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Trigonometry II Harder Exact Values and Simple Trig Equations. By Mr Porter Harder Exact Value Questions. 1) Given that tan θ = ⅗ and sin θ < 0 , find the exact value of cos θ . 3 a) Construct a right angle triangle, label Opposite (3) and Adjacent (5). c =3 +5 2 2 2 c = 34 2 c) Construct TRIG CIRCLE. c = 34 90 ° Sin(+) 180 ° d) tan θ is positive (1st and 3rd Quad.) and sin θ is negative (3rd and 4th Quad.) Hence, θ is in 3rd Quadrant. cos θ is negative. All (+) 180° — θ° 180° + θ° Tan(+) θ 0°(36 ° 360° — 0°) θ° Cos(+) 270 ° θ 5 c2 = a 2 + b2 b) Use Pythagoras Thm to find missing side. 34 By definition A H -5 cos q = 34 cos q = Harder Exact Value Questions. 2) Given that sin θ = -⅔ and cos θ < 0 , find the exact value of cot θ . 3 2 a) Construct a right angle triangle, label Opposite (2) and Hypotenuse (3). b) Use Pythagoras Thm to find missing side. 5 a 2 + b2 = c2 a +2 =3 2 2 2 By definition a 2 = 32 - 2 2 c) Construct TRIG CIRCLE. a = 5 90 ° Sin(+) d) sin θ is negative (3rd and 4th 180 Quad.) and cos θ is negative (2nd ° and 3rd Quad.) Hence, θ is in 3rd Quadrant. cot θ is positive (same as tan θ). θ All (+) 180° — θ° 180° + θ° Tan(+) θ 0°(36 ° 360° — 0°) θ° Cos(+) 270 ° 1 tan q A cot q = O 5 cot q = 2 cot q = Harder Exact Value Questions. 3) Given that cos θ = -⅜ and tan θ < 0 , find the exact value of cosec θ . 8 55 a) Construct a right angle triangle, label Adjacent (3) and Hypotenuse (8). 3 a 2 + b2 = c2 b) Use Pythagoras Thm to find missing side. a +3 =8 2 2 2 By definition a 2 = 82 - 32 c) Construct TRIG CIRCLE. a = 55 90 ° Sin(+) (2nd 3rd d) cos θ is negative and Quad.) and tan θ is negative (2nd and 4th Quad.) Hence, θ is in 2nd Quadrant. cosec θ is positive (same as sin θ). 180 ° θ All (+) 180° — θ° 180° + θ° Tan(+) θ 0°(36 ° 360° — 0°) θ° Cos(+) 270 ° 1 sin q H cosec q = O 8 cosec q = 55 cosec q = Exercise 1 a) Given that cos θ = ¼ and sin θ < 0, find the exact value of tan θ. ans : tanq = - 15 b) Given that tan θ = -⅕ and sin θ > 0, find the exact value of sec θ. ans : secq = - 26 5 c) Given that sin θ = -⅘ and cos θ > 0, find the exact value of cot θ. ans : cot q = -3 4 d) Given that cot θ = -⅞ and sec θ < 0, find the exact value of cosec θ. ans : cosecq = 113 8 Simple Trigonometric Equations & Exact Values. {Hints: Exact values triangles and trig circle.} 90 ° Sin(+) a) Solve 2sin α = 1, for 0° ≤ α ≤ 360°. 180 ° [These normally have 2 answers] 180° — θ° 180° + θ° Tan(+) Rearrange the equation, making sinα, the subject. sin a = 1 2 30° 1st Quad answer sin a = sin30° a = 30° 45 ° 2 2 1 3 1 answers are possible. θ 0°(36 ° 360° — 0°) θ° Cos(+) 270 ° Construct the trig circle and exact values triangles. From the triangles, sin30° = All (+) 60 ° 1 45 ° 1 , and RHS is positive, indicating a 1st and 2nd Quadrant 2 2nd Quad answer (180 – α) sin a = sin(180 - 30°) sin a = sin150° a = 150° Simple Trigonometric Equations & Exact Values. 90 ° Sin(+) b) Solve 2 cos α + 1 = 0, for 0° ≤ α ≤ 360°. 180 ° Rearrange the equation, making cos α, the subject. All (+) 180° — θ° 180° + θ° Tan(+) θ 0°(36 ° 360° — 0°) θ° Cos(+) 45 ° 2 1 1 45 ° 270 ° 1 cos a = - 2 Construct the trig circle and exact values triangles. 30° 2 3 1 60 ° 1 From the triangles, cos 45° = , and RHS is negative, indicating a 2nd and 3rd Quadrant 2 answers are possible. 2nd Quad answer (180 – α) cosa = cos(180° - 45°) cos a = cos135° a = 135° 3rd Quad answer (180 + α) cosa = cos(180 + 45°) cosa = cos225° a = 225° Simple Trigonometric Equations & Exact Values. c) Solve sin α = cos α, for 0° ≤ α ≤ 360°. Rearrange the equation, making single trig ratio (divide by cos α). Sin(+) 180 ° tan a = 1 Construct the trig circle and exact values triangles. From the triangles, tan 45° = 1 , and RHS is positive, 1 All (+) 180° — θ° 180° + θ° Tan(+) θ 0°(36 ° 360° — 0°) θ° Cos(+) 45 ° 30° 2 3 270 ° indicating a 1st and 3rd Quadrant answers are possible. tan a = tan 45° a = 45° 2 1 90 ° sin a cos a = =1 cos a cos a 1st Quad answer (α) 45 ° 3rd Quad answer (180 + α) tan a = tan(180 + 45°) tan a = tan 225° a = 225° 1 60 ° Exercise :The domain of θ is 0° ≤ θ ≤ 360° a) Find all solution to 2cosq = -1 b) Find all solution to 3 tanq +1 = 0 ans : q = 120° or q = 240° ans : q = 150° or q = 330° c) Find all solution to 2sinq - 3 = 0 ans : q = 60° or q = 120° d) Find all solution to sinq - 3 cosq = 0 ans : q = 60° or q = 240° e) Find all solution to 2sin 2 q - sinq = 0 ans : q = 0°,180°,360° or q = 30°,150°