How do you evaluate #sin (-8 pi / 12)#?

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Answer 1 · 2016-06-08T08:28:25

#-sqrt(3)/2#

# sin ( - (8 * pi )/12) #

# = sin ( - 120°) #

# = - sin ( 120° ) #

# = - sin ( 180° - 60° ) #

# = - sin ( 60° ) #

# = -sqrt(3)/2 #

Answer 2 · 2016-06-13T17:20:28

#-sqrt(3)/2#

# -8*pi/12 =( pi/3) -pi #

So,

# sin(-8*pi/12) = sin((pi/3)-pi) #

# sin(-8*pi/12) = sin(-( pi - pi/3))#

Knowing that :
# sin(-alpha) = - sin(alpha)#

#sin(-8*pi/12) = - sin( pi - pi/3) #

Knowing that:
#sin(pi-alpha) =sin(alpha)#

#sin(-8*pi/12) = -sin(pi/3) #

So,
#sin(-8*pi/12) = -sqrt(3)/2#