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# complex trig

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Compute $$\large \sin10^\circ \sin30^\circ \sin50^\circ \sin70^\circ$$

Jul 11, 2020

#2
+8352
+2

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

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

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

Jul 11, 2020

#1
+10600
+1

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

$$Hello\ Guest!$$

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

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

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

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

$$\large =0.0625$$

!

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Jul 11, 2020
edited by asinus  Jul 11, 2020
#2
+8352
+2

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

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

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

MaxWong Jul 11, 2020
#3
+8352
+2

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.

MaxWong  Jul 11, 2020
#4
+25598
+1

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

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

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

Jul 12, 2020
edited by heureka  Jul 13, 2020