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Q: What is cos of pi over 2?
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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 sin23A minus sin7A upon sin2A plus sin14A if A equals pi upon 21?

Using the identity, sin(X)+sin(Y) = 2*sin[(x+y)/2]*cos[(x-y)/2] the expression becomes {2*sin[(23A-7A)/2]*cos[(23A+7A)/2]}/{2*sin[(2A+14A)/2]*cos[(2A-14A)/2]} = {2*sin(8A)*cos(15A)}/{2*sin(8A)*cos(-6A)} = cos(15A)/cos(-6A)} = cos(15A)/cos(6A)} since cos(-x) = cos(x) When A = pi/21, 15A = 15*pi/21 and 6A = 6*pi/21 = pi - 15pi/21 Therefore, cos(6A) = - cos(15A) and hence the expression = -1.


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 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.


How was it determined that the tangent of pi divided by 4 is 1?

As tan(x)=sin(x)/cos(x) and sin(pi/4) = cos(pi/4) (= sqrt(2)/2) then tan(pi/4) = 1

Related questions

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 pi over 2?

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


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 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 value of cos pie4?

cos pi over four equals the square root of 2 over 2 This value can be found by looking at a unit circle. Cos indicates it is the x value of the point pi/4 which is (square root 2 over 2, square root 2 over 2)


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 sin pi over 4?

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


What is the cos of 5 pi over 6?

cos(5π//6) = -(√3)/2 ≈ -0.866


What is x y sq over cos x pi?

Again, please clarify the question. Either "x=y^2/cos(x)*pi " or "x=y^2/cos(pi)". From the question it is not possible to tell whether the second "x" is a variable, or a multiplier sign (and if it were a multiplier, you're question is omitting a variable on the cos(x)).


What is sin23A minus sin7A upon sin2A plus sin14A if A equals pi upon 21?

Using the identity, sin(X)+sin(Y) = 2*sin[(x+y)/2]*cos[(x-y)/2] the expression becomes {2*sin[(23A-7A)/2]*cos[(23A+7A)/2]}/{2*sin[(2A+14A)/2]*cos[(2A-14A)/2]} = {2*sin(8A)*cos(15A)}/{2*sin(8A)*cos(-6A)} = cos(15A)/cos(-6A)} = cos(15A)/cos(6A)} since cos(-x) = cos(x) When A = pi/21, 15A = 15*pi/21 and 6A = 6*pi/21 = pi - 15pi/21 Therefore, cos(6A) = - cos(15A) and hence the expression = -1.


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 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.