I pose a conundrum to ye

[Pollock]

If a + ß + ? = 180° and cot ? = cot a + cot ß + cot ?, 0 < ? < 90, show that sin³? = sin(a - ?)sin(ß - ?)sin(? - ?).

Any help? Can't figure this one out.

 

[Bateluer]

The answer is 3

 

[ItTheCow]

cot = cos/sin . See where that goes.

 

[Pollock]

Originally posted by: ItTheCow
cot = cos/sin . See where that goes.

Hasn't helped too much so far.. :\

 

[KarmaPolice]

because you say so

 

[AbAbber2k]

God I hated precalc.

 

[FleshLight]

I don't get how that will have anything to do with anything in calc.

 

[BigJ]

Originally posted by: FleshLight
I don't get how that will have anything to do with anything in calc.

Helps with manipulation. It can be hard or damn near impossible to do certain things by hand if you don't break them down into easier pieces to work with.

 

[chuckywang]

I solved the problem. Solution pending.

 

[chuckywang]

You really need to know your trig identities: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

I'll use a = alpha, b=beta c=gamma.

Now:

1) cot(a)+cot(b) = cot(?)-cot(c), which means cos(a)/sin(a) + cos(b)/sin(b) = cos(?)/sin(?) - cos(c)/sin(c)

Having a common denominator gives you, (cos(a)sin(b) + cos(b)sin(a))/(sin(a)sin(b)) = (cos(?)sin(c) - cos(c)sin(?))/(sin(?)sin(c)).

Therefore, sin(a+b)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c)). Since a+b+c = 180, this means that sin(a+b) = sin(c), so sin(c)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c))

2) cot(a) + cot(c) = cot(?) - cot(b). Going through the same steps as before yields sin(b)/(sin(a)sin(c)) = cos(b-?)/(sin(?)sin(b))

3) cot(b)+cot(c) = cot(?) - cot(a). This yields sin(a)/(sin(b)sin(c)) = cos(a-?)/(sin(?)sin(a))

Multiply the three bolded equations together and simply.

 

[imported_Tick]

Originally posted by: chuckywang
You really need to know your trig identities: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

I'll use a = alpha, b=beta c=gamma.

Now:

1) cot(a)+cot(b) = cot(?)-cot(c), which means cos(a)/sin(a) + cos(b)/sin(b) = cos(?)/sin(?) - cos(c)/sin(c)

Having a common denominator gives you, (cos(a)sin(b) + cos(b)sin(a))/(sin(a)sin(b)) = (cos(?)sin(c) - cos(c)sin(?))/(sin(?)sin(c)).

Therefore, sin(a+b)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c)). Since a+b+c = 180, this means that sin(a+b) = sin(c), so sin(c)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c))

2) cot(a) + cot(c) = cot(?) - cot(b). Going through the same steps as before yields sin(b)/(sin(a)sin(c)) = cos(b-?)/(sin(?)sin(b))

3) cot(b)+cot(c) = cot(?) - cot(a). This yields sin(a)/(sin(b)sin(c)) = cos(a-?)/(sin(?)sin(a))

Multiply the three bolded equations together and simply.

There has GOT to be a short-cut. There always is with trig idents.

 

[BigJ]

Originally posted by: Tick

Originally posted by: chuckywang
You really need to know your trig identities: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

I'll use a = alpha, b=beta c=gamma.

Now:

1) cot(a)+cot(b) = cot(?)-cot(c), which means cos(a)/sin(a) + cos(b)/sin(b) = cos(?)/sin(?) - cos(c)/sin(c)

Having a common denominator gives you, (cos(a)sin(b) + cos(b)sin(a))/(sin(a)sin(b)) = (cos(?)sin(c) - cos(c)sin(?))/(sin(?)sin(c)).

Therefore, sin(a+b)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c)). Since a+b+c = 180, this means that sin(a+b) = sin(c), so sin(c)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c))

2) cot(a) + cot(c) = cot(?) - cot(b). Going through the same steps as before yields sin(b)/(sin(a)sin(c)) = cos(b-?)/(sin(?)sin(b))

3) cot(b)+cot(c) = cot(?) - cot(a). This yields sin(a)/(sin(b)sin(c)) = cos(a-?)/(sin(?)sin(a))

Multiply the three bolded equations together and simply.

There has GOT to be a short-cut. There always is with trig idents.

You do realize this is a proof, right? He should be thankful something like this is all he's proving.

 

[chuckywang]

I realize it's a lot of writing, but it's actually not that hard if you know what to do.

The only trig identity you need is the sine of sums formula, ie sin(a+b) = ....

 

[Pollock]

Originally posted by: chuckywang
You really need to know your trig identities: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

I'll use a = alpha, b=beta c=gamma.

Now:

1) cot(a)+cot(b) = cot(?)-cot(c), which means cos(a)/sin(a) + cos(b)/sin(b) = cos(?)/sin(?) - cos(c)/sin(c)

Having a common denominator gives you, (cos(a)sin(b) + cos(b)sin(a))/(sin(a)sin(b)) = (cos(?)sin(c) - cos(c)sin(?))/(sin(?)sin(c)).

Therefore, sin(a+b)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c)). Since a+b+c = 180, this means that sin(a+b) = sin(c), so sin(c)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c))

2) cot(a) + cot(c) = cot(?) - cot(b). Going through the same steps as before yields sin(b)/(sin(a)sin(c)) = cos(b-?)/(sin(?)sin(b))

3) cot(b)+cot(c) = cot(?) - cot(a). This yields sin(a)/(sin(b)sin(c)) = cos(a-?)/(sin(?)sin(a))

Multiply the three bolded equations together and simply.

Kind of messed on something on this part here :

1) cot(a)+cot(b) = cot(?)-cot(c), which means cos(a)/sin(a) + cos(b)/sin(b) = cos(?)/sin(?) - cos(c)/sin(c).

Having a common denominator gives you, (cos(a)sin(b) + cos(b)sin(a))/(sin(a)sin(b)) = (cos(?)sin(c)-cos(c)sin(?))/sin(?)sin(c).

Therefore, sin(a+b)/(sin(a)sin(b)) = cos(c-?)/(sin(?)sin(c)).

Which results in sin(c-?)/(sin(?)sin(c)), but it works from there. Thanks for the help. You're the best.

 

[toolboxolio]

Originally posted by: Bateluer
The answer 2 is 3

fixededed.

 

[GasX]

Do your own homework.

QED

 

[chuckywang]

You're right OP, that's what I get when my fingers go faster than my brain.