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Use these identities: sin2(x) + cos2(x) = 1, and tan(x) = sin(x)/cos(x) For clarity, the functions are written here without their arguments (the "of x" part). (1 - sin2) = cos2 (1 + tan2) = (1 + sin2/cos2) = (cos2+sin2) / cos2 = 1/cos2 Multiply them: (cos2) times (1/cos2) = 1'QED'
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
d/dx (sin x)^2=2sinxcosx
cos(3t) = cos(2t + t) = cos(2t)*cos(t) - sin(2t)*sin(t) = [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin(t)*sin(t) = [cos2(t) - sin2(t)]*cos(t) - 2*cos(t)*sin2(t) then, since sin2(t) = 1 - cos2(t) = [2*cos2(t) - 1]*cos(t) - 2*cos(t)*[1 - cos2(t)] = 2*cos3(t) - cos(t) - 2*cos(t) + 2*cos3(t) = 4*cos3(t) - 3*cos(t)
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.
this is due to effect of sin2(theta). as sin2(theta) will repeat its value for =90-theta as here theta=30 so for 90-30=60 sin2(30)=sin2(60) so for pair of projection angles of two projectiles as(theta,90-theta) , they will have same ranges i.e theta=10 and 90-10=80 sin2(10)=sin2(80)
cos2 + cos2tan2 = cos2 + cos2*sin2/cos2 = cos2 + sin2 which is identically equal to 1. So the solution is all angles.
1-sin2 = -1
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
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