Inverse Trigonometric Functions

Transcription

1 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka Inverse Trigonometric Functions DEFINITION: The inverse sine function, denoted by sin x or arcsin x), is defined to be the inverse of the restricted sine function sin x, π x π DEFINITION: The inverse cosine function, denoted by cos x or arccos x), is defined to be the inverse of the restricted cosine function cos x, 0 x π DEFINITION: The inverse tangent function, denoted by tan x or arctan x), is defined to be the inverse of the restricted tangent function tan x, π < x < π DEFINITION: The inverse cotangent function, denoted by cot x or arccot x), is defined to be the inverse of the restricted cotangent function cot x, 0 < x < π

2 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka DEFINITION: The inverse secant function, denoted by sec x or arcsec x), is defined to be the inverse of the restricted secant function [ ] sec x, x [0, π/) [π, 3π/) or x [0, π/) π/, π] in some other textbooks DEFINITION: The inverse cosecant function, denoted by csc x or arccsc x), is defined to be the inverse of the restricted cosecant function [ ] csc x, x 0, π/] π, 3π/] or x [ π/, 0) 0, π/] in some other textbooks IMPORTANT: Do not confuse sin x, cos x, tan x, cot x, sec x, csc x with sin x, cos x, tan x, cot x, sec x, csc x FUNCTION DOMAIN RANGE sin x [, ] [ π/, π/] cos x [, ] [0, π] tan x, + ) π/, π/) cot x, + ) 0, π) sec x, ] [, + ) [0, π/) [π, 3π/) csc x, ] [, + ) 0, π/] π, 3π/]

3 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka FUNCTION DOMAIN RANGE sin x [, ] [ π/, π/] cos x [, ] [0, π] tan x, + ) π/, π/) cot x, + ) 0, π) sec x, ] [, + ) [0, π/) [π, 3π/) csc x, ] [, + ) 0, π/] π, 3π/] sin x) = sin x cos x) = cos x tan x) = tan x csc x) = csc x sec x) = sec x cot x) = cot x sinx ± π) = sin x cosx ± π) = cos x tanx ± π) = tan x secx ± π) = sec x cscx ± π) = csc x cotx ± π) = cot x EXAMPLES: a) sin = π, since sin π = and π [ π, π ] b) sin ) = π, since sin π ) = and π [ π, π ] c) sin 0 = 0, since sin 0 = 0 and 0 [ π, π ] d) sin = π 6, since sin π 6 = and π [ 6 π, π ] e) sin 3 = π 3, since sin π 3 = f) sin = π 4, since sin π 4 = EXAMPLES: 3 and π [ 3 π, π ] and π [ 4 π, π ] cos 0 = π, cos = 0, cos ) = π, cos = π 3, cos 3 = π 6, cos = π 4 tan = π 4, tan ) = π 4, tan 3 = π 3, tan 3 = π 6, tan ) = π 3 6 EXAMPLES: Find sec, sec ), and sec ) 3

4 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka FUNCTION DOMAIN RANGE sin x [, ] [ π/, π/] cos x [, ] [0, π] tan x, + ) π/, π/) cot x, + ) 0, π) sec x, ] [, + ) [0, π/) [π, 3π/) csc x, ] [, + ) 0, π/] π, 3π/] sin x) = sin x cos x) = cos x tan x) = tan x csc x) = csc x sec x) = sec x cot x) = cot x sinx ± π) = sin x cosx ± π) = cos x tanx ± π) = tan x secx ± π) = sec x cscx ± π) = csc x cotx ± π) = cot x EXAMPLES: Find sec, sec ), and sec ) Solution: We have sec = 0, sec ) = π, sec ) = 4π 3 since and Note that sec π 3 is also, but sec 0 =, sec π =, sec 4π 3 = 0, π, 4π [ 3 0, π ) [ π, 3π ) sec ) π 3 since π [ 3 0, π ) [ π, 3π ) EXAMPLES: Find tan 0 cot 0 cot sec csc csc 3 4

5 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka EXAMPLES: We have tan 0 = 0, cot 0 = π, cot = π 4, sec = π 4, csc = π 6, csc 3 = π 3 EXAMPLES: Evaluate sin arcsin π ), arcsin sin π ), and arcsin sin 8π Solution: Since arcsin x is the inverse of the restricted sine function, we have Therefore but sinarcsin x) = x if x [, ] and arcsinsin x) = x if x [ π/, π/] arcsin sin 8π 7 ) sin arcsin π ) = π 7 7 and arcsin sin π ) = π 7 7 π )) = arcsin sin 7 + π = arcsin sin π ) = arcsin sin π ) = π ) EXAMPLES: Evaluate cot arcsin ) and sec arcsin ) Solution : We have cot θ = cos θ sin θ = ± sin θ sin θ and sec θ = cos θ = ± sin θ Since π arcsin x π, it follows that cosarcsin x) 0 Therefore if θ = arcsin, then hence and cot θ = cot arcsin ) = sec arcsin ) = sin θ sin θ and sec θ = sin arcsin ) sin arcsin ) = sin arcsin sin θ ) = ) = ) = Solution : Put θ = arcsin, so sin θ = Then cot arcsin ) = cot θ = and sec arcsin ) = sec θ = θ EXAMPLES: Evaluate, if possible, cot sin ) and sin tan )

6 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka EXAMPLES: Evaluate, if possible, cot sin ) and sin tan ) We first note that sin does not exist, since [, ], that is, is not in the domain of sin x Therefore cot sin ) does not exist We will evaluate sin tan ) in two different ways: Solution : We have tan θ sin θ = ± + tan θ Since π/ < tan x < π/, it follows that cos tan x) > 0 Therefore if θ = tan, then sin θ = tan θ + tan θ hence sin tan ) = tan tan ) + tan tan ) = = + Solution : Put θ = tan = tan, so tan θ = Then sin tan ) = sin θ = EXAMPLES: Evaluate sin cot )) and cos cot )) θ 6

7 Section 3 Inverse Trigonometric Functions EXAMPLES: Evaluate sin cot )) and cos cot )) 00 Kiryl Tsishchanka Solution : We have sin θ = ± + cot θ cot θ and cos θ = ± + cot θ Since 0 < cot x < π, it follows that sincot x) > 0 Therefore if θ = cot ), then sin θ = + cot θ and cos θ = cot θ + cot θ hence and sin cot )) = cos cot )) = + cot cot cot cot )) + cot cot )) = + )) = + ) = ) = Solution : Put θ = cot ), so cot θ = = Then sin cot )) = sin θ = and cos cot )) = cos θ = - θ 7

8 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka THEOREM: We have a) sin u) = u u d) cot u) = + u u b) cos u) = u u e) sec u) = u u u c) tan u) = + u u f) csc u) = u u u Proof: a) Let y = sin u, then sin y = u Therefore sin y) = u = cos y y = u = y = u cos y Since π sin }{{ u} π, it follows that cos y 0 Hence y cos y = sin y = [sin y = u] = u = y = u cos y = u u b) Let y = cos u, then cos y = u Therefore cos y) = u = sin y y = u = y = u sin y Since 0 cos }{{ u} π, it follows that sin y 0 Hence y sin y = cos y = [cos y = u] = u = y = u sin y = u u c) Let y = tan u, then tan y = u Therefore tan y) = u = sec y y = u = y = u sec y Note, that sec y = + tan y = [tan y = u] = + u Hence y = u sec y = u + u 8

9 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka a) sin u) = u u d) cot u) = + u u b) cos u) = u u e) sec u) = u u u c) tan u) = + u u f) csc u) = u u u d) Let y = cot u, then cot y = u Therefore cot y) = u = csc y y = u = y = u csc y Note, that csc y = + cot y = [cot y = u] = + u Hence e) Let y = sec u, then sec y = u Therefore y = u csc y = u + u sec y) = u = sec y tan y y = u = y = Since sec }{{ u} [0, π/) [π, 3π/), it follows that tan y 0 Hence y sec y tan y = sec y sec y = [sec y = u] = u u = y = f) Let y = csc u, then csc y = u Therefore u sec y tan y csc y) = u = csc y cot y y = u = y = csc y cot y Since csc }{{ u} 0, π/] π, 3π/], it follows that cot y 0 Hence y csc y cot y = csc y csc y = [csc y = u] = u u = y = EXAMPLES: a) Let fx) = x tan x) Find f x) b) Let fx) = sin 4x) Find f x) c) Let fx) = sec 3x) Find f x) u sec y tan y = u u u u u csc y cot y = u u u 9

10 Section 3 Inverse Trigonometric Functions 00 Kiryl Tsishchanka a) sin u) = u u d) cot u) = + u u b) cos u) = u u e) sec u) = u u u EXAMPLES: c) tan u) = + u u f) csc u) = u u u a) Let fx) = x tan x) Then f x) = [x tan x)] = x tan x) + x[tan x)] = tan x) + x x) + x) = tan x) + x + x) ) = tan x x) + x) = tan x x) x + x b) Let fx) = sin 4x) Then [ ] f x) = sin 4x) = sin 4x) ln [sin 4x) ] = sin 4x) ln 4x) 4x) = sin 4x) ln 4 4x) = sin 4x)+ ln 4x) c) Let fx) = sec 3x) Then f x) = [sec 3x)) / ] = sec 3x)) / [sec 3x)] = sec 3x)) / 3x) 3x) 3x) = sec 3x)) / 3x) 3x) 3) 3 = 3x) 3x3x ) sec 3x) 0