2 pi
it is the same as a sin function only shifted to the left pi/2 units
I'm not entirely sure about this, but imagine that you have a coordinate system where the x-axis is in units of seconds (variable name t) and the y-axis is in units of centimeters (variable name d (for displacement)). Simple harmonic motion would indicate a simple periodic motion pattern, which would hint at a sine or cosine function. Using the above coordinate system, an unmodified cosine wave would have an amplitude of 1.0 cm and a period of 2(pi) seconds: d=cos(t) I chose cosine because cosine at t=0 (x=0) would be at whatever its amplitude is. To get the amplitude to be 2.0 cm is easy, you must simply modify the amplitude of the cosine function by multiplying it by 2. So far: d=2cos(t) To modify the period to fit the problem statement, you must start with the fact that an unmodified cosine function would have a period of 2(pi). If you were to cut the period in half, you would multiply the argument inside the function by 2, which would give it a period of (pi). This would look like: y=cos(2x) So, if multiplying by 2 changes the period by 1/2, what would we need to multiply by to get a period of 6 seconds? 6 seconds is 6/2(pi)=3/(pi) of 2(pi), to get 3/(pi) of the current period, we must multiply the interior argument by (pi)/3. This gives us a final model of: d=2cos((pi/3)t) If this model is correct, one period of the cosine wave will complete by t=6 seconds. This means that the model will return values of 2 at t=0 and t=6. Let's check: d(0)=2cos((pi/3)(0))=2cos(0)=2*1=2 d(6)=2cos((pi/3)(6))=2cos(2(pi))=2*1=2 These values check out. This means that our model works. So, the final answer is: d=2cos((pi/3)t)
The value of Pi is 3.14 so the value of Pi by 2 is 6.28.
C=2(pi)r A=(pi)r2 If C=A, then 2(pi)r=(pi)r2. Divide both sides by r: 2(pi)=(pi)r. Divide both sides by (pi): 2=r. Therefore, if the radius of a circle is 2 ft., the circumference is 2 sq. ft.
Trigonometric functions are periodic - they repeat after a period of pi, or 2 x pi.Trigonometric functions are periodic - they repeat after a period of pi, or 2 x pi.Trigonometric functions are periodic - they repeat after a period of pi, or 2 x pi.Trigonometric functions are periodic - they repeat after a period of pi, or 2 x pi.
You can invent any function, to make it periodic. Commonly used functions that are periodic include all the trigonometric functions such as sin and cos (period 2 x pi), tan (period pi). Also, when you work with complex numbers, the exponential function (period 2 x pi x i).
Because the trigonometric functions (sine and cosine) are periodic, with period 2*pi. If the argument were not restricted, you would have an infinite number of answers. You could, of course, restrict the argument to any interval of size 2*pi: 3.5pi to 5.5pi, for example.
y = sin(-x)Amplitude = 1Period = 2 pi
The period of the tangent function is PI. The period of y= tan(2x) is PI over the coefficient of x = PI/2
2*Pi
It is the same period as cosine function which is 2 pi because sec x = 1/cos x
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
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.
The six basic trigonometric functions are applicable to almost all angles. The few exceptions are tan(pi/2 + n*pi) cosec(n*pi) sec(pi/2 + n*pi) cot(n*pi) where n is an integer. This is because the functions are undefined at these values.
The period is 2*pi radians.
The period of sin + cos is 2*pi radians (360 degrees) so the period of sin(3x) + cos(3x) is 2*pi/3 radians.