Homework Log Fri 5/6 Lesson 8 – 7 Learning Objective: To find inverse trig functions Hw: #817 Pg. 534 1 – 13 odd

5/6/16 Lesson 8 – 7 Inverse Trig Functions Day 1 Advanced Math/Trig

Learning Objective To find inverse trig functions

Solve 11. cot 𝑥 2 = 3 0<𝜃<2𝜋 tan 𝑥 2 = 3 3 𝑥 2 = 𝜋 6 +𝜋𝑛 𝑥= 𝜋 3 𝑥= 𝜋 3 +2𝜋𝑛

Solve 12. sin 𝑥 − csc 𝑥 =0 0<𝜃<2𝜋 sin 𝑥 = csc 𝑥 sin 𝑥 = 1 sin 𝑥 𝑠𝑖𝑛 2 𝑥 =1 𝑠𝑖𝑛 𝑥 =±1 𝑥= 𝜋 2 , 3𝜋 2

Consider graph of y = sin x Recall from Ch 4 that an inverse function must be a one-to-one function. 1-1 function: Passes vertical line test AND horizontal line test. Is the sine function 1-1? NO!!

Inverse of y = sin x So the sine function does not have an inverse. But, what if we restricted the domain? To get an inverse function, we must restrict the domain of the sine function to the interval − 𝜋 2 ≤𝑥≤ 𝜋 2 Now, an inverse function exists! 

Inverse Trig Functions 𝑦= 𝑠𝑖𝑛 −1 𝑥 or 𝑦= arcsin 𝑥 Is equivalent to 𝑥= sin 𝑦 Where − 𝜋 2 ≤𝑦≤ 𝜋 2 1. Find the exact value of arcsin − 1 2 y= arcsin − 1 2  − 1 2 = sin 𝑦 & − 𝜋 2 ≤𝑦≤ 𝜋 2 𝜋 2 Ref ∠ is 𝜋 6 but in QIV − 𝜋 6 𝑦=− 𝜋 6 − 𝜋 2

Inverse Trig Functions We must restrict the other trig functions to get an inverse function. 𝜋 2 Function Equiv. Eqtn. Range banana − 𝜋 2 ≤𝑦≤ 𝜋 2 y = sin-1x x = sin y − 𝜋 2 (or y = arcsin x) rainbow y = cos-1x x = cos y 0≤𝑦≤𝜋 𝜋 (or y = arccos x) 𝜋 2 banana − 𝜋 2 <𝑦< 𝜋 2 y = tan-1x x = tan y (or y = arctan x) − 𝜋 2

Inverse Trig Functions 𝜋 2 Function Equiv. Eqtn. Range banana − 𝜋 2 ≤𝑦≤ 𝜋 2 y = csc-1x x = csc y − 𝜋 2 (or y = arccsc x) 𝑦≠0 rainbow y = sec-1x x = sec y 0≤𝑦≤𝜋 𝜋 (or y = arcsec x) 𝑦≠ 𝜋 2 rainbow 0<𝑦<𝜋 y = cot-1x x = cot y 𝜋 (or y = arccot x)

How to Remember Banana or Rainbow sin B e csc B cos R ight sec R CUZ tan B ack cot R It’s cold!!

Evaluate 2. 𝑐𝑜𝑠 −1 − 1 2 y= arc𝑐𝑜𝑠 − 1 2  − 1 2 = cos 𝑦 2. 𝑐𝑜𝑠 −1 − 1 2 y= arc𝑐𝑜𝑠 − 1 2  − 1 2 = cos 𝑦 cosine is a rainbow Ref ∠ is 𝜋 3 but in QII 0≤𝑦≤𝜋 𝑦= 2𝜋 3 2𝜋 3 𝜋

Evaluate 3. 𝑠𝑖𝑛 −1 2 y= 𝑠𝑖𝑛 −1 2  sin 𝑦 =2 Impossible! −1≤ sin 𝑦 ≤1 3. 𝑠𝑖𝑛 −1 2 y= 𝑠𝑖𝑛 −1 2  sin 𝑦 =2 Impossible! −1≤ sin 𝑦 ≤1 Undefined

Evaluate 4. 𝑐𝑜𝑠 −1 1 y= 𝑐𝑜𝑠 −1 1  1= cos 𝑦 0≤𝑦≤𝜋 𝑦=0 5. 𝑐𝑠𝑐 −1 2 4. 𝑐𝑜𝑠 −1 1 y= 𝑐𝑜𝑠 −1 1  1= cos 𝑦 0≤𝑦≤𝜋 𝑦=0 5. 𝑐𝑠𝑐 −1 2 y= 𝑐𝑠𝑐 −1 2  2= csc 𝑦  1 2 = sin 𝑦 𝑦= 𝜋 6 − 𝜋 2 ≤𝑦≤ 𝜋 2

Evaluate 6. 𝑠𝑖𝑛 −1 −1 y= 𝑠𝑖𝑛 −1 −1  −1= sin 𝑦 − 𝜋 2 ≤𝑦≤ 𝜋 2 𝑦=− 𝜋 2 6. 𝑠𝑖𝑛 −1 −1 y= 𝑠𝑖𝑛 −1 −1  −1= sin 𝑦 − 𝜋 2 ≤𝑦≤ 𝜋 2 𝑦=− 𝜋 2 7. 𝑎𝑟𝑐 sec (−2) y=𝑎𝑟𝑐 sec (−2)  −2= sec 𝑦  − 1 2 = cos 𝑦 𝑦= 2𝜋 3 0≤𝑦≤𝜋

Evaluate y= 𝑐𝑜𝑠 −1 − 3 2  − 3 2 = cos 𝑦 0≤𝑦≤𝜋 𝑦= 5𝜋 6 9. 𝑠𝑖𝑛 −1 3 8. 𝑐𝑜𝑠 −1 − 3 2 y= 𝑐𝑜𝑠 −1 − 3 2  − 3 2 = cos 𝑦 0≤𝑦≤𝜋 𝑦= 5𝜋 6 9. 𝑠𝑖𝑛 −1 3 y= 𝑠𝑖𝑛 −1 3  3= s𝑖𝑛 𝑦 Undefined −1≤ sin 𝑦 ≤1

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Homework #817 Pg. 534 1 – 13 odd