# Prove $\cos^2x\sin^4x = \frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))$

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.

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$\cos^2x\sin^4x=\cos^2x(1-\cos^2x)^2=\cos^6x-2\cos^4x+\cos^2x$

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)

$=32\cos^2x\sin^4x$

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This is impressive math. – Fernando Martinez Aug 8 '12 at 15:42
I like the way the identity was solved. – Fernando Martinez Aug 8 '12 at 15:47

As AMPerrine pointed out in a comment above (before deleting it), it seems you made a mistake somewhere along the way. Your expression yields $$\frac1{32}\bigl(10-\cos(2x)+4\cos(4x)+\cos(6x)\bigr),$$ which is not identical to $$\frac1{32}\bigl(2-\cos(2x)-2\cos(4x)+\cos(6x)\bigr).$$

Applying the Pythagorean identity gives us $$\cos^2x\sin^4x=\sin^4x-\sin^6x.\tag{1}$$ Consider the double angle identities $$\cos(2\theta)=1-2\sin^2\theta\tag{2}$$ and $$\cos(2\theta)=2\cos^2\theta-1.\tag{3}$$ These give us $$\sin^2\theta=\frac12\bigl(1-\cos(2\theta)\bigr),\tag{2'}$$ and $$\cos^2\theta=\frac12\bigl(1+\cos(2\theta)\bigr),\tag{3'}$$ which (applied iteratively) let us rewrite even powers of sine--as in the right-hand side of $(1)$--in terms of cosines of multiple angles--as in the right-hand side of the desired identity.

By $(2')$, $$\sin^4x=\frac14\bigl(1-2\cos(2x)+\cos^2(2x)\bigr)=\frac14\bigl(1-2\cos(2x)\bigr)+\frac14\cos^2(2x)\tag{4}$$ and similarly, $$\sin^6x=\frac18\bigl(1-3\cos(2x)\bigr)+\frac18\bigl(3-\cos(2x)\bigr)\cos^2(2x).\tag{5}$$ By $(3')$ $$\cos^2(2x)=\frac12\bigl(1+\cos(4x)\bigr),$$ so from $(4)$ and $(5)$ we get $$\sin^4x=\frac18\Bigl(2\bigl(1-2\cos(2x)\bigr)+\bigl(1+\cos(4x)\bigr)\Bigr)=\frac18\bigl(3-4\cos(2x)+\cos(4x)\bigr)\tag{4'}$$ and similarly $$\sin^6x=\frac1{16}\bigl(5-7\cos(2x)+3\cos(4x)-\cos(2x)\cos(4x)\bigr).\tag{5'}$$ Applying $(4')$ and $(5')$ to $(1)$ yields $$\cos^2x\sin^4x=\frac1{16}\bigl(1-\cos(2x)-\cos(4x)+\cos(2x)\cos(4x)\bigr),\tag{6}$$ or for simplicity, $$\cos^2x\sin^4x=\frac1{32}\bigl(2-2\cos(2x)-2\cos(4x)+2\cos(2x)\cos(4x)\bigr).\tag{6'}$$ It remains, then, only to show that $$\cos(6x)-\cos(2x)=2\cos(2x)\cos(4x)-2\cos(2x)=2\bigl(\cos(4x)-1\bigr)\cos(2x),\tag{#}$$ that is, that there is no difference between the right-hand side of the desired identity and the right-hand side of $(6')$.

Indeed, applying (i) the angle sum formula $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$, (ii) the double angle formula $\sin(2\theta)=2\sin\theta\cos\theta$, and (iii) formula $(2')$, we have

$\begin{eqnarray*} \cos(6x) & = & \cos(2x)\cos(4x)-\sin(2x)\sin(4x)\\ & = & \cos(2x)\cos(4x)-2\sin^2(2x)\cos(2x)\\ & = & \cos(2x)\cos(4x)-\bigl(1-\cos(4x)\bigr)\cos(2x), \end{eqnarray*}$

from which $(\#)$ follows readily.

Additional Calculations: I'm not entirely certain which parts you're referring to in your third comment below, so I'll go ahead and expand those that seem like what you're asking. Let me know if you have any other questions.

First, the Pythagorean identity $\cos^2x+\sin^2x=1$ gives us $\cos^2x=1-\sin^2x$, from which we have $$\cos^2x\sin^4x=(1-\sin^2x)\sin^4x=\sin^4x-\sin^6x.$$

Second, since $\cos^2(2x)=\frac12\bigl(1+\cos(4x)\bigr)$ by $(3')$, we see from $(5)$ that

$\begin{eqnarray*} \sin^6x & = & \frac18\bigl(1-3\cos(2x)\bigr)+\frac18\bigl(3-\cos(2x)\bigr)\cos^2(2x)\\ & = & \frac18\bigl(1-3\cos(2x)\bigr)+\frac1{16}\bigl(3-\cos(2x)\bigr)\bigl(1+\cos(4x)\bigr)\\ & = & \frac18\bigl(1-3\cos(2x)\bigr)+\frac1{16}\bigl(3+3\cos(4x)-\cos(2x)-\cos(2x)\cos(4x)\bigr)\\ & = & \frac5{16}-\frac7{16}\cos(2x)+\frac3{16}\cos(4x)-\frac1{16}\cos(2x)\cos(4x)\\ & = & \frac1{16}\bigl(5-7\cos(2x)+3\cos(4x)-\cos(2x)\cos(4x)\bigr). \end{eqnarray*}$

Finally, applying $(4')$ and $(5')$ to $(1)$, we have

$\begin{eqnarray*} \cos^2x\sin^4x & = & \frac18\bigl(3-4\cos(2x)+\cos(4x)\bigr) - \frac1{16}\bigl(5-7\cos(2x)+3\cos(4x)-\cos(2x)\cos(4x)\bigr)\\ & = & \frac1{16}-\frac1{16}\cos(2x)-\frac1{16}2\cos(4x)+\frac1{16}\cos(2x)\cos(4x)\\ & = & \frac1{16}\bigl(1-\cos(2x)-\cos(4x)+\cos(2x)\cos(4x)\bigr)\\ & = & \frac1{32}\bigl(2-2\cos(2x)-2\cos(4x)+2\cos(2x)\cos(4x)\bigr). \end{eqnarray*}$

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This answer is very informative as well. – Fernando Martinez Aug 8 '12 at 15:45
This answer is excellent as well upon studied closely. – Fernando Martinez Aug 8 '12 at 15:55
Well one part slightly confuses me the part on getting sin^6x and then the sin^4-sin^6. – Fernando Martinez Aug 8 '12 at 16:25
I will add some more detail at the end. – Cameron Buie Aug 9 '12 at 1:30

The equation you have written is not correct in general.

What you can do is note that there exists some $\theta \in [0, \pi)$ such that $\sin^2 x = \sin \theta$.

Then $1 - \sin^2 x = 1 - \sin \theta \Rightarrow \cos^2 x = 1 - \sin \theta$.

Thus we know that $\cos^2 x \sin^4 x = (1 - \sin \theta) \sin^2 \theta$.

Let $1 - 2 \sin \theta = T$. Then we know:

$\cos^2 x \sin^4 x = (1 - \sin \theta) \sin^2 \theta = \frac{1}{8} (1 + T)(1 - T)^2$.

Similarly,

$\cos(2x) = T,$

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

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

The sum-to-product formula tells us:

$\cos (2x) + \cos (6x) = 2 \cos(4x) \cos (8x) = 32T^6 - 48T^4 + 20T^2 - 2$.

Perhaps you can go somewhere from here.

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