complex trig

$\quad\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ)\sin(60^\circ + 10^\circ)$

Now, use the identity: $4\sin x \sin (60^\circ - x)\sin(60^\circ + x) = \sin 3x$ and substitute x = 10^o.

$\quad \sin10^\circ \sin30^\circ \sin50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\\ = \dfrac18 \left(4\sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\right)\\ = \dfrac18 \left(\sin(3\cdot 10^\circ)\right)\\ = \dfrac18 \sin 30^\circ\\ = \dfrac18 \cdot \dfrac12\\ = \dfrac1{16}$

$\large \sin10^\circ \cdot \sin30^\circ \cdot \sin50^\circ \cdot \sin70^\circ\color{blue}= 0.0625$

$Hello\ Guest!$  

${\displaystyle {sin\ x=\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} x}-\mathrm {e} ^{-\mathrm {i} x}\right)}$

$\large \sin\frac{\pi}{18} \cdot \sin\frac{3\pi}{18} \cdot \sin\frac{5\pi}{18} \cdot \sin\frac{7\pi}{18}$

 $={\displaystyle {\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{\pi}{18}}\right)}\times$${\displaystyle {\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{3\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{3\pi}{18}}\right)}$

              $\times {\displaystyle {\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{5\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{5\pi}{18}}\right)}$${\displaystyle {\times \frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{7\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{7\pi}{18}}\right)}$

$\large =0.0625$

  !

Compute  
$\large \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ)$

Formula:
$\begin{array}{|lrcll|} \hline (1) & \cos(x-y) &=& \cos(x)\cos(y)+\sin(x)\sin(y) \\ (2) & \cos(x+y) &=& \cos(x)\cos(y)-\sin(x)\sin(y)\\ \hline (1)-(2): & \mathbf{2\sin(x)\sin(y)} &=& \mathbf{\cos(x-y)-\cos(x+y)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \quad | \quad \sin(30^\circ)=\dfrac{1}{2} \\ &=& \dfrac{1}{2}* \sin10(^\circ) \sin(70^\circ) \sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(70^\circ)=\cos(60^\circ)-\cos(80^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\cos(80^\circ) \quad | \quad \cos(80^\circ)=\cos(90^\circ-10^\circ)=\sin(10^\circ) \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\sin(10^\circ) \\ &&\quad \mathbf{\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ)} \\ \\ \hline &=& \dfrac{1}{2}\left( \dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ) \right) \sin(50^\circ) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\sin(10^\circ)\sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\cos(60^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\dfrac{1}{2} \quad | \quad \cos(40^\circ)=\cos(90^\circ-50^\circ)=\sin(50^\circ) \\ &&\quad 2\sin(10^\circ)\sin(50^\circ)=\sin(50^\circ)-\dfrac{1}{2} \\ &&\quad \mathbf{\sin(10^\circ)\sin(50^\circ)=\dfrac{1}{2}*\sin(50^\circ)-\dfrac{1}{4} } \\ \\ \hline &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\left(\dfrac{1}{2}\sin(50^\circ)-\dfrac{1}{4}\right) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{8}\sin(50^\circ)+\dfrac{1}{16} \\ &=& \mathbf{\dfrac{1}{16}} \\ \hline \end{array}$