Respuesta :

Well, you're just suppose to know where 5pi/3 is on the unit circle.
That is in the fourth quadrant with coordinates (1/2, -root3/2)
So sin of 5pi/3 would be = to -root3/2
Nirina7

very simple, we use the formula sin(a+b)=sinacosb + sinbcosa and sin(20)=2sinacosa

5pi = 2pi/3+3pi/3,

First, we use sin(a+b)=sinacosb + sinbcosa

sin(5pi/3)=sin(2pi/3+3pi/3)= sin(2pi/3+pi)= sin(2pi/3)cos(pi) +sin(pi)cos(2pi/3)

but we know that sin(pi)= 0, and cos (pi) = -1, so sin(5pi/3)= - sin(2pi/3)

now, use sin(2a)=2sinacosa, sin(5pi/3)= - sin(2pi/3)= -2sin(pi/3)cos(pi/3)

sin(5pi/3)=  -2sin(pi/3)cos(pi/3)

sin(pi/3)= 0.86, cos(pi/3)=0.5, finally we have   sin(5pi/3)=  -0.86 x 0.5= -0.43

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