Review Exercise 2. , æ. ç ø. ç ø. ç ø. ç ø. = -0.27, 0 x 2p. 1 Crosses y-axis when x = 0 at sin 3p 4 = 1 2. ö ø. æ Crosses x-axis when sin x + 3p è

Transcription

1 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) ( d.p.) Þ x - p» and x - p» p Þ x.89, 5.49 ( d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1

2 a Let C be the midpoint of the line AB, then AOC is a right-angled triangle and AC cm, so sin q Þ q q 1.87 radians ( d.p.) b Use l rq So arc AB cm ( 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 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 cm 6 a arc BD cm b Area of triangle ABC 1 (1 10)sin cm (1 d.p.) Area of sector ABD cm Area of shaded area BCD cm (1 d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.

3 7 a So r cm ( d.p.) cos0.7 Area of sector OAB 1 r cm (1 d.p.) b BC AC r tan0.7 So perimeter r tan0.7 + r 1.4 ( ) + ( ) 40.cm Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.

4 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 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 p ö r 1 r 4 - p ö 1 So area U r - p 4 r - R 1- p p ö r 6 r 4 - p ö r 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

5 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 (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 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

6 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

7 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 ( d.p.) Solutions are q 41.81, , , So the solution set is: 0.9, 69.1, 00.9, 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

8 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, , 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 b ö b b 4 - b 4 b 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 sin 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

9 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 Þ tan x + 1- tan x 1 6 6tan x + 1- tan x 18 + ö tan x 1- ( )( 18 - ( )( 18 - ) 1- tan x [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

10 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- ) ( ) ( )(- - ) b tan(x + 60 ) tan x + tan60 1- tan x tan ( ) ( 4 + ) ( ) ( )( ) a sin165 sin( ) sin10 cos45 +cos10 sin Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 10

11 1 sin165 b cosec165 4 ( 6 + ) ( 6 - ) ( 6 + ) ( ) 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 Using the double-angle formula for sin gives sina sin Acos A - 7 ö ö b cosa cos A ( ) Þ tana sina cosa 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, Solution set: , - 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

12 6 a Rsin(x +a) Rsin xcosa + Rcos xsina So Rcosa, Rsina R cos a + R sin a Þ R 1 (as cos a + sin a º 1) tana Þa ( d.p.) b R 4 ( 1) since the maximum value the sin function can take is 1 c 1sin(x ) 1 sin(x ) x p , p x.7, ( 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 p, p, 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 So secq Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1

13 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 cosq cos4q 0 a Rsin(q +a) Rsinq cosa + Rcosq sina 6sinq + cosq So Rcosa 6, Rsina R cos a + R sin a Þ R 40 (as cos a + sin a º 1) tana Þ a ( 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 ö So minimum value of T is 9 - Occurs when sin pt ö -1 Þ pt p Þ t hours C ( d.p.) Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 1

14 0 d sin pt ö 14 Þ 40 sin pt ö 5 Þ sin pt ö 5 40 Þ pt ,.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 ö - 1- x ( 1- x) x 1- x ( ) ( + 1- x ) ( ) ( 1- x) x +1- x + x 1- x x +10x 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

15 b When x, y When x ln6, y eln6-6 e ln6 +1 So range is 1 7 < y < ( 6) 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 sin t x 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

16 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 < Pearson Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free. 16

17 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) ( ) t -5 ± 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 x ö 5 x x 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

18 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 ö r p 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

19 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