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Example Solution Example Solution Example Solution Example Solution Example Solution Example 4 If cos π₯ = β and tan x is positive, find the value of sin x. 5 Solution Since cos x is negative and tan x is positive, x is in the third quadrant. In right ΞPOQ, the length of Μ Μ Μ Μ ππ is 4 and of Μ Μ Μ Μ ππ is 5. It follows that the 3 length of PO is 3. Therefore, sin x = β 5 Example Solution Example Express sin2x and cos2x in terms of tan x. Solution 2 sin x = cos2x = sin2 π₯ 1 cos2 π₯ 1 = = sin2 π₯ sin2 π₯ + cos2 π₯ cos2 π₯ sin2 π₯ + cos2 π₯ = = Thus, sin2x = cos2x = tan2 π₯ tan2 π₯ + 1 1 tan2 π₯ + 1 sin2 π₯ cos2 π₯ 2 sin π₯ cos2 π₯ + cos2 π₯ cos2 π₯ cos2 π₯ cos2 π₯ 2 sin π₯ cos2 π₯ + 2 cos π₯ cos2 π₯ = = tan2 π₯ , tan2 π₯ + 1 1 tan2 π₯ + . 1 Example What is the value of sin (arctan β2 )? Solution Let π = arctan β2, then 0° < π < 90° and tan π = β2. Hence, 2 sin π = tan2 π tan2 π + 1 2 = (β2) 2 (β2) + 1 = 2 3 2 β sin π = β = 3 β6 . 3 Example Find the value of sin (arcsin 1 + arccos 1). Solution arcsin 1= 90°, arccos 1= 0°, therefore, sin (arcsin 1 + arccos 1) = sin (90° + 0°) = sin 90° = 1. Example Prove the identity: 1 + sec x β cos x β sec x cos x = tan x sin x Solution 1 + sec x β cos x β sec x cos x = (1 + sec x) β cos x (1 + sec x) = (1 + sec x) (1β cos x) = (1 + = (1 + = = = 1 cos π₯ 1 cos π₯ ) (1β cos x) ) (1β cos x) 1β cos2 π₯ cos π₯ sin2 π₯ cos π₯ sin π₯ cos π₯ β sin π₯ = tan π₯ β sin π₯ Example If p = cos 200°, express cot 70° in terms of p. Solution cos 200° = cos (270° β 70°) = βsin 70° β sin 70° = βπ, cos 70° = β1 β sin2 70° = β1 β (βπ)2 = β1 β π2 , cot 70° = cos 70° sin 70° = β β1β π2 π . sec π₯ Example The expression is equivalent to tan π₯ sin π₯ (A) tan x (B) csc 2 π₯ (C) sec 2 π₯ (D) sec 2 π₯ csc π₯ (E) cos x Solution Answer: (B) sec π₯ = tan π₯ sin π₯ 1 cos π₯ sin π₯ β sin π₯ cos π₯ = 1 cos π₯ sin2 π₯ cos π₯ 1 = cos π₯ β cos π₯ sin2 π₯ = 1 sin2 π₯ = csc2 π₯. Example Solution Example Solution Example sin 2π Express 2 as a trigonometric function of π. 2sin π Solution sin 2π 2 = 2 sin π 2 sin π cos π 2 sin2 π = cos π sin π = cot π. Example 3 If sin x = and x is an acute angle, find the value of sin 2x. 5 Solution 3 sin x = , 0° < π₯ < 90° β 5 cos x = β1 β sin2 π₯ 2 3 = β1 β ( ) = β1 β 5 3 4 5 5 sin 2x = 2 sin x cos x = 2 β β = 24 25 9 4 25 = , 5 . Example 1 π₯ If cos x = , 360° < π₯ < 450°, what is the value of sin ( ) ? 9 2 Solution π₯ π₯ 360° < π₯ < 450° β 180° < < 225° β sin ( ) < 0, 2 sin 2 π₯ ( ) 2 = 1 β cos π₯ 2 = 1β 2 1 9 = 2 4 9 π₯ 2 2 3 β sin ( ) = β . Example Prove the formulas: π₯ 1 β cos π₯ 2 sin π₯ π₯ sin π₯ 2 1 + cos π₯ tan ( ) = tan ( ) = Solution 1 β cos π₯ sin π₯ sin π₯ 1 + cos π₯ π₯ = π₯ π₯ 2 sin(2 ) cos(2) π₯ = π₯ 2sin 2 ( ) 2 π₯ 2 sin(2 ) cos(2) π₯ 2cos 2 (2) = sin( ) 2 π₯ cos(2) π₯ = sin(2) π₯ cos(2) π₯ = tan ( ), 2 π₯ = tan ( ). 2 Example Prove the formulas: π₯ sin π₯ = 2 tan(2) π₯ 1 + tan2 (2) π₯ cos x = 1β tan2 (2) π₯ 1 + tan2 (2) Solution π₯ π₯ 2 sin( ) cos( ) 2 2 2 2 sin2 (2) + cos2 (2 ) sin π₯ = 2 sin ( ) cos ( ) = = π₯ π₯ 2 sin( ) cos( ) 2 2 π₯ 2 cos ( ) 2 π₯ π₯ sin2 ( )+ cos2 ( ) 2 2 π₯ cos2 ( ) 2 cos x = cos = π₯ π₯ 2 π₯ ( ) 2 π₯ π₯ cos 2 ( ) β sin 2 ( ) 2 2 π₯ cos2 ( ) 2 π₯ π₯ 2 sin ( )+ cos2 ( ) 2 2 π₯ cos2 ( ) 2 π₯ π₯ = β sin 2 tan( ) 2 π₯ , 1 + tan2 ( ) 2 2 π₯ ( ) 2 = π₯ = π₯ 1β tan2 (2) π₯ , 1 + tan2 (2) π₯ π₯ π₯ π₯ cos 2 (2) β sin 2 (2) sin2 (2) + cos2 (2) Example 4 8 5 17 If sin x = , 90° < π₯ < 180°, and cos y = , 0° < π¦ < 90°, what is the value of cos (x + y)? Solution 4 sin x = , 90° < π₯ < 180° β 5 cos x = ββ1 β cos y = 8 sin2 π₯ 2 4 3 = ββ1 β ( ) =β , 5 5 , 0° < π¦ < 90° β 17 8 2 15 17 17 sin y = β1 β cos 2 π¦ = β1 β ( ) = cos (x + y) = cos x cos y β sin x sin y 3 8 4 15 5 17 5 17 = (β ) ( ) β ( ) ( ) =β 24 85 β 60 85 = β 84 . 85 ,