# Show that $\frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x$

The Question reads - $$\frac{\cos5x + \cos4x} {1-2\cos3x} = -\cos2x -\cos x$$

I tried using the obvious approach by converting $5x , 4x$ and $3x$ to either $2x$ or $x$ but all that seemed to do was to further complicate the fraction. Any hints would be much appreciated.

I also tried applying various identities but to no avail.

we can use product to sum formula. we have \begin{align}(1- 2\cos 3x)(-\cos 2x - \cos x) &= 2\cos 3x \cos 2x + 2 \cos 3x \cos x - \cos 2x - \cos x\\ &= (\cos 5x + \cos x) + (\cos 4x + \cos 2x) - \cos 2x - \cos x\\ &=\cos 5x+\cos 4x\end{align} dividing it out by $1- 2\cos 3x$ gives the result.

• What is the approach that you generally apply to such questions? May 18 '15 at 12:50
• @MayankJain, if all the arguments are the same and is $t$, then i use $x = \cos t, y = \sin t$ together with $x^2 + y^2 = 1.$ with that trig equations and identities become rational functions. i find them much easier to work with.
– abel
May 18 '15 at 14:05
• Hmmm... A good thought. Any tips for more complex questions like these? May 18 '15 at 19:01

First I'll multiply through by $-(1-2\cos(3x))$ to do away with the negatives on the RHS and still have an equivalent equation. So it will suffice to prove that $$\cos5x + \cos4x = (\cos2x +\cos x)(2\cos3x-1)$$ We will expand the RHS with complex exponentials. \begin{align}(\cos2x +\cos x)(2\cos3x-1) = \frac{1}{2}\left(e^{2ix}+e^{-2ix}+e^{ix}+e^{-ix}\right)\left(e^{3ix}+e^{-3ix}-1\right) \\ = \frac{1}{2}\left( e^{5ix}+e^{-ix}-e^{2ix}+e^{ix}+e^{-5ix}-e^{-2ix}+e^{4ix}+e^{-2ix}-e^{ix}+e^{2ix}+e^{-4ix}-e^{-ix}\right) \\ = \frac{1}{2}\left(e^{5ix}+e^{-5ix}+e^{4ix}+e^{-4ix}\right) \\ = \frac{e^{5ix}+e^{-5ix}}{2}+\frac{e^{4ix}+e^{-4ix}}{2} \\ = \cos(5x)+\cos(4x) \end{align}

Main idea: Stick to the $\cos x$ function and use brute force.

LHS:

The numerator $=\cos5x+\cos4x$

$=\cos3x\cos2x-\sin3x\sin2x+2\cos^22x-1$

$=(\cos2x\cos x-\sin2x\sin x)(2\cos^2x-1)-(\sin2x\cos x+\sin x\cos2x)(2\sin x\cos x)+2(2\cos^2x-1)^2-1$

$=((2\cos^2x-1)\cos x-2\sin^2 x\cos x)(2\cos^2x-1)-(2\sin x\cos^2x+\sin x(2\cos^2x-1))(2\sin x\cos x)+2(4\cos^4x-4\cos^2x+1)-1$

$=(2\cos^3 x-\cos x-2\sin^2x\cos x)(2\cos^2x-1)-(4\sin x\cos^2x-\sin x)(2\sin x\cos x)+8\cos^4x-8\cos^2x+1$

$=(2\cos^3 x-\cos x-2(1-\cos^2x)\cos x)(2\cos^2x-1)-(4(1-\cos^2x)\cos^2x-(1-\cos^2 x))2\cos x+8\cos^4x-8\cos^2x+1$

$=(4\cos^3x-3\cos x)(2\cos^2x-1)-(5\cos^2x-4\cos^4x-1)2\cos x+8\cos^4x-8\cos^2x+1$

$=8\cos^5x-6\cos^3x-4\cos^3x+3\cos x-10\cos^3x+8\cos^5x+2\cos x+8\cos^4x-8\cos^2{x}+1$

$=16\cos^5x+8\cos^4x-20\cos^3x-8\cos^2x+5\cos x+1$

$=(1+6\cos x - 8\cos^3 x)(1-\cos x - 2\cos^2 x)$

The denominator $=1-2\cos3x$

$=1-2(\cos2x\cos x - \sin2x\sin x)$

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

$=1-2(2\cos^3x-\cos x-2(1-\cos^2x)\cos x)$

$=1-2(4\cos^3x-3\cos x)$

$=1+6\cos x-8\cos^3x$

The fraction $=1-\cos x - 2\cos^2 x$

$= - (2\cos^2x-1)-\cos x$

$=-\cos2x-\cos x$

$=$ RHS

Let's apply the identity $\cos (x+y)+\cos(x-y)=2\cos x \cos y$ to the product

$$(1-2\cos 3x)(-\cos 2x-\cos x)=-\cos 2x -\cos x+2\cos 3x \cos 2x+2\cos 3x \cos x$$

Note that $2\cos 3x \cos 2x=\cos x+\cos 5x$ while $2\cos 3x \cos x=\cos 2x+\cos 4x$

Thus

\begin{align} (1-2\cos 3x)(-\cos 2x-\cos x)&=-\cos 2x -\cos x+2\cos 3x \cos 2x+2\cos 3x \cos x\\\\ &=-\cos 2x -\cos x+(\cos x+\cos 5x)+(\cos 2x +\cos 4x)\\\\ &=\cos 5x + \cos 4x \end{align}

which was to be shown!