Solve the equations for 0 ≤ θ ≤ 2 π . sin θ = 1 2

Given that, sin θ = 1 2 , 0 ≤ θ ≤ 2 π ⇒ sin θ = sin π 6 ⇒ θ = π 6 We know that, sin ( π - x ) = sin x ⇒ sin θ = sin π - π 6 ⇒ sin θ = sin 5 π 6 ⇒ θ = 5 π 6 ∴ θ = π 6 , 5 π 6

Solve the equations for 0 ≤ θ ≤ 2 π . cos θ = 2 2

Given that, cos θ = 2 2 , 0 ≤ θ ≤ 2 π ⇒ cos θ = 1 2 ⇒ cos θ = cos π 4 ⇒ θ = π 4 We know that, cos ( 2 π - x ) = cos x ⇒ cos θ = cos 2 π - π 4 ⇒ cos θ = cos π 4 ⇒ θ = 7 π 4 ∴ θ = π 4 , 7 π 4

Solve the equations for 0 ≤ θ ≤ 2 π . tan θ = 3

Given that, tan θ = 3 , 0 ≤ θ ≤ 2 π ⇒ tan θ = tan π 3 ⇒ θ = π 3 We know that, tan ( π + x ) = tan x ⇒ tan θ = tan π + π 3 ⇒ tan θ = tan 4 π 3 ⇒ θ = 4 π 3 ∴ θ = π 3 , 4 π 3

E M B I B E

Mathematics : Analysis and Approaches Standard Level Course Companion>Chapter 12 - Periodic Relationships: Trigonometric Functions>Exercise 12C

Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 12: Exercise 12C

Author:Natasha Awada, Paul La Rondie, Laurie Buchanan & Jill Stevens

Natasha Awada Mathematics Solutions for Exercise - Natasha Awada, Paul La Rondie, Laurie Buchanan and, Jill Stevens Solutions for Chapter: Periodic Relationships: Trigonometric Functions, Exercise 12: Exercise 12C

Attempt the practice questions on Chapter 12: Periodic Relationships: Trigonometric Functions, Exercise 12: Exercise 12C with hints and solutions to strengthen your understanding. Mathematics : Analysis and Approaches Standard Level Course Companion solutions are prepared by Experienced Embibe Experts.