# Download 5.6 Day 1 Inverse Trig Functions

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```AP CALCULUS AB
5.6 Inverse Trigonometric Functions
Objectives: Develop properties of the six
inverse trig functions and differentiate an
inverse trig function
Inverse Functions
β’ To have an inverse, a function must be one to one and
have a reflection across the line y = x.
β’ However, none of the 6 trig functions have an inverse
because they are periodic and not one to one.
β’ In order for the trig functions to have an inverse, the
domains and ranges must be restricted.
β’ Take a look at the Inverse Trig worksheet for the Domain
and Range of each function.
Restrictions for an Inverse
sin π¦ = π₯
πππ
π¦ = πππ sin π₯
π¦ = π ππ
or
β1
y = π ππβ1 π₯
1
π₯β
sin π₯
Example
1
π¦ = πππ sin β
2
1
sin π¦ = β
2
Sine of what angle is
1
equal to β ?
2
π
π¦=β
6
Example
π¦ = πππ cos 0
cos π¦ = 0
Cosine of what angle is equal to 0?
π
π¦=
2
Example
π¦ = πππ tan 3
π‘ππ π¦ = 3
sin π¦
3
=
cos π¦
1
3
sin π¦
= 2
1
cos π¦
2
π
π¦=
3
Example with Calculator
π¦ = πππ sin(0.3)
MODE must be in RADIANS
π¦ = π ππβ1 (0.3)
π¦ β 0.305
Example with Calculator
π¦ = πππ sin(π)
MODE must be in RADIANS
π¦ = π ππβ1 (π)
π¦ β πΈππππ
WHY?
π
π
β β€πβ€
2
2
SOH β CAH β TOA
β’ Given: π¦ = πππ sin π₯
β’ Find: cos π¦
π¦ = πππ sin π₯
π₯ πππ
πππππππ  sin π¦ = =
1 π»π¦π
β’ cos π¦ = cos(πππ sin π₯)
β’ Use Right Triangle Trig
to Solve
β’ cos π¦ =
β’ cos π¦ =
π΄ππ
π»π¦π
1βπ₯ 2
1
Opp = x
y
π₯ 2 + π2 = 12
π2 = 1 β π₯ 2
π = ± 1 β π₯2
π = 1 β π₯2
sin π¦ ππ  πππ ππ‘ππ£π π π + 1 β π₯ 2
SOH β CAH β TOA
β’ Given: π¦ = πππ sππ
5
2
β’ Find: tan π¦
tan π¦ = tan πππ sec
5
2
β’ Use Right Triangle Trig to
Solve
β’ tan π¦ =
β’ tan π¦ =
πππ
π΄ππ
1
2
5
π¦ = πππ sππ
2
5 π»π¦π
πππππππ  sππ π¦ =
=
2
π΄ππ
Opp = b
y