Q: What is cos of pi over 2?

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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

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.

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.

sin(2*pi/65537) = 0.0001 cos(2*pi/65537) = 1.0000 to 4 dp.

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

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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

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

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

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

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)

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.

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

cos(5π//6) = -(√3)/2 ≈ -0.866

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)).

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.

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

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.