ID : pk-0-trigonometry [] Grade 0 Trigonometry For more such worksheets visit www.edugain.com Answer t he quest ions () Simplif y - 2 sin 3 θ - 2 cos 3 θ (2) If secθ tan θ y, simplif y in terms of θ. (3) [sec(7 ) sec(8 ) sec(9 )... sec(83 ) ] [sin(7 ) sin(8 ) sin(9 ) ]? (4) Simplif y ( cotθ - cosecθ) ( tanθ secθ) (5) Simplif y cos 6 θ sin 6 θ 3 cos 2 θ sin 2 θ (6) Simplif y cot 2 β cosecβ. (7) Simplif y (8) Simplif y 3(sin 4 θ cos 4 θ) - 2(sin 6 θ cos 6 θ) Choose correct answer(s) f rom given choice (9) (cosecθ ) 2 (secθ ) 2? a. ( sin θ cos θ) 2 b. ( sec θ cosec θ) 2 c. ( - sin θ cos θ) 2 d. (sec θ cosec θ) 2 (0)? a. 2-2 sin 2 θ b. 2 2 sin 2 θ - c. 2 - cos 2 θ d. 2 2 sin 2 θ () Simplif y a. sin 2 θ - cos 2 θ b. c. d. 0
ID : pk-0-trigonometry [2] (2) If - 0, f ind value of tan 4 θ - sin 4 θ. a. 0.5 b. 0.25 c. 0.75 d. (3) Simplif y. a. tanθ cotθ b. sin 2 θ - cos 2 θ c. - d. tanθ cotθ (4) Simplif y a. cos 2 θ - sin 2 θ b. c. - d. tanθ - cotθ Check True/False (5) The angle of elevation of the top of a pole is 30. If the height of the pole is tripled, then the angle of elevation of its top will also be tripled. True False 206 Edugain (www.edugain.com). All Rights Reserved Many more such worksheets can be generated at www.edugain.com
Answers ID : pk-0-trigonometry [3] () -tan θ Let, S - 2 sin 3 θ - 2 cos 3 θ On taking and common, S S ( - 2 sin 2 θ) ( - 2 cos 2 θ) [( - sin 2 θ) - sin 2 θ] [( - cos 2 θ) - cos 2 θ] Using identity sin 2 θ cos 2 θ [cos 2 θ - sin 2 θ] S [sin 2 θ - cos 2 θ] S -tan θ (2)
(3) ID : pk-0-trigonometry [4] We need to f ind some of f ollowing series S (sec(7 ) sec(8 ) sec(9 )... sec(83 ) )(sin(7 ) sin(8 ) sin(9 ) ) Since we know that sec(θ)sin(θ) tan(θ), on multiplication we get S tan(7 ) tan(8 ) tan(9 )... tan 83 Lets write some more terms explicitly S tan(7 ) tan(8 ) tan(9 )... tan(44 ) tan(45 ) tan(46 )... tan(80 ) tan(8 ) tan(82 )... sin(83 ) Now re-write terms bef ore 45 as f ollowing S tan(90-83 ) tan(90-82 ) tan(90-8 )... tan(90-46 ) tan(45 ) tan(46 )... tan(80 ) tan(8 ) tan(82 )... sin(83 ) Step 5 We know that tan(90 - θ) cot(θ) /tan(θ) Now replace some of terms bef ore 45 using this equality S [/tan(83 )] [/tan(82 )] [/tan(8 )]... [/tan(46 )] tan(45 ) tan(46 )... tan(80 ) tan(8 ) tan(82 )... sin(83 ) Step 6 Now denominator of terms bef ore 45 will cancel with terms af ter 45. Theref ore, S tan(45 ) S
(4) 2 ID : pk-0-trigonometry [5] We need to f ind f ollowing product S ( cotθ - cosecθ) ( tanθ secθ) On multiplying each terms S ( tanθ secθ) cotθ ( tanθ secθ) - cosecθ ( tanθ secθ) S ( tanθ secθ) (cotθ cotθ tanθ cotθ secθ ) - (cosecθ cosecθ tanθ cosecθ secθ) Using identities cotθ tanθ, cotθ secθ cosecθ and cosecθ tanθ secθ S ( tanθ secθ) (cotθ cosecθ ) - (cosecθ secθ cosecθ secθ) Now positive cosecθ and secθ will cancel each other S ( tanθ secθ) (cotθ cosecθ ) - (cosecθ secθ cosecθ secθ) S 2 tanθ cotθ - cosecθ secθ Step 5 Using identities tanθ /, and cotθ / S 2 - cosecθ secθ S 2 sin2 θ cos 2 θ - cosecθ secθ S 2 - cosecθ secθ S 2 cosecθ secθ - cosecθ secθ S 2
(5) ID : pk-0-trigonometry [6] Expression can be rewritten as f ollowing (cos 2 θ) 3 (sin 2 θ) 3 3 cos 2 θ sin 2 θ Since x 3 y 3 ( x y ) ( x 2 y 2 - xy) (cos 2 θ sin 2 θ ) [ (cos 2 θ) 2 (sin 2 θ) 2 - cos 2 θ sin 2 θ ] 3 cos 2 θ sin 2 θ Since cos 2 θ sin 2 θ and x 2 y 2 (xy) 2-2 xy) (cos 2 θ sin 2 θ ) [ (cos 2 θ sin 2 θ) 2-2 cos 2 θ sin 2 θ - cos 2 θ sin 2 θ] 3 cos 2 θ sin 2 θ [ - 3 cos 2 θ sin 2 θ ] 3 cos 2 θ sin 2 θ (6) cosec β Let, S cot 2 β cosecβ Using identity cot 2 θ cosec 2 θ -, S cosec2 β - cosecβ Using a 2 - b 2 (ab)(a-b), (cosecβ )(cosecβ - ) S cosecβ S (cosecβ - ) S cosecβ
ID : pk-0-trigonometry [7] (7) sin 2 θ cos 2 θ We have been asked to simplif y the ( cotθ tanθ)( - ) sec 3 θ - cosec 3 θ. ( cotθ tanθ)( - ) sec 3 θ - cosec 3 θ ( )( - ) (secθ - cosecθ)(sec 2 θ secθ cosecθ cosec 2 θ) ) [Since, a 3 - b 3 (a - b)(a 2 ab b 2 ] ( ( cos2 θ sin 2 θ )( - ) - )( cos 2 θ cos 2 θ ) [Since, secθ and cosecθ ] ( )( - ) ( - )(sin 2 θ cos 2 θ) ( )(sin 2 θ cos 2 θ) ( )( - )(sin3 θ cos 3 θ) ( )( - )( ) sin 2 θ cos 2 θ Theref ore, ( cotθ tanθ)( - ) sec 3 θ - cosec 3 θ is equal to the sin 2 θ cos 2 θ.
(8) ID : pk-0-trigonometry [8] 3 (sin 4 θ cos 4 θ) - 2 [ (sin 2 θ) 3 (cos 2 θ) 3 } ] 3 (sin 4 θ cos 4 θ) - 2 [ (sin 2 θ cos 2 θ) {(sin 2 θ) 2 (cos 2 θ) 2 - sin 2 θ cos 2 θ } ] Since sin 2 θ cos 2 θ 3 (sin 4 θ cos 4 θ) - 2 {sin 4 θ cos 4 θ - sin 2 θ cos 2 θ } sin 4 θ cos 4 θ 2 sin 2 θ cos 2 θ Step 5 (sin 2 θ cos 2 θ) 2 2 (9) b. ( sec θ cosec θ) 2 (0) a. 2-2 sin 2 θ On adding two f ractions, S Let, S S Using identity sin 2 θ cos 2 θ, S S S S
() d. 0 ID : pk-0-trigonometry [9] On adding two f ractions (2) c. 0.75 It is given that - 0, which means that From above relation we can derive that tanθ We also know that, sin 2 θ /() sin 2 θ /2 Theref ore, tan 4 θ - sin 4 θ 2-0.5 2 tan 4 θ - sin 4 θ 0.75
(3) a. tanθ cotθ ID : pk-0-trigonometry [0] We know that, tanθ, cotθ Now, cotθ - tanθ tanθ - cotθ can be simplif ied as: cotθ - tanθ tanθ - cotθ - - - - cos 2 θ ( - ) sin 2 θ ( - ) cos 3 θ - sin 3 θ.( - ) ( - )(cos2 θ. sin 2 θ).( - ) cos2 θ. sin 2 θ. cos 2 θ... sin 2 θ. cotθ tanθ tanθ cotθ
(4) b. ID : pk-0-trigonometry [] Multiply numerator and denominator of f irst term by Since tanθ / (5) False