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

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 $$M \cos 10^\circ$$.

• Further hint: $\sin 70^\circ = \cos 20^\circ$. May 17, 2016 at 0:14

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 $$\sin(10^\circ) \sin(50^\circ) \sin(70^\circ) =-\sin(10^\circ) \sin(50^\circ) \sin(-70^\circ) =-(-\frac18) =\frac18.$$

• – lhf
May 17, 2016 at 0:59

$$\begin{eqnarray*}\sin(A)\sin(B)\sin(C) &=& \frac{1}{2}\left(\cos(A-B)-\cos(A+B)\right)\sin C\\&=&\frac{1}{4}\left(\sin(C+A-B)+\sin(C-A+B)-\sin(C+A+B)-\sin(C-A-B)\right)\end{eqnarray*}$$ 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:

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

• 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? May 17, 2016 at 0:30
• Also, what is the name of the identity used in the first line: sinAsinB=(cos(A-B)-cos(A+B))/2? May 17, 2016 at 0:36
• @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. May 17, 2016 at 0:36
• @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. May 17, 2016 at 0:38
• 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. Jun 24, 2019 at 12:58

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 $$\cos(10) = \sin(80)$$) $$M = 1/8$$

• Use LaTeX please. Jun 24, 2019 at 10:34
• 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. Jun 24, 2019 at 13:06
• I think you are right. In my opinion, down-voting for right solution this is a very bad think. Your solution is true. +1. Jun 24, 2019 at 15:15
• In fact this identity is a direct consequence of Morrie's law (en.wikipedia.org/wiki/Morrie%27s_law). Sep 24, 2020 at 6:07

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

$$M=\dfrac18.$$

Edit: in fact, I just discover that it is a direct consequence of Morrie's law.

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

Proof $\#1:$

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

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

• I don't understand how this (and the other answer you posted here) has to do with the question I posed. May 17, 2016 at 6:41
• @SirJony, Set $x=10^\circ$ May 17, 2016 at 6:53

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 $r=0$

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?