the same method as I do when I prove those addition- and subtraction- formulas. Together with the philosophy of first trying to find sweet solution, this is dynamite.

 

 

And for the sake of it, competing in several contests and math olympiads, I have been lucky to sharpen my skills of problem solving. The ability to find short solutions is the strongest weapon in such environments.

[Asimov Pupil]

thank you I'll remember that