Question #d07cf

Question

1571 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-11T09:25:05

$sin(-pi/12)$

$=-sin(pi/12)$

$=-sqrt(1/2(1-cos((2xxpi)/12)$

$=-sqrt(1/2(1-cos(pi/6)$

$=-sqrt(1/2(1-sqrt3/2)$

$=-sqrt(1/8(4-2sqrt3)$

$=-sqrt(1/8((sqrt3)^2+1^2-2sqrt3xx1)$

$=-sqrt(1/8((sqrt3-1)^2)$

$=-1/(2sqrt2)(sqrt3-1)$

Answer 2 · 2016-12-12T19:07:44

$(sqrt2-sqrt6)/4$

$sin(-pi/12) =sin(pi/6-pi/4)$

Now use the formula $sin(A-B) =sin A cos B-cos A sin B$ to evaluate $sin(pi/6-pi/4)$ . That is,

$sin(pi/6-pi/4)=sin (pi/6) cos (pi/4) - cos (pi/6) sin (pi/4)$

$=1/2*sqrt2/2 - sqrt3/2*sqrt2/2$

$=sqrt2/4-sqrt6/4$

$:=(sqrt2-sqrt6)/4$