Sin(3pi/2) = Sin(2pi - pi/2)
Double angle Trig. Identity.
Hence
Sin(2pi)Cos(pi/2) - Cos(2pi) Sin(pi/2)
Sin(2pi) = 0
Cos(pi/2) = 0
Cos(2pi) = 1
Sin(pi/2) = 1
Substituting
0 x 0 - 1 x 1 =
0 - 1 = -1 The answer!!!!!
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sin pi/2 =1
sin 3 pi/2 is negative 1 ( it is in 3rd quadrant where sin is negative
sin(3π/2) = -1
Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0
y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.
cos(5π//6) = -(√3)/2 ≈ -0.866
If sin (theta) is 3/5, then sin2 (theta) is (3/5)2, or 9/25.