1. A Presentation By Cody Lee & Robyn Bursch

2. y= sin x means x= sin y where -1 ≤ x ≤ 1, - π/2 ≤ y ≤ π/2 y= cos x means x= cos y where -1 ≤ x ≤ 1, 0 ≤ y ≤ π y= tan x means x= tan y where -∞ < x < ∞, 0 < y < π y= sec x means x= sec y where |x|≥1, 0 ≤ y ≤ π, y≠ π/2 y= csc x means x= csc y where |x|≥1, - π/2 ≤ y ≤ π/2, y≠0 y= cot x means x= cot y where -∞ < x < ∞, 0 < y < π See pg 489 for formulas, pg 429-442 for detailed explanation

5. tan [cos (-1/3)] Ѳ= cos (-1/3), so cos Ѳ = -1/3 and since cos Ѳ < O, Ѳ lies in quadrant II Since cos is x/r, we let x=-1 and r=3 Use pythagoream theorem to find y (-1)²+ y² = 3² » 9-1=y² » y²=8 » y=2√2 Since we have y=2√2 and r= 3, tan [cos (-1/3)] = tan Ѳ= (2√2)/-1 = - 2√2

7. Show that cos Ѳ / 1+ sin Ѳ = 1-sin Ѳ /cos Ѳ by multiplying the numerator and denominator by 1-sin Solution: cos Ѳ/ 1+ sin Ѳ = cos Ѳ/ 1+ sin Ѳ x 1-sin Ѳ/ 1- sin Ѳ = cos Ѳ (1-sin Ѳ)/1-sin² Ѳ = cos Ѳ (1-sin Ѳ)/cos² Ѳ = 1-sin Ѳ/ cos Ѳ

9. Find the exact value of cos(75°) Solution: since 75° = 45°+30°, we use the formula for cos(α+β) Cos 75° = (45°+30°) = cos 45°cos 30° - sin 45°sin 30° = (√2/2)(√3/2)- (√2/2)( ½ ) = ¼(√6-√2)

13. If sin Ѳ= 3/5 and π/2 < Ѳ < π, find the exact value of cos (2Ѳ) Solution: because we are given sin Ѳ= 3/5, we can use the formula cos (2Ѳ)= 1 - 2sin²Ѳ. cos (2Ѳ)= 1- 2(3/5)² » 1- 2(9/25) » 1- 18/25 cos (2Ѳ)= 7/25

14. Use a half-angle formula to find the exact value of: sin(-15°) Solution: We use the fact that sin(-15°)= -sin(15°) and 15°= 30°/2 Use the formula sin α/2= ± √(1 - cos α/2) sin(-15°) = -sin(30°/2) = - √(1 - cos 30°/2) » - √(1 –(√3/2)/2) » 2 (- √(1 –(√3/2)/2) » (- √(2 –√3)/4) = - √(2 –√3)/2

15. Good Luck on the Final…