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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.
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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.
If x is obtuse and sin x equals b how do you show sin 2x is less than b?
you need to explain it better but with what i got i if sin x equals b obviously sin 2x is double of b hence sin 2x is more than b. {Not obvious at all, actually. And the above is false. Sin(Pi/2) = 1 and Sin(Pi)=0. But clearly 2 is not greater than 0. Contradiction.} Obtuse means Pi/2 < x < Pi So, sin(x) = b means b>0, because sin(y) > 0 if 0<y<Pi sin(2x)=2sin(x)cos(x) cos(x) < 0 because cos(Pi/2)= 0 and the derivative is negative there. Hence, sin(2x) = 2 sin(x) cos(x) = 2*b*(-K), where K is a positive constant Since b>0, -2Kb < b
What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
What is the cos pi over 2?
The answer is:cos (pi/2) = 0
Find the amplitude and period of the functions and sketch their graphs y equals sin -x?
y = sin(-x)Amplitude = 1Period = 2 pi
