The Domain and Range are both the set of real numbers.
y=sin x y=cos x cos x sin x = cos^2 x sin x = 1-sin^2 x sin x -1 + sin^2 x = 0 sin^2 x + sin x -1 = 0 Let y=sin x y^2+y-1 = 0 This equation is of form ay^2+by+c=0 a = 1 b = 1 c = -1 y=[-b+/-sqrt(b^2-4ac)]/2a] y=[-1 +/-sqrt(1^2-4(1)(-1)]/(2)(1) discriminant is b^2-4ac =5 y=[-1 +√(5)] / 2 y=[-1 -√(5)] / 2 sin x = [-1 +√(5)] / 2 x = sin^-1 [-1 +√(5)] / 2] = 0.6662394 radians x = sin^-1 [-1 -√(5)] / 2] = sin^-1 (-1.618) -- has no solution When x = 0.6662394 radians, sin x and cos x times cos x are equal.
Domian is x>-6 Range is y> or equal to 0
x
D = {x [element of reals]}R = {y [element of reals]|y >= 4}
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].
y= sin 3x
Assuming a large enough domain, the range is -1 to 1.
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
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
Domain (input or 'x' values): -∞ < x < ∞.Range (output or 'y' values): -2 ≤ y ≤ 2.
Give the domain for
The amplitude of the wave [ y = -2 sin(x) ] is 2.
The domain of y = 2x is [0, +infinity].
The domain is the x values, so x = 0 to 10. The range is the y values, so y = 0 to 25.
D(y)= sin 2x