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

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rvc
 one year ago
Best ResponseYou've already chosen the best response.0\[\rm \Large~ Show~ That : 2^{12}\cos^6\theta~\sin^7\theta=\\sin13\theta\sin11\theta6\sin9\theta+6\sin7\theta+15\sin5\theta15\sin3\theta20\sin\theta\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \begin{gathered}  {2^{12}}{\left( {\cos \theta } \right)^6}{\left( {\sin \theta } \right)^7} = \sin \left( {13\theta } \right)  \sin \left( {11\theta } \right)  6\sin \left( {9\theta } \right) + \hfill \\ + 6\sin \left( {7\theta } \right) + 15\sin \left( {5\theta } \right)  15\sin \left( {3\theta } \right)  20\sin \theta \hfill \\ \end{gathered} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2try expanding \((\cos\theta + i\sin\theta)^{13}\) using binomial theorem and compare imaginary parts both sides

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1using the Prosthaphaeresis formulas, I got this: \[\large right\;side = 2\sin \theta \left\{ {\cos \left( {12\theta } \right)  6\cos \left( {8\theta } \right) + 15\cos \left( {4\theta } \right)  10} \right\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$(\cos\theta+i\sin\theta)^n=\sum_{k=0}^n\binom{n}ki^{nk}\cos^n\theta\sin^{nk}\theta\\\quad =\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\cos^n\theta\sin^{n2k}\theta+i\sum_{k=0}^{\lfloor n/2\rfloor} (1)^k\binom{n}{2k+1}\cos^n\theta\sin^{n2k1}\theta$$ 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: $$\sin(n\theta)=\sum_{k=0}^{\lfloor n/2\rfloor} (1)^k\binom{n}{2k+1}\cos^n\theta\sin^{n2k1}\theta$$ if we sum the right \(\sin((2k+1)\theta)\) we can make the coefficients cancel ot and isolate a particular \(\cos^m\theta\sin^{m2k1}\theta\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the power of 2 will come from the simple binomial identity: $$\sum_{k=0}^n \binom{n}k=2^n$$ from the binomial theorem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2not so sure if this helps simplifying the work here \[\cos^6\theta \sin^7\theta = \dfrac{(z+z^*)^6}{2^6}\dfrac{(zz^*)^7}{(2i)^7}\]

rvc
 one year ago
Best ResponseYou've already chosen the best response.0it is the definition of cosh right

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is a very huge computation!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Looks this simplifies nicely \[\cos^6\theta \sin^7\theta = \dfrac{(z+z^*)^6}{2^6}\dfrac{(zz^*)^7}{(2i)^7}\] \(\implies 2^{12}\cos^6\theta \sin^7\theta = (z^2z^{2^*})^6\sin\theta\) \(= \left[(z^{12}+z^{12^*}) 6(z^{8}+z^{8^*})+15(z^{4}+z^{4^*})20 \right]\sin\theta\) \(= \left[(2\cos12\theta) 6(2\cos8\theta)+15\cos4\theta 20 \right]\sin\theta\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we got the same expression @ganeshie8
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