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The given answer is $2^{-8}$ My attempt $$=\sin(10^\circ) \sin(30^\circ) \sin(50^\circ)\cdots\sin(170^\circ)$$ after this I multiply and divide to fill in the even multiples of $10^\circ$

$\dfrac{\sin(10)\sin(20)\sin(30) \sin(40)\cdots \sin(160) \sin(170)}{\sin(20)\sin(40)\sin(60)\cdots\sin(160)}$

after this I know there is some identity to simplify the numerator, and a slight modification of that also helps to solve the denominator, but I'm unsure what it is, and how to apply it.

Any help or hint will be appreciated.

Also, note that all angles I've used are measured in degrees.

emacs drives me nuts
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Centelle
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    Why is it not possible to calculate each of the nine sines and multiply them ? – Kurt G. May 27 '22 at 11:54
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    @KurtG. I believe that is not the objective of the exercise - as it doesn’t have much educational value. –  May 27 '22 at 11:55

3 Answers3

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Note that the desired product is $$\prod_{n=1}^{n=9}(\sin(20n-10^o))=\prod_{n=1}^{n=4}(\sin(20n-10^o))\cdot \sin(90^o)\cdot\prod_{n=6}^{n=9}(\sin(20n-10^o))=\left(\prod_{n=1}^{n=4}(\sin(20n-10^o))\right)^2=\sin^2(10^o)\sin^2(30^o)\sin^2(50^o)\sin^2(70^o)$$ Now note that $$\begin{align}\sin(x)\sin(60^o-x)\sin(60^o+x)&=\sin(x)\left(\frac{3\cos^2(x)}{4}-\frac{\sin^2(x)}{4}\right)\\&=\frac{1}{4}\cdot \sin(x) (3-4\sin^2(x))\\&=\frac{\sin(3x)}{4}\end{align}$$ Hence $\sin^2(10^o)\sin^2(30^o)\sin^2(50^o)\sin^2(70^o)=\sin^2(30^o)\left(\frac{\sin(30^o)}{4}\right )^2=2^{-8}$.

Wizard0001
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$$A=\sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \sin(90^\circ) \sin(110^\circ) \sin(130^\circ) \sin(150^\circ) \sin(170^\circ)$$

$$\sin(180^\circ-n)=\sin n \Rightarrow A=\sin^2(10^\circ)\sin^2(30^\circ)\sin^2(50^\circ)\sin^2(70^\circ)\sin(90^\circ)$$

$$\sin(30^\circ)=\frac12,\sin(90^\circ)=1\Rightarrow A=\frac14 \sin^2(10^\circ)\sin^2(50^\circ)\sin^2(70^\circ)$$

$$B=\sin(10^\circ)\sin(50^\circ)\sin(70^\circ)\Rightarrow A=\frac14 B^2$$

$$\sin(90^\circ-n)=\cos n\Rightarrow B=\sin(10^\circ)\cos(20^\circ)\cos(40^\circ)$$

$$2 \sin n \cos n = \sin(2n)\Rightarrow 8B\cos(10^\circ)=8 \sin(10^\circ)\cos(10^\circ)\cos(20^\circ)\cos(40^\circ)=\\ 4\sin(20^\circ)\cos(20^\circ)\cos(40^\circ)=2\sin(40^\circ)\cos(40^\circ)=\sin(80^\circ)$$

$$\sin(90^\circ-n)=\cos n \Rightarrow 8B\cos(10^\circ)=\cos(10^\circ) \Rightarrow B=\frac18$$

$$A=\frac14 B^2 = \frac1{4\cdot 8^2}=2^{-8}$$

Ivan Kaznacheyeu
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Continuing from where you left...

Step 1. Apply product of sine formula $\sin(a) *\sin(b)=\frac{1}{2}(\cos(a-b)-\cos(a+b))$. Simplify. ($\cos 180^o=-1$)

Step 2. Apply half-angle formula. $\frac{(1+\cos(x))}{2}=\cos^2(x/2)$ and $\sin(x)=\frac{1}{2}\sin(x/2)\cos(x/2)$

Step 3. Apply tan product formula. ($\tan 90^o=\infty$) (Derive from $\tan(a+b)$ formula.)

You will reach the desired answer

Aatmaj
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