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Question: $$2\cos x -\cos 3x - \cos 5x = 16\cos^3 x\sin^2 x$$

What I have tried: Using the identities, I have converted all the cos and sin so that the angle inside is only $x$. However, I couldn't proceed by simplifying and the equation got too complicated. I must be missing something, which would make the question surprisingly easy...

Please point me in the right direction!

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Using the trigonometric sum and difference angle identities we have (the last expression involves factorising out the common factor $2\cos x$) $$2\cos x -\cos 3x - \cos 5x = 2\cos x - (\cos 3x + \cos 5x )\\=2\cos x - 2\cos x\cos 4x=2\cos x(1-\cos 4x)$$ This is because (have a look at the "angle sum and difference identities" in here) we have $$\cos 5x = \cos (4x+x)=\color{green}{\cos 4x\cos x} - \sin 4x \sin x$$ $$\cos 3x = \cos (4x-x)=\color{green}{\cos 4x\cos x} + \sin 4x \sin x$$ Adding the above two equations leads to the cancellation of the sine terms, resulting in $$\cos 5x + \cos 3x=\color{green}{2\cos 4x\cos x}$$ Applying the double angle formula for $\cos 4x$ then $\sin 2x$ as below, results in $$2\cos x(1-\color{blue}{\cos 4x})=2\cos x(1-\color{blue}{(1-2\sin^22x)})=4\cos x\color{red}{\sin^22x}\\=4\cos x\color{red}{(2\sin x\cos x)^2}=16\cos^3x\sin^2x$$

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  • $\begingroup$ I don't see how you combined cox3x and cox5x. Can you please explain? $\endgroup$ – Gummy bears Jul 2 '14 at 16:54
  • $\begingroup$ I have elaborated my answer to show how to combine the $\cos 3x$ and $\cos 5x$ terms. Hope this helps. $\endgroup$ – Alijah Ahmed Jul 2 '14 at 17:04
  • $\begingroup$ I don't understand how cos3x + cos5x can become 2cosxcos4x and how that becomes 1-cos4x, guess an identity I never learnt $\endgroup$ – Gummy bears Jul 2 '14 at 17:05
  • $\begingroup$ @Cookies you hate trignometry.Don't you? You have seen the identity and its there. you are just not able to see it. $\endgroup$ – MonK Jul 2 '14 at 17:27
  • $\begingroup$ @Sid How were you able to deduce that I hate trigonometry? Well unfortunately it is true. One of the subjects that I can rarely ever understand. I find it difficult to prove anything in trigonometry. $\endgroup$ – Gummy bears Jul 3 '14 at 11:52
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Since $\cos 3x = 4\cos^3x - 3\cos x$ and $\cos 5x = 5\cos x - 20\cos^3 x + 16\cos^5 x$, substituting those on the left-hand side and simplifying gives $$16\cos^3 x - 16\cos^5 x = 16\cos^3x(1-\cos^2 x) = 16\cos^3x\sin^2 x.$$

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  • $\begingroup$ Oh... I see. But there's an identity for $\cos 5x$ ?? $\endgroup$ – Gummy bears Jul 2 '14 at 16:54
  • $\begingroup$ Just figure it out by expanding. Or, do what I did: lmgtfy.com/?q=cos+5x+in+terms+of+cos+x $\endgroup$ – rogerl Jul 2 '14 at 16:55
  • $\begingroup$ Haha I see. That's a nice website $\endgroup$ – Gummy bears Jul 2 '14 at 16:58
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    $\begingroup$ @Cookies You can derive these types of formulas by using De Moivre's formula and comparing the real and imaginary parts of both sides. $\endgroup$ – Peter Woolfitt Jul 2 '14 at 17:25
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Use Euler's formula to get $e^{ix}+e^{-ix} - {1 \over 2} (e^{i3x}+e^{-i3x} ) - {1 \over 2} (e^{i5x}+e^{-i5x} ) $ on the left hand side.

On the right hand side we have (using $\sin^2 x = 1-\cos^2 x$ to simplify): $ (e^{ix}+e^{-ix})^3 (2-{1 \over 2}(e^{ix}+e^{-ix})^2) = (e^{i3x}+ 3e^{ix} + 3 e^{-ix} + e^{-i3x} ) (1 -{1\over 2} e^{i2x} -{1\over 2} e^{-2x}) $.

Carrying out the tedious multiplication shows that they are the same.

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  • $\begingroup$ Well.... I think the others have provided a much easier method. Thanks anyways! $\endgroup$ – Gummy bears Jul 2 '14 at 16:59
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For Details check all the trignometric identities here

$2cos x-(cos 3x+cos 5x)$

$\implies 2cos x-(2cos x.cos 4x)$

$\implies 2cos x(1-cos 4x)$

$\implies 2cos x(1-(1-2sin^2 2x)$

$\implies 4cos x(sin^2 2x)$

$\implies 4cos x((2sin xcos x)^2)$

$\implies 16cos^3 xsin^2 x$

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  • $\begingroup$ Okay, I understood the process, but lost you in the identities. How did you combine cos3x and cox5x. And how did 2cosxcos4x convert to 1 - cos4x? $\endgroup$ – Gummy bears Jul 2 '14 at 17:04
  • $\begingroup$ Okay, so you can represent 5x as 4x+x and 3x as 4x-x. This gives us 2 definite A and B in the form of x and 4x. Then you can apply $cos A cos B =½[cos (A + B) + cos (A-B)]$ $\endgroup$ – MonK Jul 2 '14 at 17:09
  • $\begingroup$ Look at the link I provided and you will have a better idea. $\endgroup$ – MonK Jul 2 '14 at 17:12
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$$I=2\cos x-\cos3x-\cos5x=(\cos x-\cos3x)+(\cos x-\cos5x)$$

Using Prosthaphaeresis Formula, $\displaystyle\cos C-\cos D$,

$$I=2\sin x\sin2x+2\sin2x\sin3x=2\sin2x(\sin x+\sin3x)$$

and using $\displaystyle\sin C+\sin D$ formula, $$\sin x+\sin3x=2\sin2x\cos x$$

Finally use $\displaystyle\sin2x=2\sin x\cos x$

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  • $\begingroup$ @Cookies, How about this? $\endgroup$ – lab bhattacharjee Jul 2 '14 at 18:28

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