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Formulas: $\sin (2A) = 2\sin (A) \cos (A)$

$\cos (2A) = \cos^2 (A) − \sin^2 (A) = 1 − 2 \sin^2 (A) = 2\cos^2 (A) − 1$

$\tan (2A) = \dfrac{2\tan (A)}{1 − \tan^2 (A)}$

I'm really confused, if I'm supposed to expand $\cos(4x)$ using a double angle formula. how do I know which formula to use? This is just one of the questions, the same goes for $\sin(4x)$ and $\sin(6x)$. Is there a method I can use to find out which formula should I use?

for $\cos(4x)$ I used $2\cos^2(2x)-1$ but that formula wasn't used in the answers.

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  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Sep 24 '17 at 13:45
  • $\begingroup$ @JoséCarlosSantos thx, will keep it in mind next time. Sorry for the trouble $\endgroup$ – question Sep 24 '17 at 13:48
  • $\begingroup$ @question It´s nice that you keep it mind for next time. But the comment of José is meant as an invitation to edit your current question. $\endgroup$ – callculus Sep 24 '17 at 13:56
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    $\begingroup$ $\cos 4x=2\cos^2 2x-1=2(2\cos^2x-1)^2-1=8\cos^4x-8\cos^2x+1.$ If you're told to expand $\cos 4x$ with no further instructions, then it's impossible to know when you're finished. You can substitute $\cos x=2\cos^2(x/2)-1$ and keep going. $\endgroup$ – DanielWainfleet Sep 24 '17 at 14:10
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Whatever formula you use, the answer will be right, unless additional instructions were given.

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  • $\begingroup$ This is not a comment, it is the answer to the main question. $\endgroup$ – Yves Daoust Sep 24 '17 at 13:49
  • $\begingroup$ i used sin(2x) = 2sin(x)cos(x) for cos(4x). I did sin(2x) = 2sin(2x)cos(2x), how is that right? $\endgroup$ – question Sep 24 '17 at 13:51
  • $\begingroup$ $\sin2x=2\sin2x\cos2x$, is wrong, obviously. $\endgroup$ – Yves Daoust Sep 24 '17 at 13:53
  • $\begingroup$ @question $\sin4x=2\sin2x\cos2x$. $\endgroup$ – Michael Rozenberg Sep 24 '17 at 13:54
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you can write $$\cos(2x+2x)=\cos(2x)^2-\sin(2x)^2=(2\cos(x)^2-1)^2-4\sin(x)^2\cos(x)^2$$

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It depends on the problem, which you want to solve.

$$\cos6x=4\cos^32x-3\cos2x=2\cos^23x-1;$$ $$\sin6x=3\sin2x-4\sin^32x=2\sin3x\cos3x;$$ $$\cos4x=2\cos^22x-1=1-2\sin^22x...$$

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  • $\begingroup$ what sort of questions would you use the first formula in and the second formula? $\endgroup$ – question Sep 24 '17 at 13:47
  • $\begingroup$ @question I used $\cos3\alpha=4\cos^3\alpha-3\cos\alpha$ and $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. $\endgroup$ – Michael Rozenberg Sep 24 '17 at 13:49
  • $\begingroup$ @question use first one when it's easier to evaluate cosine, other when it's easier to evaluate sine $\endgroup$ – john doe Sep 24 '17 at 13:51
  • $\begingroup$ would i do, cos(2x) = 4 cos(2x)^3 - 3cos(2x)? to solve the first question? $\endgroup$ – question Sep 24 '17 at 13:55
  • $\begingroup$ @question $\cos6x=4\cos^32x-3\cos2x$. $\endgroup$ – Michael Rozenberg Sep 24 '17 at 14:01

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