Trigonometry and modelling 7E sinq +cosq º sinq cosa + cosq sina Comparing sin : cos Comparing cos : sin Divide the equations: sin tan cos Square and add the equations: cos sin (cos sin ) since cos sin sinq + 6 cosq cos cos sin sin Comparing sin : sin () Comparing cos : 6 cos () Divide () by () : tan 6 So. ( d.p.) sinq - cosq cos cos sin sin Comparing sin : sin () Comparing cos : cos () Divide () by () : tan So 4.8 ( d.p.) 4 a Let cosq - sinq º cos(q +a) cos cos sin sin Compare cos : cos () Compare sin : sin () Divide () by () : tan π Square and add : 4 π So cos sin cos b This is the graph of y = cosq, translated by to the left and then stretched in the π y direction by scale factor. Meets y-ais at (0, ) π 7π Meets -ais at,0,,0 6 6 a Let 7cosq - 4sinq º cos(q +a) cos cos sin sin Compare cos : cos 7 () Compare sin : sin 4 () Divide () by () : tan 7.7 ( d.p.) 4 7 Square and add: 4 7 So 7 cos 4sin cos( 7.7 ) b Graph meets y-ais where q = 0, i.e. y = 7cos0-4sin0 = 7 so coordinates are (0, 7) c Maimum value of cos(q + 7.7 ) is when cos(q + 7.7 ) = So maimum is Minimum value is (-) = - Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
d i The line y = will meet the graph twice in 0 60, so there are solutions. ii As the maimum value is it can never be 6, so there are 0 solutions. iii As - is a minimum, line y = - only meets curve once, so only solution. 6 a Let sin cos sin sin cos cos sin Comparing sin : cos () Comparing cos : sin () Divide () by () sin tan cos So 7.6 ( d.p.) cos sin (cos sin ) 0 0 0, 7.6 ( d.p.) b Use the value of to d.p. in calculating values of to avoid rounding errors 0 sin( 7.6 ) sin( 7.6 ) 0 sin 9. ( d.p.) 0 As 0 60, the interval for 7.6 is 7.6 7.6 4.6 So 7.6 80 9., and 7.6 60 9. 7.6 40.77, 99. 69., 7.7 ( d.p.) 7 a Set cos sin cos( ) cos sin cos cos sin sin Comparing sin : sin () Comparing cos : cos () Divide () by () sin tan cos So.07 ( d.p.) cos sin (cos sin ) So cos sin cos(.07) b cos(.07).. cos(.07). cos.06 ( d.p.) As 0 π, the interval for.07 is.07.07 π.07 So.07.06, π.06.07.06,.977 0.60,.44 ( d.p.) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
8 a Write 6sin +8cos in the form sin( ), where 0, 0 90 So 6sin 8cos sin cos cos sin Compare sin : cos 6 () Compare cos : sin 8 () 4 Divide () by (): tan. ( d.p.) 6 8 0 So 6sin 8cos 0sin(. ) Solve 0sin(. ), in the interval 0 60 so sin(. ). 60, 0 6.9, 66.9 ( d.p.) b Let cosq - sinq º cos(q +a) cos cos sin sin Compare cos : cos () Compare sin : sin () Divide () by () : tan 6. ( d.p.) Solve cos( 6. ), in the interval 0 90 so cos( 6. ) for 6. 6. 6. c Let 8cosq +sinq º cos(q -a) sin cos cos sin Compare cos : cos 8 () Compare sin : sin () Divide () by () : tan 8 6.9 ( d.p.) 8 7 Solve 7 cos( 6.9 ) 0, in the interval 0 60 0 So cos 6.9, 7 6.9 6.9 98.07 cos 0.97 ( d.p.) 7 So 6.9.97,.97 8.0,.9 ( d.p.) 6. 06.0,.90 49.8, 97.6 6.6, 6.9 ( d.p.) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
8 d Let sin cos sin sin cos cos sin Compare sin : cos () Compare cos : sin () Divide () by () : tan 67.8 ( d.p.) Solve sin 67.8 6., in the interval 60 60 So sin 67.8, 47.4 67.4.6 From quadrant diagram: 67.8 0, 0 8.6, 7.8 6., 74.8 ( d.p.) 9 a Set sin 4cos sin( ) sin 4cos sin cos cos sin Compare sin : cos () Compare cos : sin 4 () 4 Divide () by () : tan. ( d.p.) 4 So sin 4cos sin(. ) b The minimum value of sin 4cos is. This occurs when sin(. ). 70 07.7 ( d.p.) c sin(. ), in the interval 0 80 So sin(. ), in the interval.. 06.87..4,68.46, 7.4.6,7.9,4.6 ( d.p.) - cosq 0 a As sin q = and cos cos So sin cos 6sin cos cos cos (sincos ) cos cos sin 4cos sin Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4
0 b Write sinq - 4cosq in the form sin( ) The maimum value of sin( ) is The minimum value of sin( ) is You know that 4 so So maimum value of 4cos sin is 6 and minimum value of 4cos sin is 4 c 4cos sin sin 4cos Write sinq - 4cosq in the form sin( ) So sin( ) sin(. ) (By solving in same way as Question 9, part a) Look for solutions in the interval.. 06.87..8, 0.8 4.8,8.4 ( d.p.) a Let cosq + sinq º cos(q -a) cos cos sin sin Compare cos : cos () Compare sin : sin () Divide () by (): tan 8.4 ( d.p.) 0 0.6 Solve 0 cos( 8.4 ), in the interval 0 60 cos( 8.4 ) 0 8.4 0.77, 09. 69., 7.7 ( d.p.) b Squaring cosq = - sinq gives 9cos q = 4 + sin q - 4sinq Þ 9(- sin q) = 4 + sin q - 4sinq Þ 0sin q - 4sinq - = 0 c 0sin q - 4sinq - = 0 4 6 sin 0 4 6 For sin, sin is positive, 0 so is in the first and second quadrants. 69., 80 69. 69., 0.8 ( d.p.) 4 6 For sin, sin is negative, 0 so is in the third and fourth quadrants. 80 (. ), 60 (. )., 7.7 ( d.p.) So solutions of quadratic in (b) are 69., 0.8,., 7.7 ( d.p.) d In squaring the equation, you are also including the solutions to cosq = -( - sinq), which when squared produces the same quadratic. The etra two solutions satisfy this equation. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
a cot cosec cos sin sin Multiplying both sides by sin gives cossin b cossin Set sin cos sin( ) sin cos cos sin So cos and sin sin tan cos tan 6.7 ( d.p.) So sin( 6.7) sin( 6.7), in the interval 6.7 6.7 86.7 6.7 6.7,.4 0,6.9 ( d.p.) As both cot and cosec are undefined at 0, 6.9 is the only solution. a cos sin 4 coscos sinsin 4 4 sin sin cos sin sin sin cos sin sin sin cos sin b cos sin Set sin cos sin( ) sin cos cos sin So cos and sin sin tan cos tan π 6 4 π sin 6 π sin, in the interval 6 π π π 6 6 6 6 4 a Set 9cos 40sin cos( ) cos cos sin sin b i So cos 9 and sin 40 sin 40 tan cos 9 40 tan 9 So 77.0 ( d.p.) cos sin 40 9 (cos sin ) 68 4 So 9cos 40sin 4cos( 77.0 ) 8 g( ) 0 4cos( 77.0 ) The minimum value of g( ) is when cos( 77.0 ) 8 8 So the minimum value is 0 4 9 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
4 b ii The minimum occurs when cos( 77.0 ) 77.0 0 77.0 a Set cos sin cos( ) cos cos sin sin So cos and sin sin tan cos tan So.6 ( d.p.) cos sin (cos sin ) 69 b cos(.6 ) 6. 6. cos(.6 ), in the interval.6.6 8.6.6 0, 40 48.7, 08.7 ( d.p.) c d 4cos 0sin cos cos 4 sin cos sin a =, b = and c = 4cos 0sin cos cos sin From part (a) cos sin cos(.6 ) The minimum value is therefore when cos(.60 ) It is ( ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7