Trigonometry Help - Mathematicians needed

Skyline969, in my book they have problems similar to "sin^3(Q) = 3sin(Q)" written the same way.
I guess it's the same as x^6 or 6*x. If x = 4, the product would be the same.

[quote name='GamerzInc' post='2170082' date='Aug 5 2009, 06:24 PM']Skyline969, in my book they have problems similar to "sin^3(Q) = 3sin(Q)" written the same way.
I guess it's the same as x^6 or 6*x. If x = 4, the value would be the same.[/quote]

But, x^6 is not equal to 6*x. Let's say x = 10,
10 * 6 = 60,
10^ 6 = 10 * 10 * 10 * 10 * 10 *10 = 1000000

But I'm pretty sure sin^3(Q) = 3sin(Q).

[quote name='Skyline969' post='2170072' date='Aug 5 2009, 08:20 PM']sin^3(Q) = 3sin(Q) (I may be wrong on this one, someone correct me if I'm wrong)[/quote]

sin^3(Q)=(sin(Q))^3

Shit, WigWrm is right. I'm wrong.

EDIT: Then I got something. If you can prove A = B = C, then I think I got it.
Let's just say it is. Then you can go
sin(Q)^3 = sin(A + Q) * sin(B + Q) * sin(C + Q)
sin(Q)^3 = sin(A + Q)^3...
Then if A=0, the problem is solved. But A can't equal 0, as A + B + C = 180. If A = B = C, A would have to be 60. But, sin(60) = .86603... so we've proven that A = B = C is not true.

XD I just understood what I wrote, my bad. I feel not so smart now. Sorry Skyline949.

I updated the OP with a link to the formulas used to convert the equation.

Well, sin^3(Q) = sin^2(Q) * sin(Q)...
sin^2(Q) = (1/2) * (1-cos(2Q))
Therefore, we could have (1/2) * (1-cos(2Q)) * sin(Q)... but that does diddly shit for us....

This best explains us:

Hopefully this helps.

How do you get that A = B = C = 2Q?

Knowing how it all equals 2Q would tie this together.

Exactly, GamerzInc, but he hasn't shown us how he got that A = B = C = 2Q... you don't have that on an assignment (the how, I mean) and you'll get no marks for that question.

Because it satisfies the equality. The right hand contains a multiplication of three sins with different angles. The sin functions all contain angles (x-theta) where x is alpha beta gamma.
You make all the angles the same and it matches the left hand side. Nothing tricky to it. When you see similarities use them to your benefit.

You can see that they are equal to 2*theta because when back substituted into the earlier equation you produce the answer radical 3 for both sides.
And while this may not be a direct use of identities it does indeed prove the solution. The solution isnt proved by stating 2*theta is alpha,gamma,beta,
it is proved by the back substitution.

I get that x can be alpha, beta, or gamma, and that sin, on the right side, contains one of each. However, I still do not understand where 2Q comes from. It satisfies it yes, but not knowing where it comes from, even though I have the answer, has taught me nothing. Sorry if it seems like I'm complaining. I'm just trying to understand the how.

EDIT: This reminds me of the time the instructor put an equation in brackets and said the problem was solved. It didn't make sense until he worked it backwards.

EDIT 2:
I started on problem 2.
z = tan X/2. So
cos X = 1-Z^2/1+Z^2
cos X = 1-(tan X/2)^2/1+(tan X/2)^2
cos X = 1-tanX/1+tanX

would I be correct would the last equation? Does the exponential value 2, and the denominator of both x/2 cancel one another out and leave me with tanX?

I have thought about this problem for awhile and have to assume that my solution is correct.
The a=b=g=2*theta confirmed that theta was between 0 and 90 degrees. I am having difficulty clarifying this for you though.
Thats fine. Ill try to define it again.

Alpha Beta and Gamma are all angles in a triangle.
They may only ever equal to 180 degrees.
The requested angle theta was to be between 0 and 90 degrees.
The equation involving cotangents is provided as a requirement that must be true.

Easily enough the equation involving sines is solved by substituting the values for Alpha Beta and Gamma.
You do not know these values and it would be rather difficult to translate the left hand side in terms of such.
The only possible solution is to ensure that each sin(x-theta) function reduces to sin(theta). There is only one substitution
for alpha beta and gamma that exists to ensure this happens. It has to be 2*theta. This in turn when substituted in for x
allows each sin function to be sin(theta). You have three identical sin functions on the right hand side and this is equivalent to
the left hand side.

Now we must satisfy the constraint equations. Alpha+Beta+Gamma must equal 180. We write this as being 2theta+2theta+2theta=180
We solve for theta and get 30. We must satisfy alpha+beta+gamma=180. We said that they were all equal to 2*theta. 2theta+2theta+2theta= 2*30+2*30+2*30=180.
We have sucessfully satsified the angle sum constraint equation. Our calculated theta shall be substituted in for the cot constraint equation.
When done correctly we will see that both sides equal radical 3.

We have satisfied all contraint equations and are now done with the problem.

However I do not understandwhy your teacher could not help you with this?

I'm a 3rd year college student and have always gone to my professor's when I
didn't understand something. They were always willing to help.

I too am a third year college student and have gone to my teacher. I have also gone to the math lab. In the lab they either do it an extremely long way, spending a day or a few chalkboards doing it and never get an answer or they don't know what to do. My instructor, when asked for help on these particular problems (take home test, use whatever means to get the answers as long as you show your work and work it out), starts us off but says we should be able to finish the rest. I mean, if someone gives you the answer you'll never learn anything.

Well go ahead and check my last post that I edited and see if this now makes sense to you.

Wigwrm it now makes sense. Thank you so much for all your help tonight, as well as Skyline969 and everyone else that contributed. "The only possible solution is to ensure that each sin(x-theta) function reduces to sin(theta). There is only one substitution
for alpha beta and gamma that exists to ensure this happens. It has to be 2*theta." Those words of yours tied everything you've posted together.

Again, thank you for your help.

[quote name='GamerzInc' post='2170160' date='Aug 5 2009, 09:03 PM']I started on problem 2.
z = tan X/2. So
cos X = 1-Z^2/1+Z^2
cos X = 1-(tan X/2)^2/1+(tan X/2)^2
cos X = 1-tanX/1+tanX

would I be correct would the last equation? Does the exponential value 2, and the denominator of both x/2 cancel one another out and leave me with tanX?[/quote]

That last equation is incorrect. Because of the addition and subtraction signs in the numerator and denominator you cannot cancel terms.

hmmm... i'll try working on this, but i'm pretty rusty cos i haven't touched this in years :S

anyway, this should be a good place to start, will report back in half an hour if i can get the answer...

wigwrm's method is generally avoided for proof questions, since in such questions, the identity should hold true for all variables of theta which satisfies 0<theta<90, so assigning a value and proving that value holds true => identity is valid is not the ideal solution.

try expanding the left side completely with all the identities you can use, and then expand the right hand side completely too. you should be able to find some similarities and from there show LHS=RHS. haha of course its easier said then done... will report back if i find anything

Ok. I'm nearing the end of problem 5 and skipped problem 4. I have:
a^2 + 2ab + b^2 - c^2/4ab

I know the top becomes (a+b)^2 - c^2, but I don't remember why or know why I know it's that.

EDIT:
Progression goes well on problem V. I think I have like one or two more steps left, although, I'm not sure what to do. Here's my work on it thus far.

EDIT2:
Removed the image as I finished solving problem 5. Trying to tackle problem IV now.