Y=sin X is a function because for each value of X, there is exactly one Y value.
D(y)= sin 2x
y=3 cos(x) y' = -3 sin(x)
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
2
The range of -sin x depends on the domain of x. If the domain of x is unrestricted then the range of y is [-1, 1].
range of y=sin(2x) is [-1;1] and in generally when is y=sin(k*x) (k=....-1,0,1....) range is always [-1;1] and the period is w=(2pi)/k
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
sin x can have any value between -1 and 1; therefore, 3 sin x has three times this range (from -3 to 3).
the range is greater then -1 but less than 1 -1<r<1
Y=sin X is a function because for each value of X, there is exactly one Y value.
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
The amplitude of the wave [ y = -2 sin(x) ] is 2.
D(y)= sin 2x
It is zero.
y=3 cos(x) y' = -3 sin(x)
y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5