f(x)= sin(1/x) and g(x)=1/sin(x)
[u(v)]' = u'(v) * v', where u and v are functions
So
f'(x) = sin'(1/x) * (1/x)' = cos(x) * (-1/x2) = -cos(x)/x2
g'(x) = (1/x)' applied to sin(x) * (sin(x))' = -1/(sin2(x)) * cos(x) = -cos(x)/(sin2(x))
Sin[x] = Cos[x] + (1/3)
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
The period is the length of x over which the equation repeats itself. In this case, y=sin x delivers y=0 at x=0 at a gradient of 1. y next equals 0 when x equals pi, but at this point the gradient is minus 1. y next equals 0 when x equals 2pi, and at this point the gradient is 1 again. Therefore the period of y=sinx is 2pi.
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
1/sin x = csc x
Sin[x] = Cos[x] + (1/3)
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
It is a trigonometric equation. A = sin-1(7/25) = 0.284 radians.
either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though
13.20
x = sin-1 (4/15) ( sin -1 is [SHIFT] [sin] on a calculator ) = 15.5
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
Assuming sin equals 0.3237, the angle is in quadrant I.
It would be 1 over square root 5.
sin(pi) = 0 so 4*sin(pi) = 0 so Y = 0
Y=sin X is a function because for each value of X, there is exactly one Y value.