How do you evaluate #\sin ( 15^ { \circ } ) - \sin ( - 15^ { \circ } )#?
1 Answer
Oct 25, 2017
Explanation:
Use:
#sin(alpha-beta) = sin alpha cos beta - sin beta cos alpha#
#sin(45^@) = cos(45^@) = sqrt(2)/2#
#sin(30^@) = 1/2#
#cos(30^@) = sqrt(3)/2#
#sin(-theta) = -sin(theta)#
So:
#sin(15^@) = sin(45^@ - 30^@)#
#color(white)(sin(15^@)) = sin(45^@)cos(30^@) - sin(30^@) cos(45^@)#
#color(white)(sin(15^@)) = sqrt(2)/2 sqrt(3)/2 - 1/2 sqrt(2)/2#
#color(white)(sin(15^@)) = 1/4(sqrt(6)-sqrt(2))#
Then:
#sin(15^@)-sin(-15^@) = 2sin(15^@) = 1/2(sqrt(6)-sqrt(2))#
