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!
On
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 $$2\cos x(1-\color{blue}{\cos 4x})=2\cos x(1-\color{blue}{(1-2\sin^22x)})=4\cos x\color{red}{\sin^22x}\\=4\cos x\color{red}{(2\sin x\cos x)^2}=16\cos^3x\sin^2x$$
On
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.
On
For Details check all the trignometric identities here
$2\cos x-\left(\cos 3x+\cos 5x\right)$
$\implies 2\cos x-\left(2\cos x.\cos 4x\right)$
$\implies 2\cos x\left(1-\cos 4x\right)$
$\implies 2\cos x(1-\left(1-2\sin ^2 2x\right)$
$\implies 4\cos x\left(\sin ^2 2x\right)$
$\implies 4\cos x\left(\left(2\sin x\cos x\right)^2\right)$
$\implies 16\cos ^3 x\sin ^2 x$
On
$$I=2\cos x-\cos3x-\cos5x=(\cos x-\cos3x)+(\cos x-\cos5x)$$
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 $\displaystyle\sin2x=2\sin x\cos x$
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 $$16\cos^3 x - 16\cos^5 x = 16\cos^3x(1-\cos^2 x) = 16\cos^3x\sin^2 x.$$