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!


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 $$2\cos x(1-\color{blue}{\cos 4x})=2\cos x(1-\color{blue}{(1-2\sin^22x)})=4\cos x\color{red}{\sin^22x}\\=4\cos x\color{red}{(2\sin x\cos x)^2}=16\cos^3x\sin^2x$$

  • $\begingroup$ 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

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 $$16\cos^3 x - 16\cos^5 x = 16\cos^3x(1-\cos^2 x) = 16\cos^3x\sin^2 x.$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.

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

For Details check all the trignometric identities here

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

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

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

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

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

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

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

  • $\begingroup$ 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

$$I=2\cos x-\cos3x-\cos5x=(\cos x-\cos3x)+(\cos x-\cos5x)$$

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

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

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