3.11 - Larger Trigonometric Equations | High School

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High School
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High School Physics
Academic year:

2024
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Trigonometry Lessons 3.11 Trigonometric Equations Identities Simplify Solve Solving Trig Equations • Solving for where two functions meet Example: [0,2π] cos 𝑥 = sin 𝑥 sin 𝑥 =1 cos 𝑥 tan 𝑥 = 1 𝑥 = tan−1 1 𝑥1 = 𝜋 4 5𝜋 𝑎𝑛𝑑 𝑥₂ = 4 Ex 1: 2cos2 𝑥 − cos 𝑥 = 1, where 𝑥 ∈ 0,2𝜋 Make equation equal to zero 2cos2 𝑥 − cos 𝑥 − 1 = 0 Equation is quadratic. Can define cos 𝑥 2𝑢 − u − 1 = 0 2 𝑢 + 1)( 𝑢 − 1 = 0 2 cos 𝑥 + 1 = 0 cos 𝑥 = − 1 2 2𝜋 4𝜋 𝑥= , 3 3 =𝑢 2𝜋 4𝜋 𝑥 = {0, , , 2𝜋} 3 3 cos 𝑥 − 1 = 0 cos 𝑥 = 1 𝑥 = 0, 2𝜋 Ex 2: 3 sin2 𝑥 − cos2 𝑥 = 1, where 𝑥 ∈ 0,2𝜋 Change cos to have equation using sin only 3sin2 𝑥 − (1 − sin2 𝑥) = 1 3sin2 𝑥 − 1 + sin2 𝑥 = 1 4 sin2 𝑥 = 2 sin2 𝑥 sin 𝑥 = + 𝜋 3𝜋 𝑥= , 4 4 2 2 1 = 2 𝜋 3𝜋 5𝜋 7𝜋 𝑥={ , , , } 4 4 4 4 sin 𝑥 = − 5𝜋 7𝜋 𝑥= , 4 4 2 2 Ex 3: cos2 𝑥 − sin2 𝑥 = 2 − 5 cos 𝑥, where 𝑥 ∈ 𝜋 −2,0 Change sine to have equation using cosine only cos2 𝑥 − (1 − cos2 𝑥) = 2 − 5 cos 𝑥 cos2 𝑥 − 1 + cos2 𝑥 = 2 − 5 cos 𝑥 2 cos2 𝑥 + 5 cos 𝑥 − 3 = 0 If cos 𝑥 = 𝑢 2𝑢2 + 5u − 3 = 0 Ex 3: 2𝑢 2 + 5u − 3 = 0 2 𝑢 − 1)( 𝑢 + 3 = 0 2 cos 𝑥 − 1 = 0 cos 𝑥 = 1 2 cos 𝑥 + 3 = 0 cos 𝑥 = −3 𝜋 5𝜋 3 3 no solution 𝑥= , Subtract period (2𝜋 ) to find answers in the interval 𝜋 3 - 2𝜋 = 5𝜋 - (too small) 3 5𝜋 3 - 2𝜋 𝜋 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑥 = − 3 𝜋 =- (just right) 3 Ex 4: 2 tan 𝜃 + 4 cot 𝜃 = −5 csc 𝜃 To simplify express this function in terms of sin 𝜃 and cos 𝜃 : 2 sin 𝜃 4 cos 𝜃 −5 + = cos 𝜃 sin 𝜃 sin 𝜃 Find a common denominator 2 sin2 𝜃 4 cos2 𝜃 −5 cos 𝜃 + = sin 𝜃 cos 𝜃 cos 𝜃 sin 𝜃 cos 𝜃 sin 𝜃 Collect 2 sin2 𝜃 + 4 cos 2 𝜃 −5 cos 𝜃 = sin 𝜃 cos 𝜃 cos 𝜃 sin 𝜃 Since both sides have the same denominator .. The numerators will be equal 2 sin2 𝜃 + 4cos 2 𝜃 = −5 cos 𝜃 Using Identity 1 2 1 − cos 2 𝜃 + 4cos 2 𝜃 = −5 cos 𝜃 2 − 2 cos 2 𝜃 + 4cos 2 𝜃 = −5 cos 𝜃 Collect Like Terms 2 cos 2 𝜃 + 5 cos 𝜃 + 2 = 0 Ex 4: 2 cos2 𝜃 + 5 cos 𝜃 + 2 = 0 Factor: (2 cos 𝜃 + 1)(cos 𝜃 + 2) = 0 𝑥2 𝑥1 cos 𝜃 + 2 = 0 2 cos 𝜃 + 1 = 0 cos 𝜃 = −2 2 cos 𝜃 = −1 1 IMPOSSIBLE cos 𝜃 = − 2 2𝜋 𝜃1 = 3 4𝜋 𝜃2 = 3 2𝜋 4𝜋 θ= + 2𝜋𝑛, + 2𝜋𝑛 ; 𝑛 ∈ ℤ 3 3