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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 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 is the smallest positive value for x where y equals sin 2x reaches its maximun?
sin(theta) reach a maximum at pi / 2 + all even multiple of 2 pi. As a result, the smallest positive value of x where y = sin(2x) is maximal is pi / 2.
What is the value of pi by 2?
The value of Pi is 3.14 so the value of Pi by 2 is 6.28.
Which functions have period 2 pi?
Sin cos sec cosec
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 sin pi over 4?
sin(pi/4) and cos(pi/4) are both the same. They both equal (√2)/2≈0.7071■
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 of pi divided by 2?
Do you mean Sin(pi/2) = 1 or [Sin(pi)] /2 = 0.0274....
What is is the smallest positive value for x where y equals sin 2x reaches its maximun?
sin(theta) reach a maximum at pi / 2 + all even multiple of 2 pi. As a result, the smallest positive value of x where y = sin(2x) is maximal is pi / 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?
Sine(pi/2) = 1 This is 'Radian' measure of an angle. pi/2 radians = 90 degrees. When you see something like 'Sine(pi/2)' make sure your calculator is in RADIAN mode. ' NOT degree mode.
What is the exact values of sin 2 pi?
2*pi is one complete revolution, i.e. 360 degrees. Sin of 2*pi = sin 360º = 0
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 2 pie?
sin(2*pi) - not pie - is the same as sin(0) = 0
What is the period of y equals sin5x?
The period of y=sin(x) is 2*pi, so sin(x) repeats every 2*pi units. sin(5x) repeats every 2*pi/5 units. In general, the period of y=sin(n*x) is 2*pi/n.
What is the value of pi by 2?
The value of Pi is 3.14 so the value of Pi by 2 is 6.28.
