I need help with a problem some may consider odd but here it is.
I have the following trig identity I been working on and I managed to get it to.
$$\cos^2x\sin^4x=\frac{3}{16}-\frac{\cos(2x)}{4}+\frac{3\cos(2x)}{16}+\frac{1}{8}(1+\cos(4x))+\frac{1}{32}(\cos(6x)+\cos(2x))$$
However I am not sure how to make this simplify to $\frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))=\cos^2x\sin^4x$
and thus I am stuck.
Best Answer
$\cos^2x\sin^4x=\cos^2x(1-\cos^2x)^2=\cos^6x-2\cos^4x+\cos^2x$
Now $\cos(2x)=2\cos^2x-1$,
$\cos(4x)=\cos(2\cdot2x)=2\cos^2x-1=2(2\cos^2x-1)^2 - 1 = 8\cos^4x-8\cos^2x+1$ and
$\cos(6x)=\cos(2\cdot3x)=4\cos^3(2x)-3\cos(2x)= 32\cos^6x + 18\cos^2x - 48\cos^4x - 1 $
Let $A\cdot\cos(6x)+B\cdot\cos(4x)+C\cdot\cos(2x)+D=\cos^6x-2\cos^4x+\cos^2x$
$Or, A(32\cos^6x - 48\cos^4x + 18\cos^2x- 1)+B(8\cos^4x-8\cos^2x+1)+C(2\cos^2x-1)+D=\cos^6x-2\cos^4x+\cos^2x$
Comparing the coefficients of different powers of $\cos x$,
6th power=>$A=\frac{1}{32}$
4th power=>$-48A+8B=-2$ =>$B=-\frac{1}{16}$
2nd power=>$18A-8B+2C=1$=>$C=-\frac{1}{32}$
constants (power $0$)=$-A+B-C+D=0$=>$D=A+C-B=\frac{1}{16}$
So, $\cos^2x\sin^4x$
$=\frac{1}{32}\cdot\cos(6x) - \frac{1}{16}\cdot\cos(4x) -\frac{1}{32}\cdot\cos(2x) + \frac{1}{16} $
$=\frac{1}{32}(\cos(6x) - 2\cos(4x) -\cos(2x) + 2) $.
Alternatively, we know $e^{iy}=\cos y+i\sin y$ => $e^{-iy}=\cos y - i\sin y$
So, $\cos y=\frac{e^{iy}+e^{-iy}}{2}$ and $\sin y=\frac{e^{iy} - e^{-iy}}{2i}$
$\cos^2x\sin^4x$
$=(\frac{e^{ix}+e^{-ix}}{2})^2\cdot (\frac{e^{ix} - e^{-ix}}{2i})^4$
$=\frac{1}{64} (e^{ix}+e^{-ix})^2\cdot(e^{ix} - e^{-ix})^4$
$=\frac{1}{64} (e^{ix}+e^{-ix})^2\cdot(e^{ix} - e^{-ix})^2 \cdot(e^{ix} - e^{-ix})^2$
$=\frac{1}{64} ((e^{ix}+e^{-ix})\cdot(e^{ix} - e^{-ix}))^2 \cdot(e^{ix} - e^{-ix})^2$
$=\frac{1}{64} (e^{2ix} - e^{-2ix})^2 \cdot(e^{ix} - e^{-ix})^2$
$=\frac{1}{64} (e^{4ix} + e^{-4ix} -2 ) \cdot(e^{2ix} + e^{-2ix} - 2)$
$=\frac{1}{64} (e^{6ix} + e^{-6ix} -2(e^{4ix} + e^{-4ix}) -(e^{2ix} + e^{-2ix}) +4)$
$=\frac{1}{64} (2\cos6x -2(2\cos4x) -2\cos2x + 4)$
$=\frac{1}{32} (\cos6x -2\cos4x -\cos2x + 2)$
This is probably how the problem came to being.
But if we know the RHS, the task becomes far easier.
$\cos(6x) - 2\cos(4x) -\cos(2x) + 2$
$=\cos(6x) -\cos(2x) - 2(1-\cos(4x))$
$=-2\sin4x\sin2x -2\cdot2\sin^22x$ applying $ \cos 2C - \cos 2D=-2\cdot\sin (C+D) \sin(C-D)$ and $\cos2A=1-2\cdot \sin^2A$ formula
$= 2\cdot\sin2x(-\sin4x+2\sin2x)$
$= 2\cdot\sin2x(-2\cdot\sin2x\cos2x+2\sin2x)$ (applying $\sin2A=2\cdot\sin A\cos A$ formula)
$=4(\sin2x)^2(1-\cos2x)$
$=4(2\cdot\sin x \cos x )^2(2\sin^2x)$ (applying $\sin2A$ and $\cos 2A$ formula)
$=32\cos^2x\sin^4x$