Trigonometry II Harder Exact Values and Simple Trig Equations. By Mr Porter

Harder Exact Value Questions. 1) Given that tan θ = ⅗ and sin θ < 0, find the exact value of cos θ. a) Construct a right angle triangle, label Opposite (3) and Adjacent (5). θ 3 5 b) Use Pythagoras Thm to find missing side. c) Construct TRIG CIRCLE. Tan(+) 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+) θ°180° — θ° 180° + θ° 360° — θ° d) tan θ is positive (1 st and 3 rd Quad.) and sin θ is negative (3 rd and 4 th Quad.) Hence, θ is in 3 rd Quadrant. cos θ is negative. By definition

Harder Exact Value Questions. 2) Given that sin θ = -⅔ and cos θ < 0, find the exact value of cot θ. a) Construct a right angle triangle, label Opposite (2) and Hypotenuse (3). θ 2 3 b) Use Pythagoras Thm to find missing side. c) Construct TRIG CIRCLE. Tan(+) 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+) θ°180° — θ° 180° + θ° 360° — θ° d) sin θ is negative (3 rd and 4 th Quad.) and cos θ is negative (2 nd and 3 rd Quad.) Hence, θ is in 3 rd Quadrant. cot θ is positive (same as tan θ). By definition

Harder Exact Value Questions. 3) Given that cos θ = -⅜ and tan θ < 0, find the exact value of cosec θ. a) Construct a right angle triangle, label Adjacent (3) and Hypotenuse (8). θ 3 8 b) Use Pythagoras Thm to find missing side. c) Construct TRIG CIRCLE. Tan(+) 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+) θ°180° — θ° 180° + θ° 360° — θ° d) cos θ is negative (2 nd and 3 rd Quad.) and tan θ is negative (2 nd and 4 th Quad.) Hence, θ is in 2 nd Quadrant. cosec θ is positive (same as sin θ). By definition

Exercise 1 a) Given that cos θ = ¼ and sin θ < 0, find the exact value of tan θ. b) Given that tan θ = - ⅕ and sin θ > 0, find the exact value of sec θ. c) Given that sin θ = -⅘ and cos θ > 0, find the exact value of cot θ. d) Given that cot θ = -⅞ and sec θ < 0, find the exact value of cosec θ.

Simple Trigonometric Equations & Exact Values. {Hints: Exact values triangles and trig circle.} a)Solve 2sin α = 1, for 0° ≤ α ≤ 360°. [These normally have 2 answers] Rearrange the equation, making sinα, the subject. Construct the trig circle and exact values triangles. Tan(+) 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+) θ°180° — θ° 180° + θ° 360° — θ° 60° 30° 45° From the triangles,, and RHS is positive, indicating a 1 st and 2 nd Quadrant answers are possible. 1 st Quad answer 2 nd Quad answer (180 – α)

Simple Trigonometric Equations & Exact Values. b) Solve cos α + 1 = 0, for 0° ≤ α ≤ 360°. Rearrange the equation, making cos α, the subject. Construct the trig circle and exact values triangles. Tan(+) 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+) θ°180° — θ° 180° + θ° 360° — θ° 60° 30° 45° From the triangles,, and RHS is negative, indicating a 2 nd and 3 rd Quadrant answers are possible. 2 nd Quad answer (180 – α)3 rd Quad answer (180 + α)

Simple Trigonometric Equations & Exact Values. c) Solve sin α = cos α, for 0° ≤ α ≤ 360°. Rearrange the equation, making single trig ratio (divide by cos α). Construct the trig circle and exact values triangles. Tan(+) 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+) θ°180° — θ° 180° + θ° 360° — θ° 60° 30° 45° From the triangles,, and RHS is positive, indicating a 1 st and 3 rd Quadrant answers are possible. 1 st Quad answer (α)3 rd Quad answer (180 + α)

Exercise :The domain of θ is 0° ≤ θ ≤ 360° a) Find all solution to b) Find all solution to c) Find all solution to d) Find all solution to e) Find all solution to