How do you express #sin(pi/ 4 ) * sin( ( 5 pi) / 3 ) # without using products of trigonometric functions?

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Answer 1 · 2018-07-01T10:09:29

#color(violet)(=> (sin (pi/3) - sin (pi/6))/2#

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#sin A cos b = (1/2) (sin (A+B) + sin (A-B)), Identity#

#sin (pi/4) * sin ((5pi)/3)#

#=> (1/2) (sin (pi/4 + (5pi)/3) + sin (pi/4 - (5pi)/3))#

#=> (1/2) (sin ((8pi)/12) + sin -(pi/6))#

#=> (sin ((2pi)/3) - sin (pi/6))/2#

#color(violet)(=> (sin (pi/3) - sin (pi/6))/2#