sin x can have any value between -1 and 1; therefore, 3 sin x has three times this range (from -3 to 3).
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].
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
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)
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.
2
1.5
the range is greater then -1 but less than 1 -1<r<1
y=3 cos(x) y' = -3 sin(x)
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
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.