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

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Better than what? At least give an outline of your own solution. –  Henning Makholm Feb 26 '12 at 1:51
    
my solution is very messy using a lot of identities.. It is not elegant after all.I am thinking that there could be a better approach or perspective in solving this kind of problem. thanks –  Keneth Adrian Feb 26 '12 at 1:56
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@HenningMakholm : Better than any proof that's "too messy and very long". That's what I would presume is meant. –  Michael Hardy Feb 26 '12 at 2:20
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6 Answers

up vote 12 down vote accepted

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 $B = 5x$ leads to the result immediately.

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A. Raina: it isn't nice to make one feel old :-) I don't want even to think about how many years have lapsed since I studied prostaphaeresis: en.wikipedia.org/wiki/Prosthaphaeresis –  Francesco Feb 26 '12 at 8:06
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I'm so old-fashioned that I write "old-fashioned" with a hyphen. –  Michael Hardy Feb 27 '12 at 18:07
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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.

diagram for the above argument

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 $0/0$.

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thanks for the solution. I will recall some topics in complex analysis.. I only use trigometric identities alone, that's why it is to messy.. –  Keneth Adrian Feb 26 '12 at 2:07
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@Ken: The use of complex numbers here is really just to simplify the notation. The idea is geometrical and could be explained in $\mathbb R^2$: The two vectors $(\cos x,\sin x)$ and $(\cos 5x,\sin 5x)$ both have unit length and both make an angle of $2x$ with the $(\cos 3x,\sin 3x)$, but on different sides. So when we add them, by symmetry we must get something parallel to $(\cos 3x,\sin 3x)$. –  Henning Makholm Feb 26 '12 at 2:11
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(But in general, remembering the complex exponential has great simplifying power when manipulating complex trigonometric expressions, also when there isn't an obvious geometric interpretation). –  Henning Makholm Feb 26 '12 at 2:17
    
Yes, your solution is spectacular, using tools in complex analysis.. I will take time to fully grasp the perspective you've shown. thanks –  Keneth Adrian Feb 26 '12 at 2:28
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You keep saying that word ... but there's no complex analysis going on here. I'm just borrowing the exponential as a compact notation for rotations, and for unit vectors pointing in various directions on the plane. –  Henning Makholm Feb 26 '12 at 2:36
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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.

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+1: Cute proof! –  Aryabhata Feb 26 '12 at 2:10
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your solution is elegant.. –  Keneth Adrian Feb 26 '12 at 2:19
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For $\tan\left(\frac{\alpha+\beta}{2}\right) = \frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta}$ one can also note that the right-hand side is the slope of the diagonal in a rhombus whose sides make a angles of $\alpha$ and $\beta$ to the horizontal. And the direction of that diagonal must, by symmetry, be $\frac{\alpha+\beta}{2}$. –  Henning Makholm Feb 26 '12 at 2:39
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@HenningMakholm : That's worth making into another answer. –  Michael Hardy Feb 26 '12 at 4:11
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@HenningMakholm : At en.wikipedia.org/wiki/File:Tan.half.svg , I've credited you with acquainting me with this argument. The picture is now used at en.wikipedia.org/wiki/Tangent_half-angle_formula . –  Michael Hardy Feb 28 '12 at 19:55
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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 $$ \tan(3x)=\frac{\sin(x)+\sin(3x)+\sin(5x)}{\cos(x)+\cos(3x)+\cos(5x)}\tag{4} $$

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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 $\sin 2x$.

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No, what i did is expand the left side and use a lot of trigomertic identities, a lot of them.. it took me 4 pages of bond paper to fully prove the statement –  Keneth Adrian Feb 26 '12 at 2:10
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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!

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+1 for the color-coding - can you teach me how to do that? @kidding@ –  Jose Arnaldo Dris Oct 5 '13 at 14:04
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