1. Chapter 11 Trigonometric Functions 11.1 Trigonometric Ratios and General Angles 11.2 Trigonometric Ratios of Any Angles 11.3 Graphs of Sine, Cosine and Tangent Functions

2. Trigonometric Equations Objectives In this lesson, we will learn how to extend the definitions of sine, cosine and tangent to any angle, determine the sign of a trigonometric ratio of an angle in a quadrant, relate the trigonometric functions of any angle to that of its basic (reference) angle and solve simple trigonometric equations. 11.2 Trigonometric Ratios of Any Angles

3. Trigonometric Ratios of Any Angles The three trigonometric ratios are defined as Trigonometric Equations x y r yy r x r x PQ = y OQ = x

4. Example 4 Find the values of cos θ, sin θ and tan θ when θ = 135 0. Solution Trigonometric Equations When θ = 135 0, 180 0 – θ = 45 0. P has coordinates (1, -1) and

5. Trigonometric Equations

6. Signs of Trigonometric Ratios in Quadrants 1st quadrant Trigonometric Equations θ = α P has coordinates ( a, b ) 2nd quadrant θ = ( 180° – α ) P has coordinates ( – a, b ) 3rd quadrant θ = ( 180° + α ) P has coordinates ( – a, – b ) 4th quadrant θ = ( 360° – α ) P has coordinates ( a, – b ).

7. Trigonometric Equations For positive ratios Signs of Trigonometric Ratios in Quadrants In the four quadrants S (sin θ)A ( all ) T (tan θ) C (cos θ) The signs are summarised in this diagram.

8. Trigonometric Equations Example 6(a) Without using a calculator, evaluate cos 120°. Solution Basic angle, 120° is in the 2nd quadrant, so cosine is negative AS TC

9. Basic angle, is in the 4th quadrant, so sine is negative AS T C Example 6(c) Without using a calculator, evaluate Solution Trigonometric Equations

10. Example 8(a) Find all the values of θ between 0° and 360° such that sin θ = – 0.5. Solution For the basic angle, Since sin θ < 0, θ is in the 3rd or 4th quadrant, AS TC Basic Trigonometric Equations

11. Trigonometric Equations For the basic angle, θ is in the 1st, 2nd, 3rd or 4th quadrant, AS TC Example 9(a) Find all the values of θ between 0 and 2π such that 2sin 2 θ – 1 = 0. Solution