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What is the sin pi over 4?

Updated: 4/28/2022
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14y ago

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

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14y ago
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Q: What is the sin pi over 4?
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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 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 sin pi over 2?

sin(pi/2)=1


Y equals 4 sin x for x equals pi?

sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0


What is sin of 3 pi over 2?

sin pi/2 =1 sin 3 pi/2 is negative 1 ( it is in 3rd quadrant where sin is negative


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


Sin x - cos x 0 0?

sin x - cos x = 0sin x = cos x(sin x)^2 = (cos x)^2(sin x)^2 = 1 - (sin x)^22(sin x)^2 = 1(sin x)^2 = 1/2sin x = ± √(1/2)sin x = ± (1/√2)sin x = ± (1/√2)(√2/√2)sin x = ± √2/2x = ± pi/4 (± 45 degrees)Any multiple of 2pi can be added to these values and sine (also cosine) is still ± √2/2. Thus all solutions of sin x - cos x = 0 or sin x = cos x are given byx = ± pi/4 ± 2npi, where n is any integer.By choosing any two integers , such as n = 0, n = 1, n = 2 we can find some solutions of sin x - cos x = 0.n = 0, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(0)(pi) = ± pi/4 ± 0 = ± pi/4n = 1, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(1)(pi) = ± pi/4 ± 2pi = ± 9pi/4n = 2, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(2)(pi) = ± pi/4 ± 4pi = ± 17pi/4


What is the sin of pi?

sin(pi) = 0


What are the distinct fourth root of -1?

The four roots are cos(theta)+i*sin(theta) where theta = pi/4, 3*pi/4, 5*pi/4 and 7*pi/4.


What is the value of sin 3 pi over 2?

sin(3π/2) = -1


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 answer to the sine of 5pi 12?

Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. However, I am assuming the question is about sin (5pi/12). If not, please resubmit your question spelling out the symbols as "plus", "minus", "times" sin(5pi/12) = sin(pi/4 + pi/6) = sin(pi/4)*cos(pi/6) + cos(pi/4)*sin(pi/6) = √2/2*√3/2 + √2/2*1/2 = √2(√3 + 1)/4