Some Worked Problems on Inverse Trig Functions

arccos is defined in a range of [0, π] and domain of [ 1,1]− Find the exact value of the expression whenever it is defined − − 2 1 sin 1 − 2 2 cos 1 tan −1(−1) 2 1 arcsin − 2 3 arccos arctan (3) For the following problems: Always work from the inside out 2 2 sin (1)]arcsin cos[cos −1 tan [tan −1(5)] You will notice when

Inverse trigonometric functions (Sect 76) Review

y = arctan(x) y y = arccsc(x)-1 0 1 p / 2 - p / 2 x y = arcsec(x)-1 10 p / 2 p y x y 0 p / 2 x y = arccot(x) Review: Deﬁnitions and properties Theorem For all x ∈ [−1,1] the following identities hold, arccos(x)+arccos(−x) = π, arccos(x)+arcsin(x) = π 2 Proof: arccos(-x) q 1 y-x = cos (p-q) x = cos (q) p - q q x arccos(x) arccos(x) q

63: Inverse Trigonometric Functions

Other notations for inverse functions:sin 1(x) = arcsin(x), cos 1(x) = arccos(x) and tan 1 (x) = arctan(x) Note that the inverse function notation is only true for the restricted

CALCULUS Inverse functions

arccos, arctan, arccot EXERCISE Sin Sin : [ — 2 sin —00, oo Sin in x un arccos arcsin angles complementary ïFWfns yield Inve ext subt arccot R (O, T)

46 Inverse Trigonometric Functions

(a)arccos(p 2 2) (b)arccos(p 3 2) (c)arcsin(p 3 2) (d)arctan(1) (e)arctan(p 3) (f)arctan(p 3) Solutions (a)Since arccos(x) is the inverse function of cos(x) then we seek here an angle whose cosine is p 2 2:Since cos(ˇ 4) = p 2 2 then arccos(p 2 2) should be ˇ 4: (b)arccos( 5 p 3 2) = ˇ 6 since cos(5ˇ 6) = p 3 2: (c)arcsin(p 3 2) = ˇ 3

Inverse Trigonometric Functions

y arcsin sin 1x: y cosx: y arccos x cos 1 x: y xtanx: y arctan x tan 1: Trig function Restricted domain Inverse trig function Principle value range 2 2 S S d x d y arcsinx 2 2 S S d y d y cosx 0 xd S y Sarccosx 0 yd y tanx 2 2 S S x y arctanx 2 2 S S y 1 2 S 2 S 1 2 S 1 1 2 S 2 S S 2 S 2 S y x

Inverse trigonometric functions (Sect 76)

The derivative of arcsin is given by arcsin0(x) = 1 √ 1−x2 Proof: For x ∈ [−1,1] holds arcsin0(x) = 1 sin0 arcsin(x) = 1 cos arcsin(x) For x ∈ [−1,1] we get arcsin(x) = y ∈ hπ 2, π 2 i, and the cosine is positive in that interval, then cos(y) = + q 1−sin2(y), hence arcsin0(x) = 1 q 1−sin2 arcsin(x) ⇒ arcsin 0(x) = 1 √ 1

MATH 150 { TOPIC 15 INVERSE TRIGONOMETRIC FUNCTIONS I

and/or by plotting some points, we see that arcsin(x) has the following graph arcsin(x)orsin−1(x) 1 −1 ˇ 2 −ˇ 2 Fig 15 3 The derivations of the graphs of arccos(x) and arctan(x) are similar to that of arcsin(x): For arccos(x); we rst consider cosx where 0 x ˇ: cosx,0 x ˇ −1 1 ˇ 2 ˇ Fig 15 4