Review Exercise 1 Crosses y-axis when x 0 at sin p 4 1 Crosses x-axis when sin x + p 4 ö 0 x + p 4 -p, 0, p, p x - 7p 4, - p 4, p 4, 5p 4 So coordinates are 0, 1 ö, - 7p 4,0 ö, - p 4,0 ö, p 4,0 ö, 5p 4,0 ö a y cos x - p π ö is y cos x translated by the vector 0 b Crosses y-axis when y cos - p ö 1 Crosses x-axis when cos x - p ö 0 x - p p, p x 5p 6, 11p 6 So coordinates are 0, 1 ö, 5p 6,0 ö, 11p 6,0 ö c cos x - p ö -0.7, 0 x p cos -1 (-0.7) 1.844 ( d.p.) Þ x - p» 1.844 and x - p» p -1.844 Þ x.89, 5.49 ( d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1
a Let C be the midpoint of the line AB, then AOC is a right-angled triangle and AC cm, so sin q 5 0.6 Þ q 0.645 q 1.87 radians ( d.p.) b Use l rq So arc AB 5 1.87 6.44cm ( s.f.) 4 As ABC is equilateral, BC AC 8 cm BP AB AP 8 6 cm QC BP cm ÐBAC p, PQ 6 p p 6.8cm ( d.p.) So perimeter BC + BP + PQ + QC 18.8 cm ( d.p.) Exact answer 1 + p cm 5 a 1 (r + 10) q - 1 r q 40 Þ 0rq +100q 80 Þ rq + 5q 4 Þ r 4 q - 5 b r 4 q - 5 6q Þ 4-5q 6q Þ 6q + 5q - 4 0 Þ (q + 4)(q -1) 0 Þq - 4 or 1 But cannot be negative, so q 1, r So perimeter 0 + rq + (10 + r)q 0 + + 1 8cm 6 a arc BD 10 0.6 6cm b Area of triangle ABC 1 (1 10)sin0.6 65 0.567 6.7cm (1 d.p.) Area of sector ABD 1 10 0.6 0cm Area of shaded area BCD 6.7-0 6.7cm (1 d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
7 a So r 10 1.07cm ( d.p.) cos0.7 Area of sector OAB 1 r 1.4 119.7cm (1 d.p.) b BC AC r tan0.7 So perimeter r tan0.7 + r 1.4 ( 1.07 0.84) + (1.07 1.4) 40.cm Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
8 Split each half of the rectangle as shown. EFB is a right-angled triangle, and by Pythagoras theorem EF r Let ÐEBF q, so tanq Þq p So ÐFBC p - p p 6 Area S 1 r p 6 p 1 r Area T 1 r 1 r 8 r Þ Area R 1 r - Area S - Area T 1-8 - p ö r 1 Area of sector ACB 1 r p p 4 r Area U Area ABCD - Area sector ACB - R r - p 4 r - 1-8 - p ö r 1 r 4 - p ö 1 So area U r - p 4 r - R 1- p 4-1+ 4 + p ö r 6 r 4 - p ö r 1 1 - p So shaded area U r ( ) ( ) 6 - p Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 4
9 a sin x + 7cos x + (1- cos x) + 7cos x + -cos x + 7cos x + 6 cos x - 7cos x - 6 b cos x - 7cos x - 6 0 (cos x + )(cos x - ) 0 cos x - or cos x cannot be so cos x - x.0, p -.0,0,.98 ( d.p.) 10 a For small values of q : sin4q» 4q cos4q» 1-1 (4q)» 1-8q, tanq» q ( ) + q sin4q - cos4q + tanq» 4q - 1-8q b -1» 8q + 7q -1 11 a y 4 - cosec x is y cosec x stretched by a scale factor in the y-direction, then reflected in the x-axis and then translated by the vector 0 4 b The minima in the graph occur when cosec x 1 and y 6. The maxima occur when cosec x 1 and y. So there are no solutions for < k < 6. 1 a The graph is a translation of y secq by a. So a p b As the curve passes through (0, 4) 4 k sec p Þ k 4cos p Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 5
1 c - sec q - p ö Þ cos q - p ö - 1 Þq - p - 5p 4, - p 4 Þq - 11p 1, - 5p 1 1 a cos x 1- sin x + 1- sin x cos x º cos x + (1- sin x) cos x(1- sin x) º cos x +1- sin x + sin x cos x(1- sin x) º - sin x cos x(1- sin x) º cos x º sec x b By part a the equation becomes sec x - Þ sec x - Þ cos x - 1 x p 4, 5p 4, 11p 4, 1p 4 14 a sinq cosq + cosq sinq sin q + cos q cosq sinq 1 (using cos q + sin q º 1) sinq cosq 1 1 sinq (using double-angle formula sinq º sinq cosq) cosecq Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 6
14b The graph of y cosecq is a stretch of the graph of y cosecq by a scale factor of 1 horizontal direction and then a stretch by a factor of in the vertical direction. in the c By part a the equation becomes cosecq Þ cosecq Þ sinq, in the interval 0 q 70 Calculator value is q 41.81 ( d.p.) Solutions are q 41.81,180-41.81, 60 + 41.81, 540-41.81 So the solution set is: 0.9, 69.1, 00.9, 49.1 15 a Note the angle BDC q cosq BC Þ BC 10cosq 10 sinq BC BC Þ BD BD sinq 10cosq sinq 10cotq b 10cotq 10 Þ cotq 1, q p From the triangle BCD, cosq DC BD Þ DC BDcosq So DC 10cotq cosq 10 1 ö 1ö 5 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 7
16 a sin q + cos q º 1 Þ sin q cos q + cos q cos q º 1 cos q Þ tan q +1º sec q (dividing by cos q) b tan q + secq 1 Þ sec q - + secq 1 Þ sec q + secq - 0 Þ (secq + )(secq -1) 0 Þ secq -, secq 1 Þ cosq -, cosq 1 Solutions are 11.8, 60-11.8, 0 So solution set is: 0.0,11.8, 8. (1 d.p.) 17 a a 1 sin x 1 1 b b b 4 - b a -1 4 - b ö b -1 4 - b 4 - b 4 b -1 4 - b b b (4 - b ) b 4 - b An alternative approach is to first substitute the trigonometric functions for a and b 4 - b a -1 4-4sin x cosec x -1 4(1- sin x) cot x 4cos x cot x 4sin x b 18 a y arcsin x Þ sin y x ( ) x cos p - y Þ p - y arccos x b arcsin x + arccos x y + p - y Using sinq cos( p -q ) p Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 8
19 a arccos 1 x p Þ cos p 1 x Use Pythagoras theorem to show that opposite side of the right-angle triangle with angle p is x -1 So sin p x -1 x Þ p arcsin x -1 x b If 0 x 1 then x 1 is negative and you cannot take the square root of a negative number. 0 a y arccos x - p translated by - p is y arccos x stretched by a scale factor of in the y-direction and then in the vertical direction b arccos x - p 0 Þ arccos x p 4 Þ x cos p 4 1 Coordinates are 1, 0 ö 1 tan x + p ö 6 1 6 Þ tan x + 1- tan x 1 6 6tan x + 1- tan x 18 + ö tan x 1- ( )( 18 - ( )( 18 - ) 1- tan x 18 + 7-111 1 [using the addition formula for tan (A + B)] Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 9
a sin(x + 0 ) sin(x + 60 ) So sin xcos0 + cos xsin0 sin xcos60 cos xsin60 sin x + 1 cos x 1 sin x - cos x ö ( ) (using the addition formulae for sin) sin x + cos x sin x - cos x (multiplying both sides by ) (- + )sin x (-1- )cos x So tan x -1- - + (-1- ) - - - + ( ) ( )(- - ) + 6 + 4 + 4-8 + 5 b tan(x + 60 ) tan x + tan60 1- tan x tan60 8 + 5 + 1-8 + 5 ( ) 8 + 6-14 - 8 ( 4 + ) -7 + 4-7 - 4 ( ) ( )(-7 + 4 ) 6-8 - 1 +16 49-48 8-5 a sin165 sin(10 + 45 ) sin10 cos45 +cos10 sin45 1 + -1 1-1 6-4 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 10
1 sin165 b cosec165 4 ( 6 + ) ( 6 - ) ( 6 + ) ( ) 6-4 6 + 6 + 4 a cos A 4 Using Pythagoras theorem and noting that sin A is negative as A is in the fourth quadrant, this gives sin A - 7 4 Using the double-angle formula for sin gives sina sin Acos A - 7 ö ö 4 4-7 8 b cosa cos A-1 1 8 ( ) Þ tana sina cosa - 7 8 5 a cosx + sin x 1 1 ( 8) - 7 Þ1- sin x + sin x 1 (using double-angle formula for cosx) Þ sin x - sin x 0 Þ sin x(sin x -1) 0 Þ sin x 0, sin x 1 Solutions in the given interval are: 180, 0, 0,150,180 b sin x(cos x + cosec x) cos x Þ sin xcos x +1 cos x Þ in xcos x cos x -1 Þ 1 sinx cosx (using the double-angle formulae for sinx and cosx) Þ tanx, for - 60 x 60 So x 6.4-60, 6.4-180, 6.4, 6.4 +180 Solution set: - 148., - 58., 1.7,11.7 (1 d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 11
6 a Rsin(x +a) Rsin xcosa + Rcos xsina So Rcosa, Rsina R cos a + R sin a + 9 + 4 1 Þ R 1 (as cos a + sin a º 1) tana Þa 0.588 ( d.p.) b R 4 ( 1) 4 169 since the maximum value the sin function can take is 1 c 1sin(x + 0.588) 1 sin(x + 0.5880) 1 1 0.775 x + 0.588 p - 0.81, p + 0.81 x.7, 5.976 ( d.p.) 7 a LHS º cotq - tanq º cosq sinq - sinq cosq º cos q - sin q sinq cosq º cosq 1 sinq (using the double angle formulae for sinq and cosq) º cot q º RHS b cot q 5 Þ cot q 5 Þ tanq, for - p < q < p 5 So q 0.805 - p, 0.805 - p, 0.805, 0.805 + p Solution set: -.95, -1.8, 0.190,1.76 ( s.f.) 8 a LHS º cosq º cos(q +q) º cosq cosq - sinq sinq º (cos q - sin q)cosq - ( sinq cosq )sinq º cos q - sin q cosq º cos q - (1- cos q)cosq º 4cos q - cosq º RHS b From part a cosq 4 7 - -19 7 So secq - 7 19-7 8 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1
9 sin 4 q ( sin q )( sin q ) Use the double-angle formula to write sin q in terms of cosq 1- cosq cosq 1- sin q Þ sin q Now substitute the expression for sin q and expand the brackets 1- cosq ö 1- cosq ö So sin 4 q 1 ( 4 1- cosq + cos q ) Again use the double-angle formula to write cos q in terms of cos4q So sin 4 q 1 1+ cos4q ö 1- cosq + 4 8-1 cosq + 1 8 cos4q 0 a Rsin(q +a) Rsinq cosa + Rcosq sina 6sinq + cosq So Rcosa 6, Rsina R cos a + R sin a 6 + 6 + 4 40 Þ R 40 (as cos a + sin a º 1) tana Þ a 0.175 0. ( d.p.) 6 So 6sinq + cosq» 40 sin(q + 0.) b i 40, since the maximum value the sin function can take is 1 ii Maximum occurs when sin(q + 0.) 1 c T 9 + Þq + 0. p Þq 1.5 ( d.p.) Note that you should use a value of a to decimal places in the model and then give your answers to decimal places. 40 sin pt 1 + 0. ö So minimum value of T is 9 - Occurs when sin pt 1 + 0. ö -1 Þ pt 1 + 0. p Þ t 16.77 hours 40.68 C ( d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1
0 d 9 + 40 sin pt 1 + 0. ö 14 Þ 40 sin pt 1 + 0. ö 5 Þ sin pt 1 + 0. ö 5 40 Þ pt + 0. 0.9117,.99 1 Þ t.5, 7.9 ( d.p.) 0.5h» 15 minutes and 0.9 h» 17 minutes So times are 11:15am and 4:17 pm 1 a As 4 t ¹ 0, x ¹ 1 The equation for y can be rewritten as y t - ö So y ³ -1.5 b t 4 1- x 4 ö So y 1- x - 5 4 4 ö - 1- x +1 16 ( 1- x) - 1 1- x 1- x ( ) ( + 1- x ) ( ) ( 1- x) 16-1 +1x +1- x + x 1- x x +10x + 5 1- x ( ) ( ) So a 1, b 10, c 5 a x ln(t + ) Þ e x t + Þ t e x - y t t + ex - 6 e x +1 t > 4 Þ e x - > 4 Þ e x > 6 Þ x > ln6 So the solution is y ex - 6 e x +1, x > ln6 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 14
b When x, y When x ln6, y eln6-6 e ln6 +1 So range is 1 7 < y < ( 6) - 6 6 +1 1 7 x 1 1+ t Þ t 1 1- x -1 x x 1 y 1-1- x x x x - ( 1- x) x x -1 ( ) costcost - sintsint 4 a y cost cos t + t ( cos t -1)cost - sin t cost cos t - cost - ( 1- cos t)cost 4cos t - cost x cost Þ cost x y 4 x ö ( ) - x ö 1 x - x x x - b 0 t p So 0 cost 1 and -1 cost 1 So 0 x, -1 y 1 5 a y sin t + p ö 6 sintcos p 6 + costsin p 6 sint + 1 cost sint + 1 1- sin t x + 1 1- x As - p t p, -1 sint 1Þ -1 x 1 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 15
5 b At A, sin t + p ö 6 0 Þ t - p 6 x sin - p ö 6-1 Coordinates of A are - 1, 0 ö At B, x sint 0 Þ t 0 y sin t + p ö 6 sin p 6 1 Coordinates of B are 0, 1 ö 6 a y cost cos t -1 y x ö -1, - x b Curve is a parabola, with a minima and y-intercept at (0, 1) and x-intercepts when x ö 1Þ x ± 1 Þ x ± Coordinates -, 0 ö,, 0 ö 7 y x + c would intersect curve C if ( ) ( 4t) + c 8t t -1 16t - 0t - c 0 Using the quadratic formula, this equation has no real solutions if (-0) - 4( 16) (-c) < 0 Þ 64c < -400 Þ c < - 5 4 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 16
8 a The curve intersects the x-axis when cost +1 0 Þ cost - 1 Solutions in the interval are t p, 4p Þ x sin 4p ö, sin 8p ö So coordinates are -, 0 ö and, 0 ö b sint 1.5 Þ sint 1 In the interval p t p solutions are t 1p 6, 17p 6 Þ t 1p 1, 17p 1 9 a Find the time the ball hits the ground by solving -4.9t + 5t +50 0 ( )( 50) ( ) t -5 ± 5-4 -4.9-4.9 t ³ 0, so only valid solution is t 6.64s ( d.p.) Þ k 6.64 ( d.p.) x b t 5 x ö y 5 5-4.9 x ö 5 x - 49 18750 x + 50 + 50 Domain of the function is from where the ball is hit at x 0 to where it hits the ground when t 6.64 seconds. When t 6.64, x 5 (6.64) 87.5 (1 d.p.) So domain is 0 x 87.5 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 17
Challenge 1 Angle of minor arc p because it is a quarter circle Let the chord meet the circle at R and T. The area of P is the area of sector formed by O, R and T less the area of the triangle ORT. So area of P 1 r p - 1 r sin p r p 4-1 ö r (p - ) 4 Area of Q pr - area of P r p - p 4 + 1 ö r p 4 + 1 ö r (p + ) 4 So ratio (p - ) :(p + ) p - p + :1 a sin x b cos x c ÐCOA p - x Þ ÐCAO x OA 1 sinx cosec x d AC 1 tan x cot x e tan x f OB 1 cos x sec x Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 18
a sint x - y +1, cost 4 4 As sin t + cos t 1 x - ö 4 Þ x - + y +1 ö 4 1 ( ) + ( y +1) 16 The curve is a circle centre (, 1) and radius 4. Endpoints when t - p, x -1, y -1 and when t p, x +, y -1 4 b C is ths of a circle, radius 4 8 So length 8p p 8 Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 19