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
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))
First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i
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.
1/sin x = csc x
Sin[x] = Cos[x] + (1/3)
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
It equals 2 over 8.
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
It is a trigonometric equation. A = sin-1(7/25) = 0.284 radians.
x = sin-1 (4/15) ( sin -1 is [SHIFT] [sin] on a calculator ) = 15.5
sinx=n/1 (1)sinx=n/1(1) sin(-n)x=n(-n) six=6
x is 9.
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check:
Cos x = 1 / Sec x so 1 / Cos x = Sec x Then Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x
x equals 4 over 35