PRINTABLE VERSION Quiz 12 Question 1 Given sec(x) = 7 with 90 < x < 180. Find sin(2x) . a) 102 49 b) 83 7 c) 43 49 d) 83 49 e) 83 49 f) None of these. Question 2 Given sin(x) = a) 83 7 b) 1213 49 c) 13 2 49 d) 213 49 e) 1213 49 f) None of these. 6 with 180 < x < 270. Find sin(2x) . 7 Question 3 Given csc(x) = 8 with 270 < x < 360. Find cos(2x) . a) 31 32 b) 1 c) 2 d) 63 64 e) 37 1 8 f) None of these. Question 4 Given cos(x) = 1 with 270 < x < 360. Find tan(2x) . 5 a) 46 23 b) 26 23 c) 26 23 d) 26 25 e) 46 23 f) None of these. Question 5 Given cos(x) = a) 25 5 1 x with 180 < x < 270. Find cos( ) . 5 2 b) 10 5 c) 15 5 d) 2 e) 10 5 f) None of these. Question 6 Given sin(x) = 22 x where x is an acute angle. Find sin( ) . 3 2 a) 4 15 b) 2 3 c) 6 3 d) 3 3 e) 2 3 f) None of these. Question 7 5 using a half-angle formula. ( 12 ) Evaluate sin a) 2 3 2 b) 2 + 3 2 c) + 3 2 2 d) 3 2 2 e) 1 3 + 2 4 f) None of these. Question 8 Given that is an acute angle with tan() = a) 31 25 b) 17 25 c) 1 5 d) 1 e) 7 25 f) None of these. 4 , evaluate sin(2) cos(2). 3 Question 9 Triangle ABC with right angle C is shown below. Given that sin(B) = 1 and AD = BD , find sin(ADC) . 5 Note that the diagram may not be drawn to scale. a) 26 25 b) 24 125 c) 2 5 d) 23 25 e) 46 25 f) None of these. Question 10 20tan(5x) sin(20x) + Which of the following is equivalent to: 1 + cos(20x) 1 tan2 (5x) a) 21tan(10x) b) 11tan(10x) c) 9tan(10x) d) 12tan(10x) e) 41tan(10x) f) None of these. Question 11 ( 8 . ( 17 )) Evaluate: sin 2 arcsin a) 16 17 b) 1800 4913 c) 240 289 d) 120 289 e) 161 289 f) None of these. Question 12 ( 5 . ( 12 )) Evaluate: cos 2 arctan a) 5601 601 b) 24 13 c) 119 169 d) e) 120 169 f) None of these. 25 288 Solution Hide Steps ( ( Decimal: 0.83045... ) ( ) ) = 240 289 sin 2arcsin 8 17 Steps ( ( )) sin 2arcsin 8 17 Use the following identity: sin ( 2x ) = 2cos ( x ) sin ( x ) ( ( ) ) = 2cos (arcsin ( 178 ) )sin (arcsin ( 178 ) ) = 2cos ( arcsin ( 8 ) ) sin ( arcsin ( 8 ) ) 17 17 sin 2arcsin 8 17 Use the following identity: cos ( arcsin ( x ) ) = 1 x2 ( ) sin (arcsin ( ) ) = 2 1 8 17 2 8 17 Use the following identity: sin ( arcsin ( x ) ) = x ( )( ) = 2 1 8 17 2 8 17 ( )( ) 2 2 1 8 17 8 240 17 = 289 = 240 289 click here to practice evaluate functions Show Steps Solution Hide Steps ( ( Decimal: 0.70414... ) ( ) ) = 119 169 cos 2arctan 5 12 Steps ( ( )) cos 2arctan 5 12 ( 2 Use the following identity: cos ( 2x ) = 1 + 2cos ( x ) ) ( ( ) ) = 1 + 2cos2 (arctan ( 125 ) ) 2 = ( 1 + 2cos ( arctan ( 5 ) ) ) 12 cos 2arctan 5 12 Remove parentheses: ( a ) = a 2 ( ( )) = 1 + 2cos arctan 5 12 Use the following identity: cos ( arctan ( x ) ) = 1 + x2 1+x 2 = 1 + 2 1 + 2 ( ( ) 1 + ( 125 ) 2 2 1+ ( 5 ) 12 ( ) ( ) 5 2 12 2 1+ 5 12 1+ ) 2 Show Steps 2 = 119 169 = 119 169 click here to practice evaluate functions Solution Hide Steps ( ( Decimal: 0.70414... ) ( ) ) = 119 169 cos 2arctan 5 12 Steps ( ( )) cos 2arctan 5 12 ( 2 Use the following identity: cos ( 2x ) = 1 + 2cos ( x ) ) ( ( ) ) = 1 + 2cos2 (arctan ( 125 ) ) 2 = ( 1 + 2cos ( arctan ( 5 ) ) ) 12 cos 2arctan 5 12 Remove parentheses: ( a ) = a 2 ( ( )) = 1 + 2cos arctan 5 12 Use the following identity: cos ( arctan ( x ) ) = 1 + x2 1+x 2 = 1 + 2 1 + 2 ( ( ) 1 + ( 125 ) 2 2 1+ ( 5 ) 12 ( ) ( ) 5 2 12 2 1+ 5 12 1+ ) 2 Show Steps 2 = 119 169 = 119 169 click here to practice evaluate functions \f\fSolution Hide Steps ( ( Decimal: 0.83045... ) ( ) ) = 240 289 sin 2arcsin 8 17 Steps ( ( )) sin 2arcsin 8 17 Use the following identity: sin ( 2x ) = 2cos ( x ) sin ( x ) ( ( ) ) = 2cos (arcsin ( 178 ) )sin (arcsin ( 178 ) ) = 2cos ( arcsin ( 8 ) ) sin ( arcsin ( 8 ) ) 17 17 sin 2arcsin 8 17 Use the following identity: cos ( arcsin ( x ) ) = 1 x2 ( ) sin (arcsin ( ) ) = 2 1 8 17 2 8 17 Use the following identity: sin ( arcsin ( x ) ) = x ( )( ) = 2 1 8 17 2 8 17 ( )( ) 2 2 1 8 17 8 240 17 = 289 = 240 289 click here to practice evaluate functions Show Steps Solution Hide Steps ( ( Decimal: 0.70414... ) ( ) ) = 119 169 cos 2arctan 5 12 Steps ( ( )) cos 2arctan 5 12 ( 2 Use the following identity: cos ( 2x ) = 1 + 2cos ( x ) ) ( ( ) ) = 1 + 2cos2 (arctan ( 125 ) ) 2 = ( 1 + 2cos ( arctan ( 5 ) ) ) 12 cos 2arctan 5 12 Remove parentheses: ( a ) = a 2 ( ( )) = 1 + 2cos arctan 5 12 Use the following identity: cos ( arctan ( x ) ) = 1 + x2 1+x 2 = 1 + 2 1 + 2 ( ( ) 1 + ( 125 ) 2 2 1+ ( 5 ) 12 ( ) ( ) 5 2 12 2 1+ 5 12 1+ ) 2 Show Steps 2 = 119 169 = 119 169 click here to practice evaluate functions Solution Hide Steps ( ( Decimal: 0.70414... ) ( ) ) = 119 169 cos 2arctan 5 12 Steps ( ( )) cos 2arctan 5 12 ( 2 Use the following identity: cos ( 2x ) = 1 + 2cos ( x ) ) ( ( ) ) = 1 + 2cos2 (arctan ( 125 ) ) 2 = ( 1 + 2cos ( arctan ( 5 ) ) ) 12 cos 2arctan 5 12 Remove parentheses: ( a ) = a 2 ( ( )) = 1 + 2cos arctan 5 12 Use the following identity: cos ( arctan ( x ) ) = 1 + x2 1+x 2 = 1 + 2 1 + 2 ( ( ) 1 + ( 125 ) 2 2 1+ ( 5 ) 12 ( ) ( ) 5 2 12 2 1+ 5 12 1+ ) 2 Show Steps 2 = 119 169 = 119 169 click here to practice evaluate functions \f\f