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I came across an expression: $$(\sin10)^3+(\sin50)^3-(\sin70)^3$$

All the angles are in degrees. Can we calculate it without using calculator by using the trigonometric identities?? I have tried to manipulate this expression such as using sum/difference formulas, but it didn't help. Any help is appreciated.

resilient
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    Hint: $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$. – Sammy Black Nov 18 '24 at 17:43
  • Are the $10,50, \text{ and }70$ degrees? You should say so. The default is radians. If I were doing it I would get the angles close to $0$ or $90$ degrees, then use Taylor's series. – Ross Millikan Nov 18 '24 at 17:59
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    $(\sin x)^3$ can be expressed in terms of $\sin 3x$ and $\cos 3x$. Since all the angles in the original expression are multiples of $10°$, all the arguments in the rewritten expression will be multiples of $30°$. – MJD Nov 18 '24 at 18:24
  • The comment of Sammy Black allows you to convert the original expression into an expression of linear terms $~\sin(a) + \sin(b) + \cdots .~$ At this point, my next (experimental) step, which might well fail to solve the problem is to notice that $~\sin(a+b) + \sin(a - b) = 2\sin(a)\cos(b).~$ I would strive to associate $~(a+b)~$ (et al) with special angles, like $~30^\circ, 45^\circ, 60^\circ.~$ As I say, this approach might well fail. – user2661923 Nov 18 '24 at 18:26
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    $$(\sin\frac{\pi}{18})^3+(\sin\frac{5\pi}{18})^3-(\sin\frac{7\pi}{18})^3=-\frac{3}{8}$$ – JJacquelin Nov 18 '24 at 18:45

3 Answers3

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Let $S$ be the sum of $\pm$ cubes of sine values to be computed. We use $4\sin^3 t=3\sin t -\sin (3t)$ for $t$ among $10^\circ$, $50^\circ$, $70^\circ$. Then: $$ \begin{aligned} S&:=\sin ^3 10^\circ + \sin ^3 50^\circ - \sin ^3 70^\circ\ ,\\ 4S &= 4\sin ^3 10^\circ + 4\sin ^3 50^\circ - 4\sin ^3 70^\circ\\ &=3(\sin 10^\circ + \sin 50^\circ - \sin 70^\circ)\\ &\qquad -\sin 30^\circ -\underbrace{\sin 150^\circ}_{=\sin 30^\circ} +\underbrace{\sin 210^\circ}_{=-\sin 30^\circ} \\ &=-3(\sin 70^\circ-\sin 50^\circ+\sin 30^\circ-\sin 10^\circ) \\ &=-6(\cos 60^\circ\sin 10^\circ+\cos 20^\circ\sin 10^\circ) \\ &=-6\sin 10^\circ(\cos (40^\circ+20^\circ)+\cos (40^\circ-20^\circ)) \\ &=-12\sin 10^\circ\cos 20^\circ\cos 40^\circ \\ &=-\frac 6{\cos 10^\circ}\cdot \underbrace{2\sin 10^\circ\cos 10^\circ}_{\sin 20^\circ} \cdot \cos 20^\circ\cos 40^\circ \\ &=-\frac 3{\cos 10^\circ}\cdot \underbrace{2\sin 20^\circ\cos 20^\circ}_{\sin 40^\circ} \cdot \cos 40^\circ \\ &=-\frac 3{2\cos 10^\circ}\cdot \underbrace{2\sin 40^\circ\cos 40^\circ}_{\sin 80^\circ} \\ &=-\frac 32\cdot\frac{\sin80^\circ}{\cos 10^\circ}=-\frac 32\ . \end{aligned} $$ From here we obtain the value $S=-3/8$ for the given sum.

dan_fulea
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$\sin10^\circ$, $\sin50^\circ$ and $-\sin70^\circ=\sin(-70)$ are the roots of $8x^3-6x+1$ (see this). Therefore, $$8((\sin10^\circ)^3+(\sin50^\circ)^3-(\sin70^\circ)^3)=6((\sin10^\circ)+(\sin50^\circ)-(\sin70^\circ))-3= -3$$ and so the answer is $-3/8$ because the sum of the three roots is zero.

lhf
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Using the identity $\sin(\alpha)-\sin(\beta)=2\sin\left(\dfrac{\alpha-\beta}2\right)\cos\left(\dfrac{\alpha+\beta}2\right)$, we can prove that $\sin(10^{\circ})+\sin(50^{\circ})-\sin(70^{\circ})=0$. Note that for $a,b,c\in\mathbb{R}$, $a+b+c=0$ implies $a^3+b^3+c^3=3abc$. Hence the expression can be written as follows,

$$\begin{aligned}(\sin10^{\circ})^3+(\sin50^{\circ})^3-(\sin70^{\circ})^3&= -3\sin(10^{\circ})\sin(50^{\circ})\sin(70^{\circ})\\&=-3\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})\end{aligned}$$

Now multiply the expression by $\dfrac{\sin(20^{\circ})}{\sin(20^{\circ})}$. Can you take it from here?

user1380196
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