Efficient Method for Finding Constants in Trig Function Power Problem

AI Thread Summary
To find constants a, b, and c in the equation sin(∅)^5 = asin∅ + bsin3∅ + csin5∅, one method involves expressing sin in its complex form and using the binomial theorem to simplify the expansion. An alternative approach is to utilize the identity for sin(A+B) to express sin 3∅ and sin 5∅ in terms of sin∅ and cos∅, then convert everything into a polynomial in sin∅. This method may provide a quicker solution than multiplying out the brackets multiple times. The discussion highlights the effectiveness of both techniques and the importance of understanding the binomial theorem in such problems. Overall, using these mathematical identities can streamline the process of finding the constants.
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Homework Statement


Find constants a, b and c such that
sin(∅)^5=asin∅+bsin3∅+csin5∅



Homework Equations





The Attempt at a Solution


I expressed sin in its complex form (if that's what its called) and put it to the power 5, and then multiplied each bracket out one at a time, and got the correct answers, as checked by wolfram, but I was wondering if there is a quicker way to find the bracket to the power 5 other than just multiplying each bracket out 5 times with its self? This is probably a simple question but I never seem to be able to get the simple stuff. Any help is greatly appreciated. Thanks.

Sorry for the wording; I don't know how to use latex, and it's really ugly just typed out.
 
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Using the exponential form of sin is probably the fastest way to work this out. Expanding (e^{i\phi} - e^{-i\phi})^5 might be a bit simpler using the binomial theorem: http://en.wikipedia.org/wiki/Binomial_theorem

An alternative to using complex numbers is to use the identity for \sin(A+B) to write \sin 3\phi and \sin 5\phi in terms of \sin\phi and \cos\phi. Then use \cos^2\phi = 1-\sin^2\phi to write the whole thing as a polynomial in \sin\phi.
 
Ah, thanks for clearing that up. My tutor used the binomial theorem, but didn't mention it, probably because he's been doing it for so long. Thanks for your help :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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