Find the reference angle for −11π12 15?

　
−11π/12 is in Q3

So reference angle
= difference between −11π/12 and negative x-axis (−π)
= −11π/12 − (−π)
= π/12 = (15 in degrees)...Show more

−11π/12 + 2π = 13π/12 .Quadrant III where the reference angle is θ − π. 13π/12 − π = π/12

sin(−11π/12) = -sin(π/12) = -sin((π/6)/2) = -√([1 − cos(π/6)]/2) = -√([1 − √(3)/2]/2) = -√([2 − √3]/4) = -√(2 − √3)/2 [sine is negative in Quadrant III]

cos(−11π/12) = -cos(π/12) = -cos((π/6)/2) = -√([1 + cos(π/6)]/2) = -√([1 + √(3)/2]/2) = -√([2 + √3]/4) = -√(2 + √3)/2 [cosine is negative in Quadrant III]

tan(−11π/12) = tan(π/12) = tan((π/6)/2) = [1 − cos(π/6)]/sin(π/6) = [1 − √(3)/2)]/(1/2) = 2 − √3 [tangent is positive in Quadrant III]...Show more

pi/12 radians
15 degrees...Show more