'Y' varies between -4 and +4. Viewed as a wave, its amplitude is 4.
4
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
cos A=3/5 sin=square root of (1-cos2) sin=square root of (1-3/52) sin=square root of (1-9/25) sin=square root of (16/25) sin=4/5 csc=1/sin csc=1/(4/5) csc=5/4
The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.
'Y' varies between -4 and +4. Viewed as a wave, its amplitude is 4.
4
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 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.
x = sin-1 (4/15) ( sin -1 is [SHIFT] [sin] on a calculator ) = 15.5
it equals 4
0.5
cos A=3/5 sin=square root of (1-cos2) sin=square root of (1-3/52) sin=square root of (1-9/25) sin=square root of (16/25) sin=4/5 csc=1/sin csc=1/(4/5) csc=5/4
0.75
4 sin(x) - 3 = 0 Therefore sin(x) = 3/4 And so the primary solution is x = sin-1(3/4) = 49 deg The second solution in the domain is 180 - 49 = 131 deg.
The proof that sin2A plus sin2B plus sin2c equals 4sinAsinBsinC lies in the fact that (sin 2A + sin 2B + sin 2C) = 4 sinA.sinB.sinC.
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))