Q: What is the sin pi over 6?

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Sin(2*pi/6) = sin(pi/3) which, by definition, is 0.5 If you wish, you can calculate y/1! - y^3/3! + y^5/5! - y^7/7! + ... where y = pi/3.

11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4

1/2

Negative 0.5 is.

sin(pi) = 0

Related questions

cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2

sin(pi/2)=1

Sin(2*pi/6) = sin(pi/3) which, by definition, is 0.5 If you wish, you can calculate y/1! - y^3/3! + y^5/5! - y^7/7! + ... where y = pi/3.

sin pi/2 =1 sin 3 pi/2 is negative 1 ( it is in 3rd quadrant where sin is negative

11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4

sin(pi/4) and cos(pi/4) are both the same. They both equal (√2)/2≈0.7071■

1/2

Negative 0.5 is.

sin(pi) = 0

sin(3π/2) = -1

Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. However, I am assuming the question is about sin (5pi/12). If not, please resubmit your question spelling out the symbols as "plus", "minus", "times" sin(5pi/12) = sin(pi/4 + pi/6) = sin(pi/4)*cos(pi/6) + cos(pi/4)*sin(pi/6) = âˆš2/2*âˆš3/2 + âˆš2/2*1/2 = âˆš2(âˆš3 + 1)/4

Pi