1/sin x = csc x
x = sin-1 (4/15) ( sin -1 is [SHIFT] [sin] on a calculator ) = 15.5
-1
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:
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].
Tan(x) = Sin(x) / Cos(x) Hence Sin(x) / Cos(x) = Cos(x) Sin(x) = Cos^(2)[x] Sin(x) = 1 - Sin^(2)[x] Sin^(2)[x] + Sin(x) - 1 = 0 It is now in Quadratic form to solve for Sin(x) Sin(x) = { -1 +/-sqrt[1^(2) - 4(1)(-1)]} / 2(1) Sin(x) = { -1 +/-sqrt[5[} / 2 Sin(x) = {-1 +/-2.236067978... ] / 2 Sin(x) = -3.236067978...] / 2 Sin(x) = -1.61803.... ( This is unresolved as Sine values can only range from '1' to '-1') & Sin(x) = 1.236067978... / 2 Sin(x) = 0.618033989... x = Sin^(-1) [ 0.618033989...] x = 38.17270765.... degrees.
Sin[x] = Cos[x] + (1/3)
x = sin-1 (4/15) ( sin -1 is [SHIFT] [sin] on a calculator ) = 15.5
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))
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
It would be 1 over square root 5.
Cos (x) = -Sin(x) 1 = -Sin(x) / Cos (x) 1 = -Tan(x) Tan(x) = -1 x = Tan^-1(-1( x = -45 degrees = - pi /4 , 3pi/4, 5pi/4 ....
-1
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:
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.
13.20
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.