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Prove that $$16\cos^5A-20\cos^3A+5\cos A=\cos5A$$

My solution begins here; $$ \begin{align} \text{RHS} & =\cos5A \\ & =\cos(A+4A) \\ & =\cos A\cos4A-\sin A\sin4A \\ & =\cos A(2\cos^2 2A-1)-\sin A(2\sin2A\cos2A) \\ & =2\cos A\cos^2 2A-\cos A-2\sin A\sin2A\cos2A \\ & =2\cos A\cos^2 2A-\cos A-2\sin A(2\sin A\cos A)\cos2A \end{align} $$ Now how do I move on?

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  • $\begingroup$ Your question has been answered very adequately below, but you should try using De Moivre's Theorem - it's a lot easier! $\endgroup$
    – David Quinn
    Feb 5, 2016 at 18:44

4 Answers 4

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Now use double angle formulae to write $\cos 2A$ in terms of $\cos A$, then turn any $\sin^2 A$ into $1-\cos^2 A$

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Use $$ 2\cos x=e^{ix}+e^{-ix} $$ Then $$ 32\cos^5x=e^{5ix}+5e^{3ix}+10e^{ix}+10e^{-ix}+5e^{-3ix}+e^{-5ix}= 2\cos5x+10\cos3x+20\cos x $$ Moreover $$ 8\cos^3x=e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}= 2\cos3x+6\cos x $$ Therefore \begin{align} 16\cos^5x-20\cos^3x+5\cos x &=(\cos5x+5\cos3x+10\cos x)-5(\cos3x+3\cos x)+5\cos x\\ &=\cos5x \end{align}

This is actually backwards, so here's a different proof: \begin{align} \cos 5x+i\sin 5x &=(\cos x+i\sin x)^5\\ &=\cos^5x+5i\cos^4x\sin x-10\cos^3x\sin^2x\\ &\qquad-10i\cos^2x\sin^3x+5\cos x\sin^4x +i\sin^5x \end{align} Taking the real part $$ \cos5x=\cos^5x-10\cos^3x(1-\cos^2x)+5\cos x(1-\cos^2x)^2 $$ and you can finish.

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You'll need to replace each instance of $\sin^2 A$ with $1-\cos^2A$ before doing the last simplifications.

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Calculating the Chebyschev polynomial of the first kind by the recurrence

$$T_0(A)=1\\T_1(A)=A\\T_{n+1}(A)=2AT_n(A)-T_{n-1}(A)$$ you have the known result $$T_n(A)=\cos nA$$ You can verify at once for $n=5$ your formule.

Comment for laamuser user: Chebyschev polynomial of the first kind is some "magic" to find $\cos nx$. For $n=5$ you have successively $$T_2(A)=2A\cdot A-1=2A^2-1$$ $$T_3(A)=2A(2A^2-1)-A=4A^3-3A$$ $$T_4(A)=2A(4A^3-3A)-(2A^2-1)=8A^4-8A^2+1$$ $$T_5(A)=2A(8A^4-8A^2+1)-(4A^3-3A)$$ i.e. $$T_5(A)=16A^5-20A^3+5A$$ and you have $$T_5(\cos x)=\cos 5x$$

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  • $\begingroup$ How does this help the OP with "Now how do I move on?" $\endgroup$
    – Simply Beautiful Art
    Feb 5, 2016 at 22:12
  • $\begingroup$ @ piquito, I could not understand it. Could you please put it simply.. $\endgroup$
    – Iaamuser user
    Feb 6, 2016 at 10:14
  • $\begingroup$ @laamuser user: read the edit please. $\endgroup$
    – Piquito
    Feb 6, 2016 at 20:47
  • $\begingroup$ @SimpleArt: I wanted to give additional information concerning $\cos nx$ in function of $\cos x$ to laamuser user. (I'm a writer and I love too much art like you). $\endgroup$
    – Piquito
    Feb 6, 2016 at 21:05
  • $\begingroup$ @Piquito Sadly, I've found making an answer like this can get negative attention or positive attention, so be careful. (I like the art of math.) $\endgroup$
    – Simply Beautiful Art
    Feb 7, 2016 at 16:34

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