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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₯ πππ π¦ = πππ sin π₯ π¦ = π ππ or β1 y = π ππβ1 π₯ 1 π₯β sin π₯ Example 1 π¦ = πππ sin β 2 1 sin π¦ = β 2 Sine of what angle is 1 equal to β ? 2 π π¦=β 6 Example π¦ = πππ cos 0 cos π¦ = 0 Cosine of what angle is equal to 0? π π¦= 2 Example π¦ = πππ tan 3 π‘ππ π¦ = 3 sin π¦ 3 = cos π¦ 1 3 sin π¦ = 2 1 cos π¦ 2 π π¦= 3 Example with Calculator π¦ = πππ sin(0.3) MODE must be in RADIANS π¦ = π ππβ1 (0.3) π¦ β 0.305 Example with Calculator π¦ = πππ sin(π) MODE must be in RADIANS π¦ = π ππβ1 (π) π¦ β πΈπ π ππ WHY? π π β β€πβ€ 2 2 SOH β CAH β TOA β’ Given: π¦ = πππ sin π₯ β’ Find: cos π¦ π¦ = πππ sin π₯ π₯ πππ πππππππ sin π¦ = = 1 π»π¦π β’ cos π¦ = cos(πππ sin π₯) β’ Use Right Triangle Trig to Solve β’ cos π¦ = β’ cos π¦ = π΄ππ π»π¦π 1βπ₯ 2 1 Opp = x y Adj = b π₯ 2 + π2 = 12 π2 = 1 β π₯ 2 π = ± 1 β π₯2 π = 1 β π₯2 sin π¦ ππ πππ ππ‘ππ£π π π + 1 β π₯ 2 SOH β CAH β TOA β’ Given: π¦ = πππ sππ 5 2 β’ Find: tan π¦ tan π¦ = tan πππ sec 5 2 β’ Use Right Triangle Trig to Solve β’ tan π¦ = β’ tan π¦ = πππ π΄ππ 1 2 5 π¦ = πππ sππ 2 5 π»π¦π πππππππ sππ π¦ = = 2 π΄ππ Opp = b y Adj = 2 22 + π2 2 = 5 4 + π2 = 5 π2 = 1 π = ±1 sππ π¦ ππ πππ ππ‘ππ£π π π π = +1 Formative Assessment β’ Pg. 377 (5-19) odd