Trigonometry I Angle Ratio & Exact Values. By Mr Porter.

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Presentation transcript:

Trigonometry I Angle Ratio & Exact Values. By Mr Porter

Basic Definitions Hypotenuse Adjacent Opposite θ New definition (Reciprocals)

Exact Values (0°, 30°, 45°, 60°, 90°) You need to know the construction of the following two triangles for exact values! 60° 30° 45° Angle θ = 0° Sin 0° = 0 Cos 0° = 1 Tan 0° = 0 Angle θ = 90° Sin 90° = 1 Cos 90° = 0 Tan 90° = UNDEFINED

Exact Values (0°, 30°, 45°, 60°, 90°) You need to know the construction of the following two triangles for exact values! 60° 30° 45° The two triangles allow you to derive the exact values.

Angles greater than 90°. Sign of the trigonometric ratios. 0°(360° ) 90° 270° 180° 1 st Quadrant 2 nd Quadrant 3 rd Quadrant 4 th Quadrant Sin (+) All (+) Cos (+) Tan (+) θ°180° — θ° 180° + θ°360° — θ° 1 st Quadrant 0°≤ angle ≤ 90 All trig ratios are positive. 2 nd Quadrant 90°< angle ≤ 180° Sin θ is positive ONLY. 3 rd Quadrant 180°< angle ≤ 270° Tan θ is positive ONLY. 4 th Quadrant 270° < angle ≤ 360° Cos θ is positive ONLY. Important: You need to write the angle α (not acute) in terms of angle θ (acute). 90° < α ≤ 180°  α = 180 – θ2 nd Quadrant 180° < α ≤ 270°  α = θ3 rd Quadrant 270° < α ≤ 360°  α = 360 – θ4 th Quadrant Learn to construct the circle!

Practice Constructing these diagrams! 0°(360° ) 90° 270° 180° Sin (+) All (+) Cos (+) Tan (+) θ°180° — θ° 180° + θ°360° — θ° Learn to construct the circle! 60° 30° 45° Learn to construct the 2 triangles!

Examples A a) Find the exact value of tan 120° (i) Construct the circle! 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° (ii) Which Quadrant? 2 nd Quad.  (–)tan … α = 180 – θ, (θ acute) (iii) tan 120° = tan (180 – θ) = tan (180 – 60) = - tan 60 (iv) Construct the triangle! 60° 30° (v) tan 120° = – tan 60° = b) Find the exact value of cos 300° (i) Construct the circle! 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° (ii) Which Quadrant? 4 th Quad.  (+)cos … α = 360 – θ, (θ acute) (iii) Cos 300° = cos (360 – θ) = cos (360 – 60) = + cos 60 (iv) Construct the triangle! 60° 30° (v) cos 300° = + cos 60° =

Examples B c) Find the exact value of sin 135° (i) Construct the circle! 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° (ii) Which Quadrant? 2 nd Quad.  (+)sin … α = 180 – θ, (θ acute) (iii) sin 135° = sin (180 – θ) = sin (180 – 45) = + sin 45 (iv) Construct the 45° isosc. triangle! (v) sin 135° = + sin 45° = d) Find the exact value of cos 210° (i) Construct the circle! 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° (ii) Which Quadrant? 3 rd Quad.  (–)cos … α = θ, (θ acute) (iii) cos 210° = cos (180 + θ) = cos ( ) = – cos 30 (iv) Construct the triangle! 60° 30° (v) cos 210° = – cos 30° = 45°

Exercise Hint: Construct the circle and 2 triangle for each question. a)Find the exact value of sin 315° b)Find the exact value of tan 240° c)Find the exact value of cos 150° d)Find the exact value of sin 120° e)Find the exact value of cos 240° f)Find the exact value of tan 210° g)Find the exact value of sin 330°