Group Thinking – CIC Problem How much money was spent at lunch yesterday?
Section 5.1, 5.4-5.6 Day 2 AP Calculus AB
Learning Targets Define Inverse Trig Functions Evaluate Inverse Trig Functions Properties of Inverse Trig Functions Solve Inverse Trig Equations Derivatives of Inverse Trig Equations
Inverse Trig Functions: Definitions Domain Range y = arcsin(x) y = sin-1(x) −1 ≤𝑥≤1 − 𝜋 2 ≤𝑦≤ 𝜋 2 y = arccos(x) y = cos-1(x) 0≤𝑦≤𝜋 y = arctan(x) y = tan-1(x) −∞ ≤𝑥≤∞ y = arccot(x) y = cot-1(x) y = arcsec(x) y = sec-1(x) 𝑥 ≥1 0≤𝑦≤𝜋, 𝑦 ≠ 𝜋 2 y = arccsc(x) y = csc-1(x) − 𝜋 2 ≤𝑦≤ 𝜋 2 , 𝑦 ≠0
Evaluating Inverse Trig Functions Find sin −1 − 1 2 =− 𝜋 6 Find cos −1 0 = 𝜋 2 Find tan −1 3 = 𝜋 3 Find sin −1 (0.3) =0.305
Properties of Inverse Trig Functions sin arcsin 𝑥 =x tan arctan 𝑥 =𝑥 cos arccos 𝑥 =𝑥 arcsin sin 𝑦 =𝑦 arccos cos 𝑦 =𝑦 arctan tan 𝑦 =𝑦 These properties hold for the other trig functions as well
Solving Trig Equations: Example 1 Solve arctan (2𝑥 −3) = 𝜋 4 2𝑥−3= tan 𝜋 4 2𝑥−3=1 𝑥=2
Example 2 Given 𝑦= arcsin 𝑥 where 0<𝑦< 𝜋 2 , find cos 𝑦 Draw the triangle knowing sin 𝑦 = 𝑥 1 Then solve for the 3rd side using Pythagorean theorem: 1− 𝑥 2 Thus cos 𝑦 = 1− 𝑥 2 1
Derivatives of Inverse Trig Functions 𝑑 𝑑𝑥 sin −1 u = u ′ 1− 𝑢 2 𝑑 𝑑𝑥 cos −1 u = − u ′ 1− 𝑢 2 𝑑 𝑑𝑥 tan −1 u = u ′ 1+ 𝑢 2
Derivatives of Inverse Trig Functions 𝑑 𝑑𝑥 csc −1 𝑢 = − 𝑢 ′ 𝑢 𝑢 2 −1 𝑑 𝑑𝑥 sec −1 𝑢 = 𝑢 ′ 𝑢 𝑢 2 −1 𝑑 𝑑𝑥 cot −1 𝑢 =− 𝑢 ′ 1+ 𝑢 2
Example 3 𝑑 𝑑𝑥 arcsin 2𝑥 𝑢 ′ 1− 𝑢 2 = 2 1−4 𝑥 2
Example 4 𝑑 𝑑𝑥 [ arctan 3𝑥 ] 𝑢 ′ 1+ 𝑢 2 = 3 1+9 𝑥 2
Example 5 𝑑 𝑑𝑥 [ arctan 𝑒 𝑥 ] 𝑢 ′ 1+ 𝑢 2 = 𝑒 𝑥 1+ 𝑒 2𝑥
Example 6 𝑑 𝑑𝑥 arccos 𝑒 2𝑥 − 𝑢 ′ 1− 𝑢 2 =− 2 𝑒 2𝑥 1− 𝑒 4𝑥
Example 7 𝑑 𝑑𝑥 𝑥 2 sin −1 𝑥 𝑥 2 1 1− 𝑥 2 + sin −1 𝑥 𝑥 2
Example 8 𝑑 𝑑𝑥 𝑥 ta n −1 ln 𝑥 + 3𝑥−1 4 𝑥 1 𝑥 1+ ln 𝑥 2 + tan −1 ( ln 𝑥 ) +4 3𝑥−1 3 (3) = 1 1+ ln 𝑥 2 + tan −1 ( ln 𝑥 ) +12 3𝑥−1 3
Summary of Derivatives Power Rule Constant Rule Product Rule Quotient Rule Chain Rule Trigonometric Rules Implicit Differentiation Exponential Differentiation Logarithmic Differentiation Inverse Trigonometric Differentiation