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What is the value of sin 3 pi over 2?
sin(3π/2) = -1
How do you find the amplitude maximum minimum and period for y equals -1 plus 3sin4x?
y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.
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 of 5 pi over 6?
cos(5π//6) = -(√3)/2 ≈ -0.866
How do you solve sin2 theta if sin theta equals 3 over 5 and theta is in the first quadrant?
If sin (theta) is 3/5, then sin2 (theta) is (3/5)2, or 9/25.
What is the value of sin 3 pi over 2?
sin(3π/2) = -1
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
Is sin3A plus sinA equal to 0?
No, for example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero. It can't be proved because the statement is false. For example if A = Pi/3. Then sin3A = sin pi = 0, but sinA = sin Pi/3 = 1/2. So for A = Pi/3, the sum is 1/2, not zero.
Why does sin 2pi over 6 radians equal .5?
Sin(2*pi/6) = sin(pi/3) which, by definition, is 0.5 If you wish, you can calculate y/1! - y^3/3! + y^5/5! - y^7/7! + ... where y = pi/3.
What is the sin of pi divided by 3?
The question is ambiguous and the two possible answers are: sin(pi)/3 = 0 and sin(pi/3) = sqrt(3)/2 It is assumed, of course, that since the angles are given in terms of pi, they are measured in radians and not degrees!
How do you solve sin2x plus sinX equals 0?
sin(2x) + sin(x) = 0 2sin(x)cos(x) + sin(x) = 0 sin(x)[2cos(x) + 1] = 0 sin(x) = 0 OR 2cos(x) + 1 = 0 sin(x) = 0 OR cos(x) = -1/2 x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)] Note that n may be any integer. The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
What is the period for y-3 sin x?
y = 3 sin x The period of this function is 2 pi.
Find the values of θ in the range 0 to 2π for cos θ plus cos 3θ equals sin θ plus sin 3θ?
The angle can be 0, pi/2, pi, 3*pi/2 or 2*pi radians.
What is the period of y equals 3 sin pi x?
2
What are the trigonometric functions for 240?
It is easiest to find these using the unit circle. Assuming you want exact values for sin, cos, and tan.240 degrees is equal to 4[Pi]/3 radians.cos(4[Pi]/3) and sin(4[Pi]/3) are easy to find using the unit circle,cos(4[Pi]/3) = -1/2sin(4[Pi]/3) = -(Sqrt[3])/2To find tan, you will need to do a little arithmetic. We know that tan(x) = sin(x)/cos(x), so,tan(4[Pi]/3) = sin(4[Pi]/3)/cos(4[Pi]/3)tan(4[Pi]/3) = (-(Sqrt[3])/2) / (-1/2)tan(4[Pi]/3) = ((Sqrt[3])/2) * (2/1)tan(4[Pi]/3) = ((Sqrt[3])*2) / (2*1)tan(4[Pi]/3) = (Sqrt[3]) / 1tan(4[Pi]/3) = Sqrt[3]Find the other 3 using reciprocal. This is just a little arithmetic and up to you. You know that cossec(x) = 1/sin(x), sec(x) = 1/cos(x) and cot(x) = cos(x)/sin(x).
Which uses single frequency digital signals?
I was told it is x(n) = 10*sin(2*pi*fo*Ts(n-1)) + 8*sin(2*pi*2*fo*Ts(n-1)) + 6*sin(2*pi*3*fo*Ts(n-1))
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
