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August 2022
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[tex] \sqrt{2}cos \ pi /12 + \sqrt{6}sin \ pi /12 [/tex]
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Zhiraffe
=кореньиздвух*(1*cos(pi/12)+кореньизтрех*sin(pi/12)=
=2*кореньиздвух*(1/2*cos(pi/12)+кореньизтрех/2*sin(pi/12)=
=2*кореньиздвух*(sin(pi/6)*cos(pi/12)+cos(pi/6)*sin(pi/12))=
=2*кореньиздвух*sin(pi/6+pi/12)=
=2*кореньиздвух*sin(pi/4)=2*кореньиздвух*кореньиздвух/2=
=2*2/2=2
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Answers & Comments
=2*кореньиздвух*(1/2*cos(pi/12)+кореньизтрех/2*sin(pi/12)=
=2*кореньиздвух*(sin(pi/6)*cos(pi/12)+cos(pi/6)*sin(pi/12))=
=2*кореньиздвух*sin(pi/6+pi/12)=
=2*кореньиздвух*sin(pi/4)=2*кореньиздвух*кореньиздвух/2=
=2*2/2=2