Question

sin ^3 10+sin^3 250-sin^3 310=​

Answer 1

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Answer 2

The answer is $\frac{-3}{8}$

We use identity $(sinx)^3=(3sinx)-(sin(3x))/4$

All the angles below are in degrees.

$0.25(3 sin10-sin 30+3sin 50-sin 150-3sin 70+sin 210)=0.25(3sin10-0.5+2sin50-0.5-3sin70-0.5)=0.25(3sin10+3sin50-3sin70-1.5)=0.75(sin10+sin50-sin70-0.5)$

We now apply another identity $sinx+siny=2$

$sin(x+y)/2 cos(x-y)/2$ to

sum$(sin10+sin 50)$

$sin10+sin50=2sin30^\circ cos20=cos 20=sin 70$

The expression then becomes

$0.75(sin 70-sin 70-0.5)-0.75\times 0.5=-0/375$

$=-3/8$

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