# Show that $\tan 3x =\frac{ \sin x + \sin 3x+ \sin 5x }{\cos x + \cos 3x + \cos 5x}$

I was able to prove this but it is too messy and very long. Is there a better way of proving the identity? Thanks.

#### Solutions Collecting From Web of "Show that $\tan 3x =\frac{ \sin x + \sin 3x+ \sin 5x }{\cos x + \cos 3x + \cos 5x}$"

In old fashioned courses in trigonometry, students were required to remember
the identities
$$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$
and
$$\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$

Applying these formulae in the numerator and denominator, choosing
$A = x$ and $B = 5x$

You want to prove
$$\frac{\sin 3x}{\cos 3x}=\frac{\sin x+\sin 3x+\sin 5x}{\cos x+\cos 3x+\cos 5x}$$
Or, in other words, that the two vectors $(\cos3x,\sin3x)$ and $(\cos x+\cos 3x+\cos 5x,\sin x+\sin 3x+\sin 5x)$ are parallel. The latter is the sum of $(\cos x,\sin x)$, $(\cos 3x,\sin 3x)$ and $(\cos 5x,\sin5x)$.

Now, $(\cos x,\sin x)$ and $(\cos 5x,\sin5x)$ both have unit length, so by the parallelogram rule, $(\cos x,\sin x)+(\cos 5x,\sin5x)$ is the diagonal of a rhombus, and by symmetry the direction of the diagonal must be halfway between the angles of the sides — that is $\frac{x+5x}{2}=3x$. So $(\cos x,\sin x)+(\cos 5x,\sin5x)$ lies even with the $(\cos3x,\sin3x)$ term and the sum of all three vectors is parallel to $(\cos3x,\sin3x)$, as required.

This geometric argument mostly closes the case, but note (because that’s how I wrote it at first) that it can be made to look slick and algebraic by moving to the complex plane. Then saying that the two vectors are parallel is is the same as saying that $e^{3xi}$ and $e^{xi}+e^{3xi}+e^{5xi}$ are real multiples of each other.

But $e^{xi}+e^{5xi}=e^{3xi}(e^{-2xi}+e^{2xi})$ and the factor in the parenthesis is real because it is the sum of a number and its conjugate. In particular, by Euler’s formula,
$$e^{xi}+e^{3xi}+e^{5xi} = e^{3xi}(1+2\cos 2x)$$
and the two vectors are indeed parallel and your identity holds — except when $\cos 2x=-\frac 12$, in which case the fraction to the right of your desired identity is $0/0$.

Notice that $\tan 3x = \sin 3x/\cos 3x$, and if $a/b=c/d$ then $a/b=c/d=(a+c)/(b+d)$, so it’s enough to prove
$$\tan 3x = \frac{\sin x + \sin 5x}{\cos x + \cos 5x}.$$
So generally, how does one prove
$$\tan\left(\frac{\alpha+\beta}{2}\right) = \frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}\ ?$$

One can say that
$$\sin \alpha = \sin\left( \frac{\alpha+\beta}{2} + \frac{\alpha-\beta}{2} \right)$$
$$\sin \beta = \sin\left( \frac{\alpha+\beta}{2} – \frac{\alpha-\beta}{2} \right)$$
and do the same for the two cosines, then apply the formulas for sine of a sum and cosine of sum. After that, it’s trivial simplification.

The identities for the sum of sines and the sum of cosines yield
$$\frac{\sin(x)+\sin(y)}{\cos(x)+\cos(y)}=\frac{2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}{2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}=\tan\left(\frac{x+y}{2}\right)\tag{1}$$
Equation $(1)$ implies that
$$\frac{\sin(x)+\sin(5x)}{\cos(x)+\cos(5x)}=\tan(3x)=\frac{\sin(3x)}{\cos(3x)}\tag{2}$$
We also have that if $b+d\not=0$, then
$$\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\tag{3}$$
Combining $(2)$ and $(3)$ yields
$$\tan(3x)=\frac{\sin(x)+\sin(3x)+\sin(5x)}{\cos(x)+\cos(3x)+\cos(5x)}\tag{4}$$

This is may be what you came up with, but I don’t personally think it’s all that bad:
Cross-multiply and cancel $\sin3x\cos3x$ from each side. You have
$$\cos3x\sin x+\cos3x\sin5x=\sin3x\cos x+\sin3x\sin5x$$
$$\sin3x\cos x – \cos3x\sin x=\cos3x\sin5x – \sin3x\cos 5x$$
By the angle addition/subtraction formulas, both sides are equal to $\sin 2x$.

More generally, for any arithmetic sequence, denoting $z=\exp(i x)$ and $2\ell=an+2b$, we have

$$\begin{array}{c l} \blacktriangle & =\frac{\sin(bx)+\sin\big((a+b)x\big)+\cdots+\sin\big((na+b)x\big)}{\cos(bx)+\cos\big((a+b)x\big)+\cdots+\cos\big((na+b)x\big)} \\[2pt] & \color{Red}{\stackrel{1}=} \frac{1}{i}\frac{z^b\big(1+z^a+\cdots+z^{na}\big)-z^{-b}\big(1+z^{-a}+\cdots+z^{-na}\big)}{z^b\big(1+z^a+\cdots+z^{na}\big)+z^{-b}\big(1+z^{-a}+\cdots+z^{-na}\big)} \\[2pt] & \color{LimeGreen}{\stackrel{2}=}\frac{1}{i}\frac{z^b-z^{-b}z^{-na}}{z^b+z^{-b}z^{-na}} \\[2pt] & \color{Blue}{\stackrel{3}=}\frac{(z^\ell-z^{-\ell})/2i}{(z^\ell+z^{-\ell})/2} \\[2pt] & \color{Red}{\stackrel{1}{=}}\frac{\sin (\ell x)}{\cos(\ell x)}. \end{array}$$

Hence $\blacktriangle$ is $\tan(\ell x)$ – observe $\ell$ is the average of the first and last term in the arithmetic sequence.

$\color{Red}{(1)}$: Here we use the formulas $$\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i} \qquad \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}.$$

$\color{LimeGreen}{(2)}$: Here we divide numerator and denominator by $1+z^a+\cdots+z^{na}$.

$\color{Blue}{(3)}$: Multiply numerator and denominator by $z^{na/2}/2$.

Note: there are no restrictions on $a$ or $b$ – they could even be irrational!