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3 Answers
- ?Lv 74 years ago
−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)
- RogueLv 74 years ago
−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]
