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What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
What is the exact value of the expression cos 7pi over 12 cos pi over 6 -sin 7pi over 12 sin pi over 6?
cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2
What is the ratio of the circumferences for two circles with areas 6 pi m2 and 150 pi m2?
area = 2pi*r2circumference = 2pi*rThe ratio of their circumferences will be 2pi*r1 /2pi*r2 = r1/r21) r2 = 6pi/2pi = 3. r1 = root(3).2) r = 150pi/2pi = 75. r2 = root(75)So the ratio of their circumferences isr1/r2 = root(3)/root(75)
Is 2Pi r the same as Pi r squared?
No. 2Pi r is equal to Pi d, though.
What is the usual period for secant?
2pi
