tan(A) = 1/2 sin(A)/cos(A) = 1/2 sin2(A)/cos2(A) = 1/4 sin2(A)/[1 - sin2(A)] = 1/4 sin2(A) = 1/4*[1 - sin2(A)] 5/4*sin2(A) = 1/4 sin2(A) = 1/5 sin(A) = ±sqrt(1/5) = ±sqrt(5)/5
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
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1-sin2 = -1
No.Remember: sin2 + cos2 = 1So, in place of (1 - 2 sin2) we can write (sin2 + cos2 - 2 sin2).Massage that around slightly: (sin2 + cos2 - 2 sin2) = cos2 - sin2That's not equal to (sin2 - cos2), which is the original question we were asked.
tan(A) = 1/2 sin(A)/cos(A) = 1/2 sin2(A)/cos2(A) = 1/4 sin2(A)/[1 - sin2(A)] = 1/4 sin2(A) = 1/4*[1 - sin2(A)] 5/4*sin2(A) = 1/4 sin2(A) = 1/5 sin(A) = ±sqrt(1/5) = ±sqrt(5)/5
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
tan = sin/cos Now cos2 = 1 - sin2 so cos = +/- sqrt(1 - sin2) In the second quadrant, cos is negative, so cos = - sqrt(1 - sin2) So that tan = sin/[-sqrt(1 - sin2)] or -sin/sqrt(1 - sin2)
-cos2(x)1 = sin2(x) +cos2(x)1 - cos2(x) = sin2(x)-cos2(x) = sin2(x) - 1
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the Laplace transform of sin2 3thttp://www7.0zz0.com/2009/12/30/19/748450027.gif
sin2csc2-sin2 (using the fact that the sin is the reciprocal of csc) = 1-sin2
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
sin2 + cos2 = 1 So, (1 - 2*cos2)/(sin*cos) = (sin2 + cos2 - 2*cos2)/(sin*cos) = (sin2 - cos2)/(sin*cos) = sin2/(sin*cos) - cos2/(sin*cos) = sin/cos - cos-sin = tan - cot
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