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How would I prove the following?

$$\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$$

I do not know how to do do the problem I do know $\sin(3x)$ can be $\sin(2x+x)$ and such yet I am not sure how to commence.

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up vote 4 down vote accepted

Using $\sin 2\theta=2\sin\theta\cos\theta$ and $\sin^2\theta=(1-\cos2\theta)/2$ we get

$$\cos^2x ~\sin^3x=(\cos x\sin x)^2\sin x=\left(\frac{\sin 2x}{2}\right)^2\sin x=\frac{1}{4}\frac{1-\cos 4x}{2}\sin x \tag{$\circ$}$$

With the sum rule we have $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha.$ Therefore we have the identity


Apply with $\alpha=4x$ and $\beta=x$ to $(\circ)$.

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If you know that $$\sin(x+y)=\sin x\cos y+\cos x\sin y\tag{1}$$ you can apply that to get $\sin (2x)\cos y + \cos(2x)\sin y$.

Then use $(1)$ again to get $\sin(2x)=\sin(x+x)=\cdots$ (I'll let you fill in the blanks.

Then recall that $\cos(x+y)=\cos x\cos y-\sin x\sin y$, and apply that to get $\cos(2x)=\cos(x+x)=\cdots$ (fill in the blanks here as well).

Then $\sin(5x) = \sin(3x+2x)=\cdots$ (and apply $(1)$ again here), and then you have to work with $\sin(3x)$ and $\sin(2x)$ again, and with $\cos(3x)$ and $\cos(2x)$. Use the identities for sine of a sum and cosine of a sum to reduce those further.

Eventually the only thing you'll be taking sines and cosines of is $x$, and then you should be done.

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This is not the fastest way to do it but I think it is helpful first use $$ \cos ^2 x=1- \sin ^2 x$$

Then convert the other side to powers of $\sin x$ this can be done using the identity $$ e^{inx} = \cos nx +i \sin nx = (e^{ix} ) ^n $$ and then take imaginary parts so $$\sin nx = \Im (e^{ix} ) ^n $$ where $\Im$ is the imaginary part.

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$2\sin x+\sin3x-\sin5x$

$=2\sin x-2\cos4x\sin x$ applying $\sin2C-\sin2D=2\sin(C-D)\cos(C+D)$ formula

$=2\sin x(1-\cos4x)$

$=4\sin x \cdot \sin^22x$, applying $\cos2A=1-2\sin^2A$

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


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 $\sin y=\frac{e^{iy} - e^{-iy}}{2i}$


$=(\frac{e^{ix}-e^{-ix}}{2i})^3(\frac{e^{ix} + e^{-ix}}{2})^2$

$=\frac{-1}{32i}(e^{ix}-e^{-ix})^2(e^{ix} + e^{-ix})^2(e^{ix}-e^{-ix})$

$=\frac{-1}{32i}((e^{ix}-e^{-ix})(e^{ix} + e^{-ix}))^2(e^{ix}-e^{-ix})$


$=\frac{1}{32i}(2(e^{ix}-e^{-ix}) + (e^{3ix}-e^{-3ix}) - (e^{5ix}-e^{-5ix}))$

$=\frac{1}{32i}(2(2i\sin x) + (2i\sin3x) - (2i\sin5x))$

$=\frac{2\sin x+\sin3x-\sin5x}{16}$

Alternatively, $\sin^3x\cos^2x$


$=\frac{1}{8}(3\sin x-\sin3x)(1+\cos2x)$ as $\sin3A=3\sin A-4\sin^3A$ and $\cos2A=2\cos^2A-1$

$=\frac{1}{8}(3\sin x-\sin3x+3\sin x\cos2x-\sin3x\cos2x)$

$=\frac{1}{16}(6\sin x-2\sin3x+6\sin x\cos2x-2\sin3x\cos2x)$

$=\frac{1}{16}(6\sin x-2\sin3x+3(\sin3x-\sin x)-(\sin5x+\sin x))$ (applying $2\sin A\cos B=\sin(A+B)+\sin(A-B))$

$=\frac{1}{16}(2\sin x+\sin3x-\sin5x)$

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I like this response. – Fernando Martinez Aug 6 '12 at 17:01
Very good work. – Fernando Martinez Aug 6 '12 at 18:01

We know $e^{iy}=\cos y+i\sin y$

So, $\cos 5x+i\sin 5x=e^{i5x}=(e^{ix})^5=(\cos x+i \sin x)^5$

Using binomial theorem & equating imaginary parts,

$sin 5x = sin^5x − 10\cdot sin^3x cos^2x + 5\cdot sin x cos^4x$

$=sin^5x − 10\cdot sin^3x(1 - sin^2x) + 5\cdot sin x (1 - sin^2x)^2$

So, $sin 5x= 16\cdot\sin^5x − 20\cdot\sin^3x + 5\cdot\sin x$

We know, $\sin3x=3\sin x -4sin^3x $


Let $A\sin5x+B\sin3x+C\sin x=\sin^3x-\sin^5x$

Or, $A(16\cdot\sin^5x − 20\cdot\sin^3x + 5\cdot\sin x) + B(3\sin x -4sin^3x)+C\sin x$ $=\sin^3x-\sin^5x$

Comparing the coefficients of different powers of sinx,

5th power=>16A=-1=>A=$-\frac{1}{16}$

3rd power=>-20A-4B=1=>B=$\frac{1}{16}$

1st power=>5A+3B+C=0=>C=-(5A+3B)=$\frac{1}{8}$

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