Simplifying $\frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)}$

Question

To simplify: $$\frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)}.$$

I know that in the denominator, $\sin(2x)$ can be re-written as $2\sin x\cos x$ using the double-angle identity.

I also know that for the numerator $\sin(3x)+\sin(5x)=2\sin(4x)\cos x$ and likewise for the denominator $\sin x+\sin(3x)=2\sin(2x)\cos x$.

After that, I get: $$\frac{2\sin(4x)\cos x +\sin(4x)}{2\sin(2x)\cos x + 2\sin x\cos x}.$$

If necessary, the $\sin(4x)$ in the numerator can also become $2\sin(2x)\cos(2x)$. This would make the fraction:

$$\frac{2\sin(4x)\cos x + 2\sin(2x)\cos(2x)}{2\sin(2x)\cos x + 2\sin x\cos x}.$$

From here, one can factor out a two from both numerator and denominator and thereby cancel them out (though I'm not sure this is helpful in the long run):

$$\frac{\sin(4x)\cos x +\sin(2x)\cos(2x)}{\sin(2x)\cos x +\sin x\cos x}.$$

And then, stuck. No like terms that I can see and which identity is helpful here, I do not know. Can someone help with some hints for the next steps in order to simplify?

Answer 1

$$\frac{\sin(3x) + \sin(4x) + \sin(5x)}{\sin(x) + \sin(2x) + \sin(3x)}$$

$$= \frac{\sin(4x) + (\sin(5x) + \sin(3x))}{\sin(2x) + (\sin(3x) + \sin(x))}$$

$$=\frac{\sin(4x) + 2\sin(4x)\cos(x)}{\sin(2x) + 2\sin(2x)\cos(x)}$$

$$=\frac{\sin(4x)(1+2\cos(x))}{\sin(2x)(1+2\cos(x))}$$

$$=\frac{2\sin(2x)\cos(2x)}{\sin(2x)}$$

$$= 2\cos(2x)$$

Answer 2

Let $z=e^{ix}.$

$$\frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)}=$$ $$\frac{z^3-z^{-3}+z^4-z^{-4}+z^5-z^{-5}}{z-z^{-1}+z^2-z^{-2}+z^3-z^{-3}}=$$$$z^2+z^{-2}=2\cos(2x).$$

(For the central step, simply check that the denominator times $z^2+z^{-2}$ equals the numerator.)

Answer 3

An approach with vectors:

Let \begin{align} v_1 &= (\cos x, \sin x), \\ v_2 &= (\cos (2x), \sin (2x)), \\ v_3 &= (\cos (3x), \sin (3x)), \\ v_4 &= (\cos (4x), \sin (4x)), \\ v_5 &= (\cos (5x), \sin (5x)). \end{align}

The vectors $v_1$ and $v_3$ are symmetric around $v_2$, so their sum is a vector along the same line as $v_2$. You can work out geometrically that the sum is

$$ v_1 + v_3 = (2\cos x) v_2. $$

Therefore

$$ v_1 + v_2 + v_3 = (1 + 2\cos x) v_2. $$

The vectors $v_3$ and $v_5$ likewise are symmetric around $v_4$ and their sum is along the same line as $v_4$,

$$ v_3 + v_5 = (2\cos x) v_4. $$

and therefore

$$ v_3 + v_4 + v_5 = (1 + 2\cos x) v_4. $$

So we have $v_1 + v_2 + v_3 = r v_2$ and $v_3 + v_4 + v_5 = r v_4$ with $r = 1 + 2\cos x$.

Intuitively, it is not even necessary to work out the exact value of $r$ as long as you recognize that the three vectors $v_3,v_4,v_5$ have the exact same geometry as $v_1,v_2,v_3$ except that the entire set of vectors is rotated, and therefore whatever multiple of $v_2$ we get from $v_1 + v_2 + v_3$, we get the same multiple of $v_4$ from $v_3 + v_4 + v_5$.

Now if we take the second component of each vector on each side of each equation $v_1 + v_2 + v_3 = r v_2$ and $v_3 + v_4 + v_5 = r v_4$, we get

\begin{align} \sin(x)+\sin(2x)+\sin(3x) &= r\sin(2x), \\ \sin(3x)+\sin(4x)+\sin(5x) &= r\sin(4x), \end{align} which we know is true because we can do the vector addition componentwise.

It follows that $$ \frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)} = \frac{r \sin(4x)}{r \sin(2x)}, $$ and you can finish simplifying it from there.

Answer 4

As other answers have shown, you have already bypassed the quick solution methods.

But if you continue from here: $$ \frac{\sin(4x)\cos x +\sin(2x)\cos(2x)}{\sin(2x)\cos x +\sin x\cos x} $$

by repeatedly applying the double-angle formula for sine, you get

$$ \frac{2\sin(2x)\cos(2x)\cos x + 2\sin x\cos x\cos(2x)}{(2\sin x\cos x)\cos x +\sin x\cos x}, $$

then

$$ \frac{(4\sin x\cos x)\cos(2x)\cos x + 2\sin x\cos x\cos(2x)}{(2\sin x\cos x)\cos x +\sin x\cos x}. $$

Now you have a factor of $\sin x$ and also a factor of $\cos x$ in each term. Cancel them out: $$ \frac{4\cos(2x)\cos x + 2\cos(2x)}{2\cos x + 1}. $$

At this point, think about factoring the numerator. You have an obvious factor of $\cos(2x)$ and a factor of $2$. Once you factor those out of the numerator, the answer should be easy to find.

Answer 5

Similar to @Anne Bauval's answer, but with slightly different intermediate steps.

Let $z = e^{ix} = \cos(x) + i\sin(x)$. Then

$$z^n = e^{inx} = \cos(nx) + i\sin(nx)$$ $$z^{-n} = e^{-inx} = \cos(nx) - i\sin(nx)$$

so $\cos(nx) = \frac{z^n + z^{-n}}{2}$ and $\sin(nx) = \frac{z^n - z^{-n}}{2i}$. Thus:

$$\frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)}$$ $$=\frac{\frac{z^3 - z^{-3}}{2i}+\frac{z^4 - z^{-4}}{2i}+\frac{z^5 - z^{-5}}{2i}}{\frac{z - z^{-1}}{2i}+\frac{z^2 - z^{-2}}{2i}+\frac{z^3 - z^{-3}}{2i}} $$ $$=\frac{z^5 + z^4 + z^3 - z^{-3} - z^{-4} - z^{-5}}{z^3 +z^2 + z - z^{-1} - z^{-2} - z^{-3}}$$ $$=\frac{z^{10} + z^9 + z^8 - z^{2} - z - 1}{z^8 +z^7 + z^6 - z^{4} - z^{3} - z^{2}}$$

Doing polynomial long division, and factoring the $z^2$ out of the denominator, simplifies this to:

$$=z^2 + \frac{z^{6} + z^{5} + z^{4} - z^{2} - z - 1}{z^2(z^6 +z^5 + z^4 - z^{2} - z - 1)}$$ $$=z^2 + \frac{1}{z^2}$$ $$=2\cos(2x)$$

Answer 6

For brevity, let $s = \sin(x)$ and $c = \cos(x)$.

Recall DeMoivre's Formula:

$$\cos(nx) + i\sin(nx) = (c + is)^n$$

We can expand the RHS using the Binomial Theorem:

$$\cos(2x) + i\sin(2x) = c^2 + 2ics - s^2$$ $$\cos(3x) + i\sin(3x) = c^3 + 3ic^2s - 3cs^2 - is^3$$ $$\cos(4x) + i\sin(4x) = c^4 + 4ic^3s - 6c^2s^2 - 4ics^3 + s^4$$ $$\cos(5x) + i\sin(5x) = c^5 + 5ic^4s - 10c^3s^2 - 10ic^2s^3 + 5cs^4 + is^5$$

Taking the imaginary parts of each equation gives us formulas for $\sin(nx)$. (And taking the real parts gives us formula's for $\cos(nx)$.)

$$\sin(2x) = 2cs$$ $$\sin(3x) = 3c^2s - s^3$$ $$\sin(4x) = 4c^3s - 4cs^3$$ $$\sin(5x) = 5c^4s - 10c^2s^3 + s^5$$

Plugging these expressions into the one from your question gives:

$$\frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)}$$ $$=\frac{3c^2s - s^3+4c^3s - 4cs^3+5c^4s - 10c^2s^3 + s^5}{s+2cs+3c^2s - s^3}$$ $$=\frac{3c^2 - s^2 + 4c^3 - 4cs^2 + 5c^4 - 10c^2s^2 + s^4}{1+2c+3c^2 - s^2}$$

At this point, all of the powers of $s$ are even, so let's make the Pythagorean substitution $s^2 = 1 - c^2$, so that we only have one variable to deal with.

$$=\frac{3c^2 - (1 - c^2) + 4c^3 - 4c(1 - c^2) + 5c^4 - 10c^2(1 - c^2) + (1 - c^2)^2}{1+2c+3c^2 - (1 - c^2)}$$ $$=\frac{3c^2 - 1 + c^2 + 4c^3 - 4c + 4c^3 + 5c^4 - 10c^2 + 10c^4 + 1 - 2c^2 + c^4}{1+2c+3c^2 - 1 + c^2}$$ $$=\frac{16c^4 + 8c^3 - 8c^2 - 4c}{4c^2 + 2c}$$ $$=\frac{2(2c^2 - 1)(2c + 1)}{2c + 1}$$ $$=2 (2c^2 - 1)$$ $$=2\cos(2x)$$

Answer 7

Noting that $$ \begin{aligned} 2 \cos (2 x)[\sin x+\sin (2 x)+\sin (3 x)] = & {[\sin (3 x)-\sin x]+\sin (4 x) } +[\sin (5 x)+\sin x] \\ = & \sin (3 x)+\sin (4 x)+\sin (5 x), \end{aligned} $$ we have $$\frac{\sin(3x)+\sin(4x)+\sin(5x)}{\sin(x)+\sin(2x)+\sin(3x)}=2\cos(2x)$$