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Mathematics 20 Online
OpenStudy (anonymous):

sin 15 degree=^3-1/2^2.prove it

OpenStudy (anonymous):

cos2x=1-2sin^(2)x put x=15 cos30=1-2sin^2{15} 2sin^(2){15}=1-c0s30

OpenStudy (anonymous):

\[2\sin^215=1-\sqrt3/2=\frac{2-\sqrt3}{2}\]\[\sin^215=\frac{2-\sqrt3}{4}\]

OpenStudy (anonymous):

\[\sin15=\frac {\sqrt{2-\sqrt3}}2\]

OpenStudy (anonymous):

sin(15) = sin(45 - 30) = sin45cos30 - cos45sin30 = (1/sqrt2)(sqrt3/2) - (1/sqrt2)(1/2) = (sqrt3-1)/(2sqrt2) = sqrt2(sqrt3 - 1)/4

OpenStudy (anonymous):

\[\sin15=\frac{\sqrt{4-2\sqrt3}}{2\sqrt2}=\frac{\sqrt{(\sqrt3-1)^2}}{2\sqrt2}=\frac{\sqrt3-1}{2\sqrt2}\]

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