0
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 else can I help you with?
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.
What is sin of pi over 12 on a radian circle?
The sine of (\frac{\pi}{12}) radians (which is equivalent to 15 degrees) can be calculated using the sine subtraction formula: (\sin(a - b) = \sin a \cos b - \cos a \sin b). By letting (a = \frac{\pi}{4}) (45 degrees) and (b = \frac{\pi}{3}) (60 degrees), we find that (\sin\left(\frac{\pi}{12}\right) = \sin\left(\frac{\pi}{4} - \frac{\pi}{3}\right) = \sin\frac{\pi}{4} \cos\frac{\pi}{3} - \cos\frac{\pi}{4} \sin\frac{\pi}{3}). This evaluates to (\frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}).
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 3 pi?
The sine of (3\pi) is 0. This is because (3\pi) corresponds to a point on the unit circle where the angle is a multiple of (2\pi) (specifically, (3\pi = 2\pi + \pi)), and the sine of any integer multiple of (2\pi) is always 0. Thus, (\sin(3\pi) = 0).
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!
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.
How do you solve sin2 theta equals 0.75?
To solve the equation (\sin^2 \theta = 0.75), first take the square root of both sides to get (\sin \theta = \pm \sqrt{0.75} = \pm \frac{\sqrt{3}}{2}). Then, find the angles (\theta) for which (\sin \theta = \frac{\sqrt{3}}{2}) and (\sin \theta = -\frac{\sqrt{3}}{2}). The solutions are (\theta = \frac{\pi}{3} + 2k\pi) and (\theta = \frac{2\pi}{3} + 2k\pi) for the positive case, and (\theta = \frac{7\pi}{6} + 2k\pi) and (\theta = \frac{4\pi}{3} + 2k\pi) for the negative case, where (k) is any integer.
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).
