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Question Number 116768 by bemath last updated on 06/Oct/20

$$\sin \left({4\sin }^{-1}\left(\mathrm{x}\right)\right)=\sin \left({2\sin }^{-1}\left(\mathrm{x}\right)\right)$$

Answered by bobhans last updated on 06/Oct/20

$$\mathrm{letting}{\sin }^{-1}\left(\mathrm{x}\right)=\mathrm{z}\Rightarrow \sin \mathrm{z}=\mathrm{x}$$$$\Rightarrow \sin \left(4\mathrm{z}\right)=\sin \left(2\mathrm{z}\right)$$$$\Rightarrow 2\sin \left(2\mathrm{z}\right)\cos \left(2\mathrm{z}\right)-\sin \left(2\mathrm{z}\right)=0$$$$\Rightarrow \{\begin{array}{l}\sin \left(2\mathrm{z}\right)=0\mathrm{or}\\ \cos \left(2\mathrm{z}\right)=\frac{1}{2}\end{array}$$$$\mathrm{case}\left(1\right)\Rightarrow \sin 2\mathrm{z}=0$$$$\Rightarrow 2\sin \mathrm{z}.\cos \mathrm{z}=0;2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}=0$$$$\mathrm{we}\mathrm{get}\mathrm{x}=0\mathrm{or}\mathrm{x}=\pm 1$$$$\mathrm{case}\left(2\right)\Rightarrow \cos \left(2\mathrm{z}\right)=\frac{1}{2}$$$$\Rightarrow 1-{2\sin }^2\mathrm{z}=\frac{1}{2};\frac{1}{2}-{2\mathrm{x}}^2=0$$$$\Rightarrow {\left(\frac{1}{2}\right)}^2-{\mathrm{x}}^2=0\Rightarrow \mathrm{x}=\pm \frac{1}{2}$$$$\mathrm{therefore}\mathrm{we}\mathrm{get}\mathrm{solution}\mathrm{set}\mathrm{is}$$$$\mathrm{x}\in \{0,\pm \frac{1}{2},\pm 1\}$$

Answered by MJS_new last updated on 06/Oct/20

$$A=\sin \left(4\arcsin x\right)=4x(1-2x^2)\sqrt{1-x^2}$$$$B=\sin \left(2\arcsin x\right)=2x\sqrt{1-x^2}$$$$A-B=0$$$$2x(1-4x^2)\sqrt{1-x^2}=0$$$$\Rightarrow x=0\vee x=\pm \frac{1}{2}\vee x=\pm 1$$

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