Solve the equation sin θ - ϕ = 1 2 and cos θ + ϕ = 1 2 .

Given equation sin θ - ϕ = 1 2 & cos θ + ϕ = 1 2 ⇒ sin θ - ϕ = sin π 6 & cos θ + ϕ = cos π 3 ⇒ θ - ϕ = nπ + - 1 n π 6 , n ∈ I . . . 1 & θ + ϕ = 2 m π ± π 3 , m ∈ I . . . 2 From equation 1 + 2 , we get ⇒ θ - ϕ + θ + ϕ = nπ + - 1 n π 6 + 2 mπ ± π 3 ⇒ 2 θ = nπ + - 1 n π 6 + 2 mπ ± π 3 ⇒ θ = nπ 2 + mπ + - 1 n π 12 ± π 6 ⇒ θ = n 2 + m π + - 1 n π 12 ± π 6 , m , n ∈ I From 2 - 1 , we get ⇒ θ + ϕ - θ - ϕ = 2 mπ ± π 3 - nπ + - 1 n π 6 ⇒ 2 ϕ = 2 mπ - nπ ± π 3 - - 1 n π 6 ⇒ ϕ = mπ - nπ 2 ± π 6 + - 1 n + 1 π 12 ⇒ ϕ = m - n 2 π ± π 6 + - 1 n + 1 π 12

Solve the equation cos 2 x + 3 y = 1 2 and cos 3 x + 2 y = 3 2 .

Given equation cos 2 x + 3 y = 1 2 & cos 3 x + 2 y = 3 2 ⇒ cos 2 x + 3 y = cos π 3 & cos 3 x + 2 y = cos π 6 ∴ cos θ = cos α ⇒ θ = 2 n π ± α , n ∈ I . ⇒ 2 x + 3 y = 2 nπ ± π 3 . . . 1 & 3 x + 2 y = 2 mπ ± π 6 . . . 2 . Multiplying equation 1 , by 3 and equation 2 , by 2 , then subtract, we get ⇒ 9 x + 6 y - 4 x + 6 y = 6 mπ ± π 2 - 4 nπ ± 2 π 3 ⇒ 5 x = 6 mπ ± π 2 - 4 nπ ∓ 2 π 3 ⇒ x = 1 5 6 mπ - 4 nπ ± π 2 ∓ 2 π 3 , m , n ∈ I . Multiplying equation 2 , by 3 and equation 1 , by 2 , then subtract, we get ⇒ 6 x + 9 y - 6 x + 4 y = 6 nπ ± π - 4 mπ ± π 3 ⇒ 5 y = 6 nπ - 4 mπ ± π ∓ π 3 ⇒ y = 1 5 6 nπ - 4 mπ ± π ∓ π 3 .

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Plane Trigonometry Part 1>Chapter 6 - General Expressions for All Angles Having a Given Trigonometrical Ratio>EXAMPLES XII

S L Loney Solutions for Chapter: General Expressions for All Angles Having a Given Trigonometrical Ratio, Exercise 2: EXAMPLES XII

Author:S L Loney

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: General Expressions for All Angles Having a Given Trigonometrical Ratio, Exercise 2: EXAMPLES XII

Attempt the practice questions on Chapter 6: General Expressions for All Angles Having a Given Trigonometrical Ratio, Exercise 2: EXAMPLES XII with hints and solutions to strengthen your understanding. Plane Trigonometry Part 1 solutions are prepared by Experienced Embibe Experts.