Q: What is the range of y equals -3 sin x?

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A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.

the range is greater then -1 but less than 1 -1<r<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

sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].

The multiplication by 3 increases the magnitude, and the + 3 shifts the graph upward to be "centred" at y = 3. The graph now oscillates between 6 and 0 (3 + 3, 3 - 3).

Related questions

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].

Sin[x] = Cos[x] + (1/3)

1.5

2 sin(x) - 3 = 0 2 sin(x) = 3 sin(x) = 1.5 No solution. The maximum value of the sine function is 1.0 .

Assuming a large enough domain, the range is -1 to 1.

A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).

2

the range is greater then -1 but less than 1 -1<r<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

y=3 cos(x) y' = -3 sin(x)

If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²