Study guides

☆☆

Q: What is the period of y equals 3 sin pi x?

Write your answer...

Submit

Still have questions?

Continue Learning about Other Math

y = 3 sin x The period of this function is 2 pi.

Two thirds pi, or rather 2pi/3.

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.

2 sin(x) - 3 = 0 2 sin(x) = 3 sin(x) = 1.5 No solution. The maximum value of the sine function is 1.0 .

1.5

Related questions

y = 3 sin x The period of this function is 2 pi.

Two thirds pi, or rather 2pi/3.

The period of sin + cos is 2*pi radians (360 degrees) so the period of sin(3x) + cos(3x) is 2*pi/3 radians.

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

The angle can be 0, pi/2, pi, 3*pi/2 or 2*pi radians.

The complex roots are 0.5 +/- i/2*sqrt(3) or cos(pi/3) +/- i sin(pi/3)where the angles are measured in radians.

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.

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

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!

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.

the period is 2pi. period is 2pi/b and the formula is y=AsinBx.

sin(60) or sin(PI/3) = sqrt(3)/2 cos(60) or cos(PI/3)=1/2 tan(60) or tan(PI/3) = sin(60)/cos(60)=sqrt(3) But we want tan for -sqrt(3). Tangent is negative in quadrant II and IV. In Quadrant IV, we compute 360-60=300 or 2PI-PI/3 =5PI/3 tan(5PI/3) = -sqrt(3) Tangent is also negative in the second quadrant, so we compute PI-PI/3=2PI/3 or 120 degrees. tan(t)=-sqrt(3) t=5PI/3 or 2PI/3 The period of tan is PI The general solution is t = 5PI/3+ n PI, where n is any integer t = 2PI/3+ n PI, where n is any integer

People also asked