+9519 \(\quad\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ)\sin(60^\circ + 10^\circ)\)
Now, use the identity: \(4\sin x \sin (60^\circ - x)\sin(60^\circ + x) = \sin 3x\) and substitute x = 10o.
\(\quad \sin10^\circ \sin30^\circ \sin50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\\ = \dfrac18 \left(4\sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\right)\\ = \dfrac18 \left(\sin(3\cdot 10^\circ)\right)\\ = \dfrac18 \sin 30^\circ\\ = \dfrac18 \cdot \dfrac12\\ = \dfrac1{16}\)
+14913 \(\large \sin10^\circ \cdot \sin30^\circ \cdot \sin50^\circ \cdot \sin70^\circ\color{blue}= 0.0625 \)
\(Hello\ Guest!\)
\({\displaystyle {sin\ x=\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} x}-\mathrm {e} ^{-\mathrm {i} x}\right)}\)
\(\large \sin\frac{\pi}{18} \cdot \sin\frac{3\pi}{18} \cdot \sin\frac{5\pi}{18} \cdot \sin\frac{7\pi}{18}\)
\(={\displaystyle {\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{\pi}{18}}\right)}\times\)\({\displaystyle {\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{3\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{3\pi}{18}}\right)}\)
\(\times {\displaystyle {\frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{5\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{5\pi}{18}}\right)}\)\({\displaystyle {\times \frac {1}{2\mathrm {i} }}\left(\mathrm {e} ^{\mathrm {i} \frac{7\pi}{18}}-\mathrm {e} ^{-\mathrm {i} \frac{7\pi}{18}}\right)}\)
\(\large =0.0625\)
!
+9519 \(\quad\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin 50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ)\sin(60^\circ + 10^\circ)\)
Now, use the identity: \(4\sin x \sin (60^\circ - x)\sin(60^\circ + x) = \sin 3x\) and substitute x = 10o.
\(\quad \sin10^\circ \sin30^\circ \sin50^\circ \sin 70^\circ\\ = \dfrac12 \sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\\ = \dfrac18 \left(4\sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)\right)\\ = \dfrac18 \left(\sin(3\cdot 10^\circ)\right)\\ = \dfrac18 \sin 30^\circ\\ = \dfrac18 \cdot \dfrac12\\ = \dfrac1{16}\)
+9519 Appendix: Why is \(4 \sin x \sin(60^\circ - x) \sin(60^\circ + x) = \sin 3x\\ \)?
Proof:
Using compound angle formula,
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = 4\sin x \left(\sin 60^\circ \cos x - \cos 60^\circ \sin x\right)\left(\sin 60^\circ \cos x + \cos 60^\circ \sin x\right)\)
Simplifying,
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = 4\sin x \left(\dfrac{\sqrt 3}2 \cos x - \dfrac12 \sin x\right)\left(\dfrac{\sqrt 3}2 \cos x + \dfrac12 \sin x\right) \)
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = \sin x \left(\sqrt 3 \cos x - \sin x\right)\left(\sqrt 3 \cos x + \sin x\right)\)
Now, by the identity \(a^2 - b^2 = (a - b)(a + b)\),
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = \sin x (3\cos^2 x - \sin^2 x)\)
Starting from the right hand side, expanding using triple angle formula,
(if you haven't learnt that yet, you can try expanding it with compound angle formula and double angle formula.)
\(\sin 3x = 3\sin x - 4\sin^3 x\)
Now, it suffices to show that \(\sin x (3\cos^2 x -\sin^2 x) = 3\sin x - 4\sin^3 x\)
To do so, we use the identity \(\cos^2 x = 1-\sin^2 x\).
\(\sin x (3\cos^2 x -\sin^2 x) = \sin x(3(1 - \sin^2 x) - \sin^2 x) = \sin x (3 - 3\sin^2 x - \sin^2 x) = \sin x (3 - 4\sin^2 x) = 3\sin x - 4\sin^3 x\)
Therefore, the original identity \(4 \sin x \sin(60^\circ - x) \sin(60^\circ + x) = \sin 3x\\ \) is true.
+26367 Compute
\(\large \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ)\)
Formula:
\(\begin{array}{|lrcll|} \hline (1) & \cos(x-y) &=& \cos(x)\cos(y)+\sin(x)\sin(y) \\ (2) & \cos(x+y) &=& \cos(x)\cos(y)-\sin(x)\sin(y)\\ \hline (1)-(2): & \mathbf{2\sin(x)\sin(y)} &=& \mathbf{\cos(x-y)-\cos(x+y)} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline && \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \quad | \quad \sin(30^\circ)=\dfrac{1}{2} \\ &=& \dfrac{1}{2}* \sin10(^\circ) \sin(70^\circ) \sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(70^\circ)=\cos(60^\circ)-\cos(80^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\cos(80^\circ) \quad | \quad \cos(80^\circ)=\cos(90^\circ-10^\circ)=\sin(10^\circ) \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\sin(10^\circ) \\ &&\quad \mathbf{\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ)} \\ \\ \hline &=& \dfrac{1}{2}\left( \dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ) \right) \sin(50^\circ) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\sin(10^\circ)\sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\cos(60^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\dfrac{1}{2} \quad | \quad \cos(40^\circ)=\cos(90^\circ-50^\circ)=\sin(50^\circ) \\ &&\quad 2\sin(10^\circ)\sin(50^\circ)=\sin(50^\circ)-\dfrac{1}{2} \\ &&\quad \mathbf{\sin(10^\circ)\sin(50^\circ)=\dfrac{1}{2}*\sin(50^\circ)-\dfrac{1}{4} } \\ \\ \hline &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\left(\dfrac{1}{2}\sin(50^\circ)-\dfrac{1}{4}\right) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{8}\sin(50^\circ)+\dfrac{1}{16} \\ &=& \mathbf{\dfrac{1}{16}} \\ \hline \end{array} \)
