The expression: 8sin(\({\pi \over 3}\))cos(\({\pi \over 3}\))
I used the double angle identity sin2A = 2sinAcosA but I'm not sure how I could get the textbook answer of 4sin(\({2 \pi \over 3}\))?
+36916 8 sin pi/3 cos pi/3 a = pi/3
4 * 2 sin pi/3 cos pi/3 2 a will equal 2pi/3
In your identity Sin2a = 2 sin a cos a but your example is multiplied by 4
4 sin 2a = 4 * 2 sina cos a See where the 4 comes from now?
4 sin (2pi/3)
