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Notes 4.7 pre-calculus Inverse Trig Functions If domain is restricted of y = sin x to 2 , 2 the inverse sine 1 function y sin x is the inverse of the restricted sine function. Inverse sine function (ArcSine function): The unique angle y in the interval 2 , 2 such that 1 sin y = x is the inverse sine (or arcsine) of x, denoted sin arcsin x. 1 y sin x is [-1,1] and the range is , . The domain of 2 2 2 f x = sin -1x 5 -5 -2 Inverse cosine function (ArcCosine function): The unique angle y in the interval 0, such that cos y = x is the inverse cosine (or arccosine) of x denoted cos 1 x or arccos x. x or 1 The domain of y cos x is [-1,1] and the range 0, gx = cos- 1x 2 -5 5 -2 Inverse Tangent Function(Arctangent Function): The unique angle y in the interval 2 , 2 such that tan y = x is the 1 inverse tangent (or arctangent) of x, denoted tan x or arctan x. 1 The domain y tan x is (-,) and the range is 2 , 2 . 2 hx = tan- 1x -5 5 -2 Always true: sin(sin 1 x) x cos(cos 1 x) x tan(tan 1 x) x (x here relates to a coordinate) True on restricted domain: sin 1 (sin x) x cos1 (cos x) x tan 1 (tan x) x (x here relates to an angle) Examples: Find the exact value of each expression without a calculator. a) 1 sin 1 2 Remember, on the unit circle the y coordinate is sin . So, we are looking for the angle where we have a y coordinate of 1 . 2 Also, the range for the inverse sin function is restricted to [ , 2 ]. So, the angle we are looking for is 6 . 2 b) sin 1 sin 9 Remember on the restricted domain sin 1 (sin x) x So, cos1 (cos x) x tan 1 (tan x) x sin 1 sin 9 9 Evaluating trig functions with a calculator: Find the approximate value, express answer in degrees. a) cos1(0.76) 139.46 b) arcsin(0.89) 62.87 Evaluating Inverse Trig Functions without a calculator: Find the exact value of the expression without a calculator. a) cos1 cos 1.1 Draw an angle of -1.1 in standard position (notice that this angle is not in the interval [0, ]) and mark its x-coordinate on the x-axis. The angle in the interval [0, ] whose cosine is this number is 1.1. So, cos1 cos 1.1 = 1.1 b) 1 sin cos 3 First we need to find the value of cos . 3 cos = 1 3 2 1 Now we find sin 1 . 2 1 sin 1 2 6 So, 1 sin cos = 3 6