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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
What s the amplitude and period of y cos 8x?
y = A sin Bx y = A Cos BxAmplitude = APeriod = 2pi/Bso....Amplitude = 1Period = 2pi/8
How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
How do you find the complex cube roots of 1 plus 0i?
z = 1 + 0i So |rz| = 1 and az = 0 radians. which allows you to write z = rz*cos(az) + i*sin(az) Then, if y = z1/3 then |y| = |z1/3| = |11/3| = 1 and ay is the angle in [0, 360) such that 3*ay = 0 mod(2*pi) that is, ay = 0, 2pi/3 and 4pi/3 And therefore, Root 1 = cos(0) + i*sin(0) Root 2 = cos(2pi/3) + i*sin(2pi/3) and Root 3 = cos(4pi/3) + i*sin(4pi/3).
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 reciprocal of 2pi?
Reciprocal of 2pi = 1/(2pi) = 0.159155(rounded)
What are the exact values of the sine and cosine functions of 2pi over 65537?
sin(2*pi/65537) = 0.0001 cos(2*pi/65537) = 1.0000 to 4 dp.
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
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
What are all the exact values for which tan t equals sqrt3?
sin(60) or sin(PI/3) = sqrt(3)/2 cos(60) or cos(PI/3)=1/2 tan(60) or tan(PI/3) = sin(60)/cos(60)=sqrt(3) But we want tan for -sqrt(3). Tangent is negative in quadrant II and IV. In Quadrant IV, we compute 360-60=300 or 2PI-PI/3 =5PI/3 tan(5PI/3) = -sqrt(3) Tangent is also negative in the second quadrant, so we compute PI-PI/3=2PI/3 or 120 degrees. tan(t)=-sqrt(3) t=5PI/3 or 2PI/3 The period of tan is PI The general solution is t = 5PI/3+ n PI, where n is any integer t = 2PI/3+ n PI, where n is any integer
What is pi added to pi over two?
(pi^2+2pi)/2pi
What is the circumference of a circle with a 2pi radius?
The circumference of a circle with a 2pi radius is: 39.58
Is 2pi rational?
NO. pi is an irrational number, therefore, 2pi is also an irrational number/
