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Question Number 157644 by Ar Brandon last updated on 26/Oct/21

Prove (1/2)(√(2−(√3)))=(((√6)−(√2))/4)

$$\mathrm{Prove}\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Answered by ajfour last updated on 26/Oct/21

v=(1/2)(√(2−(√3)))  v^2 =(1/2)−((√3)/4)=((2−(√3))/4)      =((8−(√(48)))/(16))=((((√6)−(√2))/4))^2     v=(((√6)−(√2))/4)

$${v}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${v}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{8}−\sqrt{\mathrm{48}}}{\mathrm{16}}=\left(\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$\:\:{v}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$ \\ $$

Commented by Ar Brandon last updated on 26/Oct/21

thanks

$$\mathrm{thanks} \\ $$

Answered by mindispower last updated on 26/Oct/21

(1/2)(√(2−(√3)))=(√((1−(((√3)/2)))/2))=(√((1+cos(((5π)/6)))/2))  cos(2x)=2cos^2 (x)−1 used  =∣cos(((5π)/(12)))∣=cos((π/6)+(π/4))=((√3)/2).((√2)/2)−(1/2).((√2)/2)=(((√6)−(√2))/4)

$$\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\sqrt{\frac{\mathrm{1}−\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{2}}}=\sqrt{\frac{\mathrm{1}+{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)}{\mathrm{2}}} \\ $$$${cos}\left(\mathrm{2}{x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}\:{used} \\ $$$$=\mid{cos}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\mid={cos}\left(\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{4}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

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