Use addition formulas for sine and cosine to simplify the trig expression.
sin[B + (pi/6)] - sin[B - (pi/6)]
sin[B + (pi/6)] - sin[B - (pi/6)]
-
sin (a + b) = sin a cos b + cos a sin b
sin (a - b) = sin a cos b - cos a sin b
sin pi/6 = 1/2
cos pi/6 = sqrt(3) / 2
sub in to give:
sin [B + pi/6] = sin B cos pi/6 + cos B sin pi/6 = sqrt(3) / 2 sin B + 1/2 cos B
sin [B - pi/6] = sin B cos pi/6 - cos B sin pi/6 = sqrt(3) / 2 sin B - 1/2 cos B
so sin [B + pi/6] - sin [B - pi/6] = sqrt(3) / 2 sin B + 1/2 cos B - [sqrt(3) / 2 sin B - 1/2 cos B]
= cos B
sin [B + pi/6] - sin [B - pi/6] = sin [B + pi/6] = cos B
sin (a - b) = sin a cos b - cos a sin b
sin pi/6 = 1/2
cos pi/6 = sqrt(3) / 2
sub in to give:
sin [B + pi/6] = sin B cos pi/6 + cos B sin pi/6 = sqrt(3) / 2 sin B + 1/2 cos B
sin [B - pi/6] = sin B cos pi/6 - cos B sin pi/6 = sqrt(3) / 2 sin B - 1/2 cos B
so sin [B + pi/6] - sin [B - pi/6] = sqrt(3) / 2 sin B + 1/2 cos B - [sqrt(3) / 2 sin B - 1/2 cos B]
= cos B
sin [B + pi/6] - sin [B - pi/6] = sin [B + pi/6] = cos B
-
Oops, typo on the last line -- that last sin [B + pi/6] shouldn't be there...
Report Abuse