Trigonometry Help - Mathematicians needed

[quote name='GamerzInc' post='2169796' date='Aug 5 2009, 10:15 PM']If alpha + beta + gamma = 180 (degrees) and cot Q(theta) = cot alpha + cot beta + cot gamma, 0 < Q < 90(degrees)
Show that: sin^3Q = sin(alpha - Q)sin(beta - Q)sin(gamma - Q)[/quote]

I need a drink

Ah, proofs. I hated these in my AP math class. If my old math teacher wasn't on holidays now, I'd show him this question and you'd have an answer in like 10 seconds.

I know one thing though... since sin(a)sin(b) = sin(a+b), couldn't you rewrite it as:
sin(alpha - Q + beta - Q + gamma - Q)? Logically that would work out to:
sin (alpha + beta + gamma - 3Q),
which would work out to:
sin(180 - 3Q)... now you figure out the rest.

Thanks Skyline969. I'll post my work in about an hour.

You're welcome GamerzInc. I hope I'm right so far on that equation.

I'm not a trig master by any means but it seems to me that there should be some more information provided. It is obvious that any 3 angles of a triangle must equal 180 degrees. I am surprised that no information is given on what the values of alpha, beta , and gamma are. Q is some angle smaller than pi/2 and the fact that it is multiplied by a 5th variable theta only adds to the complexity of the statement. You are only given 1 equation that expresses an equality between 5 variables (2 on the left and 3 on the right). From this it is not obvious in any way what theta may be or the angle variables.

However it seems that if alpha,beta, and gamma are all equal to 2Q your 2nd expression is correct. Then that would mean that alpha=beta=gamma=60 degrees "pi/3 rad". alpha+beta+gamma=180=60+60+60 Q therefore would be 30 degrees "pi/6 rad". This is acceptable as 0<Q<90 or 0<pi/6<pi/2

cot(Q*theta)=cot(alpha)+cot(beta)+cot(gamma) however alpha=beta=gamma and Q=pi/6 so that the statement is equal to cot(pi/6*theta)=cot(60)*3=radical(3)

cot(pi/6*theta)=radical(3)
pi/6*theta must be equal to pi/3
so theta must be 2.
cot(pi/6*2)=cot(pi/3)
radical(3)=radical(3)

Another way to state it below

(sin(Q))^3=sin(alpha-Q)*sin(beta-Q)*sin(gamma-Q) , alpha=beta=gamma=2Q
(sin(Q))^3=(sin(Q))^3

cot(Q*theta)=cot(alpha)+cot(beta)+cot(gamma) , alpha=beta=gamma=2Q
cot(Q*theta)=3(cot(2Q)) , Q=30degrees 2Q=60degrees alpha+beta+gamma=180=60+60+60=6Q
cot(Q*theta)=cot(2Q)*3=radical(3) , theta=2
cot(60degrees)=cot(60degrees)

Honestly I enjoy doing most math, but proofs? I won't touch it with a ten foot pole. It's annoying as heck.

WigWrm, we're not trying to find their values but prove that the equation on the right, equals the one on the left. This is becoming crazy. We took up two full chalkboards earlier today trying to solve this. Even the people in the Math Learning Center weren't much help.
I've been trying to work out
sin(180-3Q)
sin(180) = 0 , so I removed the 180 and was left with "sin(-3Q)". I'm also probably wrong in doing it but I then went to:
sin^3(Q). The only problem with that is that I'm supposed to have a negative value. A negative cubed is still a negative. Also, why is "cotQ = cot alpha + cot beta + cot gamma" made known if it isn't used?

(sin(Q))^3=sin(alpha-Q)*sin(beta-Q)*sin(gamma-Q) , alpha=beta=gamma=2Q
(sin(Q))^3=(sin(Q))^3

cot(Q*theta)=cot(alpha)+cot(beta)+cot(gamma) , alpha=beta=gamma=2Q
cot(Q*theta)=3(cot(2Q))
cot(Q*theta)=cot(2Q)*3=radical(3) , theta=1
radical(3)=radical(3)

This seems adequate enough in proving the statement.
Use of trig identities probably does lend its hand here but this solution
is also valid.

This may be relatively useless, but wouldn't sin(-3Q) equal -sin(3Q)?

sin(alpha-Q)*sin(beta-Q)*sin(gamma-Q)
would be the same as saying (sin(Q)) * (sin(Q))^2 because alpha = beta = gamma = 2Q?

never thought id see something like this on gbatemp...

I took trig my senior year in high school. For my college class I only have to take basic algebra. Go figure.





...I've been working this equation out for about five minutes. I can't get it either x.x

[quote name='Skyline969' post='2169986' date='Aug 5 2009, 07:45 PM']This may be relatively useless, but wouldn't sin(-3Q) equal -sin(3Q)?[/quote]

Yes since sine is one of the four odds.

And Chrisman, you're extremely lucky. I'm taking Calculus in my senior year of high school, as a REQUIREMENT to graduate. I'm probably the dumbest smart kid in the class too.

[quote name='GamerzInc' post='2169988' date='Aug 5 2009, 07:46 PM']sin(alpha-Q)*sin(beta-Q)*sin(gamma-Q)
would be the same as saying (sin(Q)) * (sin(Q))^2 because alpha = beta = gamma = 2Q?[/quote]

sin(2Q-Q)*sin(2Q-Q)*sin(2Q-Q)=sin^3(Q)
My last post was editied theta should be 1.

alpha+beta+gamma=180 and this is always true.
You said Q was an angle between 0 and 90.
Your question was to prove the equality of the cubed sin function.
It was easy enough to create the equality by substituting 2Q for alpha beta and gamma.
Since all are equal they must be 60 degrees.
Q is then only 30 degrees and it stays true to your initial statement that q must be between 0 and 90 degrees.

I'm attaching an image in case what I originally wrote didn't make a lot of sense.

Wait wait wait, where do you get alpha = beta = gamma = 2Q? I'm not sure I see where you're getting that from.

If that were true... alpha + beta + gamma = 180... so 2Q + 2Q + 2Q = 180... 6Q = 180, therefore Q = 30... but that's got nothing to do with proving that sin^3Q = sin(alpha - Q)sin(beta - Q)sin(gamma - Q). But is the left side for sure sin^3Q? That would be sin to the power of 3Q... wouldn't the left side actually be sin(3Q)? I dunno, it's whatever is in your textbook. I'm just thinking that sin^3Q doesn't make much sense to me.

EDIT: If it is sin(3Q) = sin(alpha - Q)sin(beta - Q)sin(gamma - Q), then I think I got it.

A little prep-work:
sin(3Q) = sin(alpha - Q + beta - Q + gamma - Q)
sin(3Q) = sin(alpha + beta + gamma - 3Q)

Now then, if alpha + beta + gamma = 180 and alpha = beta = gamma = 2Q, then...
sin(3Q) = sin(2Q + 2Q + 2Q - 3Q)
sin(3Q) = sin(6Q - 3Q)
sin(3Q) = sin(3Q)

...Provided the left side is not sin^3Q which makes no sense to me and is actually sin(3Q)

EDIT(2): Oh, sin^3(Q)... nevermind.
Dammit, I wish I didn't throw out my old trig proof notes from my AP math class! I'd have this one figured out in a heartbeat, as something similar was on my final exam.

In the original post you made the theta a Q.

I will try the problem again.

Wigwrm, after rereading your previous post and your edited one it makes sense. Thank you a lot. I need to reread everything and keep working it out. The members of this forum have actually been more help then the people that are teaching me (since my teacher can't give me the answers and the tutors know nothing).

The only thing I don't understand is why "alpha=beta=gamma=2Q".

EDIT: Well, back to the grind. I see everyone is editing their answers.

LOL I goofed up: I took trigonometry Junior year, then Precalculous senior year. Two months out of high school and it's already a blur

http://www.youtube.com/watch?v=ry4iwzS4Na0

Click to expand...

What's THAT got to do with anything?


I'm giving up. 2 months out of precalc has completely reformatted my brain.

Okay, to restate the problem... kinda just make it more forum-friendly here:
Let's say alpha = A, beta = B, and gamma = C... Theta = Q still as that's what everyone seems to be working with.

Now then... if A + B + C = 180 degrees,
cot(Q) = cot(A) + cot(B) + cot[C], (Had to use square brackets there because it rewrote [C] as © when I didn't use them)
0 < Q < 90 degrees (Q is in the first quadrant, meaning all trig values are positive),
Prove sin^3(Q) = sin(A - Q) * sin (B - Q) * sin (C - Q)

I don't know why I rewrote it, maybe it'll help someone though.

Some things that are common knowledge:
sin(A - Q) * sin (B - Q) * sin (C - Q) = sin(A - Q + B - Q + C - Q) = sin(A + B + C - 3Q)
sin^3(Q) = 3sin(Q) (I may be wrong on this one, someone correct me if I'm wrong)