• rvc

Difficult question. We have to use De Moivre's theorem at least once in the given question.

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  • rvc

Difficult question. We have to use De Moivre's theorem at least once in the given question.

Mathematics
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  • rvc
\[\rm \Large~ Show~ That : -2^{12}\cos^6\theta~\sin^7\theta=\\sin13\theta-\sin11\theta-6\sin9\theta+6\sin7\theta+15\sin5\theta-15\sin3\theta-20\sin\theta\]
\[\large \begin{gathered} - {2^{12}}{\left( {\cos \theta } \right)^6}{\left( {\sin \theta } \right)^7} = \sin \left( {13\theta } \right) - \sin \left( {11\theta } \right) - 6\sin \left( {9\theta } \right) + \hfill \\ + 6\sin \left( {7\theta } \right) + 15\sin \left( {5\theta } \right) - 15\sin \left( {3\theta } \right) - 20\sin \theta \hfill \\ \end{gathered} \]
try expanding \((\cos\theta + i\sin\theta)^{13}\) using binomial theorem and compare imaginary parts both sides

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I think that we have to expand all the terms sin(n\theta) at the right side, using the triangle of Tartaglia, nevertheless it is a very long computation
  • rvc
its too long @ganeshie8
using the Prosthaphaeresis formulas, I got this: \[\large right\;side = 2\sin \theta \left\{ {\cos \left( {12\theta } \right) - 6\cos \left( {8\theta } \right) + 15\cos \left( {4\theta } \right) - 10} \right\}\]
$$(\cos\theta+i\sin\theta)^n=\sum_{k=0}^n\binom{n}ki^{n-k}\cos^n\theta\sin^{n-k}\theta\\\quad =\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\cos^n\theta\sin^{n-2k}\theta+i\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n}{2k+1}\cos^n\theta\sin^{n-2k-1}\theta$$ while we also know by de Moivre's theorem: $$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$ so it follows: $$\sin(n\theta)=\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k\binom{n}{2k+1}\cos^n\theta\sin^{n-2k-1}\theta$$ if we sum the right \(\sin((2k+1)\theta)\) we can make the coefficients cancel ot and isolate a particular \(\cos^m\theta\sin^{m-2k-1}\theta\)
hint: since the left side is a term like this: \[{\left( {\cos \theta } \right)^6}{\left( {\sin \theta } \right)^7}\] one method is to get the expressions for sin(n\theta) using the binomial theorem odr triangle of Tartaglia, then we have to raise the grade in order to get a similar term
and the power of 2 will come from the simple binomial identity: $$\sum_{k=0}^n \binom{n}k=2^n$$ from the binomial theorem
not so sure if this helps simplifying the work here \[\cos^6\theta \sin^7\theta = \dfrac{(z+z^*)^6}{2^6}\dfrac{(z-z^*)^7}{(2i)^7}\]
  • rvc
it is the definition of cosh right
for example if we start from the last term, namely sinx, I can write this: sinx=sinx(cosx^2+ sinx^2)= =sinx cosx^2+ sinx^3 and every time I ultiply the result by 1= cosx^2+ sinx^2, so after more steps, I can write this identity: \[\large \begin{gathered} \sin x = {\left( {\sin x} \right)^{13}} + 6{\left( {\sin x} \right)^{11}}{\left( {\cos x} \right)^2} + 15{\left( {\sin x} \right)^9}{\left( {\cos x} \right)^4} + \hfill \\ \hfill \\ + 20{\left( {\sin x} \right)^7}{\left( {\cos x} \right)^6} + 15{\left( {\sin x} \right)^5}{\left( {\cos x} \right)^8} + 6{\left( {\sin x} \right)^3}{\left( {\cos x} \right)^{10}} + \hfill \\ \hfill \\ + \sin x{\left( {\cos x} \right)^{12}} \hfill \\ \end{gathered} \] and we have to do the same computation for other terms
it is a very huge computation!!
Looks this simplifies nicely \[\cos^6\theta \sin^7\theta = \dfrac{(z+z^*)^6}{2^6}\dfrac{(z-z^*)^7}{(2i)^7}\] \(\implies -2^{12}\cos^6\theta \sin^7\theta = (z^2-z^{2^*})^6\sin\theta\) \(= \left[(z^{12}+z^{12^*}) -6(z^{8}+z^{8^*})+15(z^{4}+z^{4^*})-20 \right]\sin\theta\) \(= \left[(2\cos12\theta) -6(2\cos8\theta)+15\cos4\theta -20 \right]\sin\theta\)
we got the same expression @ganeshie8

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