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Prerequisite

This problem is found in "Trigonometry" by I. M. Gelfand [in English].

It is asked in the section "Double the angle". So, assume that I know the sin/cos angle additions [i.e.: $\sin(A + B) = \sin A \cos B + \cos A \sin B$, etc.] as well as everything learned prior.

I've check other sources and they say to use Morrie's Law, however I have not actually learned it in the book.

Problem

Find the numerical value of $\sin 10^\circ \sin 50^\circ \sin 70^\circ$.

Hint: If the value of the given expression is $M$, find $M \cos 10^\circ$.

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    $\begingroup$ Further hint: $\sin 70^\circ = \cos 20^\circ$. $\endgroup$ – Noam D. Elkies May 17 '16 at 0:14
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$$\begin{eqnarray*}\sin(A)\sin(B)\sin(C) &=& \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\sin C\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)-\sin(C+A+B)-\sin(C-A-B)\right)\end{eqnarray*}$$ hence:

$$\begin{eqnarray*} \sin(10^\circ)\sin(50^\circ)\sin(70^\circ)&=&\frac{1}{4}\left(\sin(110^\circ)+\sin(30^\circ)+\sin(-10^\circ)-\sin(130^\circ)\right) \\&=&\frac{1}{4}\left(\cos(20^\circ)+\sin(30^\circ)-\sin(10^\circ)-\cos(40^\circ)\right)\end{eqnarray*}$$ but: $$ \cos(20^\circ)-\cos(40^\circ) = 2\sin(30^\circ)\sin(10^\circ) = \sin(10^\circ)$$ hence:

$$ \sin(10^\circ)\sin(50^\circ)\sin(70^\circ) = \color{red}{\frac{1}{8}}.$$

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  • $\begingroup$ Thank you! Although this is true, and I'll probably "check" this as the right answer, what is the purpose of Gelfand providing the hint? How could one solve it by taking it into consideration? $\endgroup$ – Fine Man May 17 '16 at 0:30
  • $\begingroup$ Also, what is the name of the identity used in the first line: sinAsinB=(cos(A-B)-cos(A+B))/2? $\endgroup$ – Fine Man May 17 '16 at 0:36
  • $\begingroup$ @SirJony: you may notice I used the $\sin$/$\cos$ addition formulas twice, but if you multiply your expression by $\cos(10^\circ)$ and "couple" the right terms in a slick way, you get the answer with less computations. However, my approach or Gelfand's intended one are more or less the same. $\endgroup$ – Jack D'Aurizio May 17 '16 at 0:36
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    $\begingroup$ @SirJony: I think the formula has a different name depending on your location: Briggs, Werner or prostapheresis (en.wikipedia.org/wiki/Prosthaphaeresis). I just call it an instance of the addition formulas. $\endgroup$ – Jack D'Aurizio May 17 '16 at 0:38
  • $\begingroup$ This is a nice solution but I don't think it's the one the problem-setter had in mind, since it doesn't make use of the hint given. $\endgroup$ – Jack Kinsella Jun 24 at 12:58
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From the triple-angle formula $\sin (3\theta) = - 4\sin^3\theta + 3\sin\theta$ when $\sin (3\theta) = 1/2$, we get that $\sin(10^\circ)$, $\sin(50^\circ)$, $\sin(-70^\circ)$ are the roots of $8x^3-6x+1$. Therefore $$ \sin(10^\circ) \sin(50^\circ) \sin(70^\circ) =-\sin(10^\circ) \sin(50^\circ) \sin(-70^\circ) =-(-\frac18) =\frac18. $$

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We can prove $$4\sin(60^\circ-x)\sin x\sin(60^\circ+x)=\sin3x$$

Proof $\#1:$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin(60^\circ-x)\sin(60^\circ+x)=\sin^260^\circ-\sin^2x?$$

and $\sin3x=3\sin x-4\sin^3x$

Proof$\#2:$

If $\sin3x=\sin3A, 3x=180^\circ n+(-1)^n3A$ where $n$ is any integer

$\implies x=60^\circ n+(-1)^nA$ where $n\equiv-1,0,1\pmod3$

$\implies x=-(60^\circ+A), A,60^\circ-A$

As $\sin3x=3\sin x-4\sin^3x,3\sin x-4\sin^3x=\sin3A\iff4\sin^3x-3\sin x-\sin3A=0$

Using Vieta's formula, $$\prod_{n=-1}^1\sin\left(60^\circ n+(-1)^nA\right)=\dfrac{\sin3A}4$$

Use $\sin\left\{-(60^\circ+A)\right\}=-\sin(60^\circ+A)$

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  • $\begingroup$ I don't understand how this (and the other answer you posted here) has to do with the question I posed. $\endgroup$ – Fine Man May 17 '16 at 6:41
  • $\begingroup$ @SirJony, Set $x=10^\circ$ $\endgroup$ – lab bhattacharjee May 17 '16 at 6:53
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Using $\sin(90^\circ-x)=\cos x$

$$\sin10^\circ\sin50^\circ\sin70^\circ=\cos20^\circ\cos40^\circ\cos80^\circ$$

Now $S=\cos x\cos2x\cos4x=\dfrac{\sin2x\cos2x\cos4x}{2\sin x}$ if $\sin x\ne0\iff x\ne180^\circ m$ where $m$ is any integer

$\implies8\sin x\cdot S=\sin8x$

If $\sin8x=\sin x\iff8S=1$

We need $8x=n180^\circ+(-1)^nx$ where $n$ is any integer

If $n$ is even, $=2m$(say), $7x=360^\circ m$ where $7\nmid m$

If $n$ is odd, $=2m+1$(say), $x=(2r+1)20^\circ$ where $r$ is any integer , but $2r+1\not\equiv9\pmod{18}\iff r\not\equiv4\pmod9$

In this problem $r=0$

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Here is a method that uses Gelfand's hint.

Let us first transform all the sines into cosines of the complementary angle:

$\tag{1}M:=\sin(10^\circ)\sin(50^\circ)\sin(70^\circ)=\cos(80^\circ)\cos(40^\circ)\cos(20^\circ)$

Let us multiply LHS and RHS of $(1)$ by $\sin(20^\circ):$

$M \sin(20^\circ)=\cos(80^\circ)\cos(40^\circ)(\cos(20^\circ)\sin(20^\circ).$

$=\cos(80^\circ)\cos(40^\circ)\frac12\sin(40^\circ).$

(applying the so useful formula $\sin(a)\cos(a)=\frac12\sin(2a).$)

Using the same trick again, we get:

$M \sin(20^\circ)=\frac14\cos(80^\circ)\sin(80^\circ).$

And once more:

$M \sin(20^\circ)=\frac18\sin(160^\circ).$

But $\sin(160^\circ)=\sin(20^\circ).$

We can thus conclude that

$M=\dfrac18.$

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I would use a variation of Jean Marie's approach to this. Let $$ 𝑀 =\sin(10)\sin(50)\sin(70) $$ (multiply both sides by $\cos(10)$ as per the hint) $$ \cos(10)M = \cos(10)\sin(10)\sin(50)\sin(70) $$ (apply double angle rule $\sin(a)\cos(a)=1/2\sin(2𝑎)$) $$ \cos(10)M = 1/2\sin(20)\sin(50)\sin(70) $$ (apply $\sin(x) = \cos(90-x)$) $$ \cos(10)M = 1/2\sin(20)\sin(50)\cos(20) $$ (rearrange) $$ \cos(10)M = 1/2 \sin(20)\cos(20)\sin(50) $$ (double angle rule) $$ \cos(10)M = 1/4 \sin(40) \sin(50) $$ (complimentary cosine and double angle again) $$ \cos(10)M = 1/8 \sin(80) $$ (cancel out because $\cos(10) = \sin(80)$) $$ M = 1/8 $$

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    $\begingroup$ Use LaTeX please. $\endgroup$ – Michael Rozenberg Jun 24 at 10:34
  • $\begingroup$ To whoever down-voted: Can you comment on why this is wrong? This is my first contribution and it's quite demotivating to be told the community disapproves of my attempt without knowing why. I realize I should have used latex, but formatting is orthogonal to correctness. $\endgroup$ – Jack Kinsella Jun 24 at 13:06
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    $\begingroup$ I think you are right. In my opinion, down-voting for right solution this is a very bad think. Your solution is true. +1. $\endgroup$ – Michael Rozenberg Jun 24 at 15:15
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Here's the strategy that I would use.

Note that $$\sin 10^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ}{\sin 30^\circ}$$

And that $$\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ = \frac{\sin 10^\circ\sin 20^\circ\sin 30^\circ\sin 40^\circ \sin 50^\circ\sin 60^\circ\sin 70^\circ\sin 80^\circ}{\sin 20^\circ\sin 40^\circ\sin 60^\circ\sin 80^\circ}$$

See where this is going?

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