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Given the question:

Use De Moivre’s formula to find a formula for $\cos(3x)$ and $\cos(4x)$ in terms of $\cos(x)$ and $\sin(x)$. Then use the identity $\cos^2(x) + \sin^2(x) = 1$ to express these formulas only in terms of $\cos(x)$.

I started out by rewriting $\cos(3x)$: $\cos(3x)$+$i \sin(3x)$=($\cos(x)$+$i \sin(x)$)$^3$

This could then be written into
$\cos(3x) = \cos^3(3x)-3 \cos(x) \sin^2(x)$
or
$\sin(3x) = \cos^2(x) \sin(x)- \sin^3(x)$

Then to use the identity I would substitute $1-\cos^2(x)$ for the $\sin^2(x)$
and in the second I would substitute $\sin(x)$ for $\sqrt{1 - cos(x)}$ right and would need to separate the $\sin^3(x)$ into $\sin^2(x) * \sin(x)$ and substitute from there. I'm a lot less confident about the second equation substitution. Would this be the right way to go about doing this problem?

mathminutemaid
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    So, you want to express $\sin(3x)$ also in terms of $\cos(x)$ only? Your original question only asks for $\cos(3x)$ and $\cos(4x)$ to be expressed in such a fashion. The method is right, anyway. –  Nov 22 '17 at 11:30
  • I believe so since in the second part formulas was plural so it seemed to refer to both the formulas in terms of cos(x) and the sin(x) one. Thanks for the answer – mathminutemaid Nov 22 '17 at 11:42
  • You should not hope to express $\sin(nx)$ in terms of cosines, ’cause the sine function is odd and any polynomial in cosine will be even. – Lubin Nov 24 '17 at 20:30

2 Answers2

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$${ \left( \cos { x } +i\sin { x } \right) }^{ 3 }=\cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \\ \cos ^{ 3 }{ x } +3i\cos ^{ 2 }{ x\sin { x } -3\cos { x\sin ^{ 2 }{ x } -i\sin ^{ 3 }{ x } = } } \cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \\ \\ \cos { \left( 3x \right) =\cos ^{ 3 }{ x } -3\cos { x } \sin ^{ 2 }{ x } =\cos ^{ 3 }{ x } -3\cos { x } \left( 1-\cos ^{ 2 }{ x } \right) =4\cos ^{ 3 }{ x-3\cos { x } } } \\ \sin { \left( 3x \right) =3\cos ^{ 2 }{ x } \sin { x } -\sin ^{ 3 }{ x } } =3\left( 1-\sin ^{ 2 }{ x } \right) \sin { x } -\sin ^{ 3 }{ x } =3\sin { x } -4\sin ^{ 3 }{ x } \\ $$

haqnatural
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  • I got the same thing for the cos(3x) written only in terms of cos(x) but it also wanted the sin(3x) one in terms of cos(x) only, and with that I'm not sure if I can take the square root of a negative to find what sin(x) is equal too. Thank you – mathminutemaid Nov 22 '17 at 11:46
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You seem to have misunderstood what is meant by "formulas," based on your approach and comments. That said, the problem given to you is poorly phrased (perhaps due to a typo), so this is understandable. The problem should read as follows (with emphasis to make the adjustment clear):

Use DeMoivre's formula to find formulas for $\cos(3x)$ and $\cos(4x)$ in terms of $\cos(x)$ and $\sin(x).$ Then use the identity $\cos^2(x)+\sin^2(x)=1$ to express these formulas only in terms of $\cos(x).$

Your approach is just right for finding $\cos(3x).$ Unfortunately, without more information, it is impossible to unambiguously define $\sin(3x)$ in terms of $\cos(x)$ only, since $$\sqrt{1-\cos^2(x)}=\sqrt{\sin^2(x)}=\left|\sin(x)\right|,$$ so your substitution $\sin(x)=\sqrt{1-\cos^2(x)}$ needn't be correct. Fortunately, you aren't being asked to do such a thing. Rather, it remains only to find a formula for $\cos(4x)$ in terms of $\cos(x)$ and $\sin(x),$ then use $\cos^2(x)+\sin^2(x)=1$ to express the found formula only in terms of $\cos(x).$

Cameron Buie
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