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What is cos pi over 3?

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∙ 14y ago

Updated: 4/28/2022

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What is the cos of 5 pi over 6?

cos(5π//6) = -(√3)/2 ≈ -0.866

What is the integral from 0 to pi over 6 sine 2x dx?

Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0

What is sin of 3 pi over 2?

Sin(3pi/2) = Sin(2pi - pi/2) Double angle Trig. Identity. Hence Sin(2pi)Cos(pi/2) - Cos(2pi) Sin(pi/2) Sin(2pi) = 0 Cos(pi/2) = 0 Cos(2pi) = 1 Sin(pi/2) = 1 Substituting 0 x 0 - 1 x 1 = 0 - 1 = -1 The answer!!!!!

What is the sin pi over 4?

sin(pi/4) and cos(pi/4) are both the same. They both equal (√2)/2≈0.7071■

1 over tan x equals what?

1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)

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Related Questions

What is the cosine of negative pi over 3?

Cos(Pi/3) is 1/2 so Cos(-Pi/3) ould be flipped over the x-axis. The answer is still 1/2.

What is the exact value using a sum or difference formula of the expression cos 11pi over 12?

11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4

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 cos of 5 pi over 6?

cos(5π//6) = -(√3)/2 ≈ -0.866

What is the integral from 0 to pi over 6 sine 2x dx?

Integral from 0 to pi 6sin2xdx: integral of 6sin2xdx (-3)cos2x+c. (-3)cos(2 x pi) - (-3)cos(2 x 0) -3 - -3 0

What is the cos pi over 2?

The answer is:cos (pi/2) = 0

What is x equals y squared over co sign times pi?

Either you mean "cos(x) multiplied by pi", (i.e pi*cos(x)) or "cos(pi)" (i.e cosine of pi), but it is unclear which you mean from the question. Please clarify.

What is sin of 3 pi over 2?

Sin(3pi/2) = Sin(2pi - pi/2) Double angle Trig. Identity. Hence Sin(2pi)Cos(pi/2) - Cos(2pi) Sin(pi/2) Sin(2pi) = 0 Cos(pi/2) = 0 Cos(2pi) = 1 Sin(pi/2) = 1 Substituting 0 x 0 - 1 x 1 = 0 - 1 = -1 The answer!!!!!

What is x equal y sq over cos x pi?

Can you please claify if you mean x=y^2/ pi*cos(x) , or x=y^2/cos(pi), since they are very different sums.

What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?

(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)

What is the equation of the line tangent to the curve y equals cos x at the point x equals pi over 6?

y = 2(x) - (pi/3) + (sqrt(3)/2)

What are the 3 complex roots of -1 using the DeMoivre's theorem?

Problem: find three solutions to z^3=-1. DeMoivre's theorem is that (cos b + i sin b)^n = cos bn + i sin bn So we can set z= (cos b + i sin b), n = 3 cos bn + i sin bn = -1. From the last equation, we know that cos bn = -1, and sin bn = 0. Three possible solutions are bn=pi, bn=3pi, bn=5pi. This gives three possible values of b: b=pi/3 b=pi b = 5pi/3. Now using z= (cos b + i sin b), we can get three possible cube roots of -1: z= (cos pi/3 + i sin pi/3), z= (cos pi + i sin pi), z= (cos 5pi/3 + i sin 5pi/3). Working these out gives -1/2+i*sqrt(3)/2 -1 -1/2-i*sqrt(3)/2

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