Trigonometric Functions Mixed Exercise

Transcription

1 Trigonometric Functions Mied Eercise tan = cot, Þ tan = tan Þ tan = Þ tan = ± Calculator value for tan = + is 54.7 ( d.p.) 4 a i cosecq = cotq, 0 <q <80 Þ sinq = cosq sinq Þ cosq = Þ cosq = Þ q = 60 ii cot q = 7cosecq -8, 0 <q <80 Þ (cosec q -) = 7cosecq - 8 Þ cosec q - 7cosecq + 6 = 0 Þ (cosecq - )(cosecq - ) = 0 Þ cosecq = or cosecq = Solutions are required in the st, rd and 4th quadrants. Solution set is: -5., , 54.7 ( d.p.) So sinq = or sinq = p = secq Þ secq = p q = 4cosq Þ cosq = q 4 secq = cosq Þ p = q 4 = 4 q Þ p = 8 q p = sinq Þ p = sinq = cosecq q = 4cotq Þ cotq = q 4 Using + cot q cosec q Þ + q 6 = p (multiply by 6 p ) Þ 6 p + p q = 6 Þ p q = 6-6 p = 6(- p ) Solutions are a and (80 -a) where a is the calculator value. sinq = Calculator value is 4.8 ( d.p.) Solutions are 4.8, 8. sinq = Calculator value is 0 ( d.p.) Solutions are 0, 50 Solution set is: 0, 4.8, 8., 50 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

2 4 b i sec(q -5 ) = cosec5, 0 q 60 Þ cos(q - 5 ) = cosec5 = sin5 = Solve cos(q -5 ) =, in the interval -5 q The calculator value is 45 cos is positive, so (q -5 ) is in the st and 4th quadrants. c i cosec + p ö 5ø = -, 0 p Þ sin + p ö 5ø = - Calculator value is - p 4 sin + p ö 5ø is negative, so + p is in rd and 4th quadrants. 5 So + p 5 = 5p 4, 7p 4 So (q -5 ) = 45, 5, 405, 675 Þ q = 60, 0, 40, 690 Þ q = 0, 65, 0, 45 ii sec q + tanq =, 0 q 60 Þ + tan q + tanq = Þ tan q + tanq - = 0 Þ (tanq -)(tanq + ) = 0 Þ tanq = or tanq = - tanq = Þ q = 45, , i.e. 45, 5 tanq = -, calculator value is ( d.p.) Þ q = 80 + (-6.4 ) = 6.6 q = 60 + (-6.4 ) = 96.6 Solution set is: 45, 6.6, 5, 96.6 Þ = 5p 4 - p 5, 7p 4 - p 5 = 75p - 4p, 60 = 7p 60, 0p 60 ii sec = 4, 0 p Þ cos = 4 Þ cos = ± 05p - 4p 60 Calculator value for cos = + is p 6 As cos is ±, is in all four quadrants.solution set is: = p 6, p - p 6, p + p 6, p - p 6 = p 6, 5p 6, 7p 6, p 6 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

3 5 5sin cos y + 4cos sin y = 0 5sin cos y 4cos sin y Þ + sin sin y sin sin y = 0 (divide by sin sin y) Þ 5cos y sin y + 4cos sin = 0 So 5cot y + 4cot = 0 As cot = 5cot y + 8 = 0 5cot y = -8 cot y = a LHS (tanq + cotq)(sinq + cosq) sinq cosq + cosq ö sinq ø (sinq + cosq) sin q + cos q ö cosq sinq ø (sinq + cosq) ö cosq sinq ø (sinq + cosq) sinq cosq sinq + cosq cosq sinq cosq + sinq secq + cosecq RHS cosec b LHS cosec - sin sin sin - sin sin - sin sin sin sin - sin - sin cos (using sin + cos ) sec RHS c LHS (- sin )(+ cosec) - sin + cosec - sin cosec - sin + cosec - as cosec = ö sin ø cosec - sin sin - sin - sin sin cos sin cos sin cos cos cot RHS cot d LHS cosec - - cos + sin cos sin - cos sin - + sin cos sin - cos - sin + sin sin cos - sin - cos + sin cos (+ sin ) - cos (- sin ) (- sin )(+ sin ) cos sin - sin cos sin cos sin cos tan RHS Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

4 6 e LHS cosecq - + cosecq + (cosecq +) + (cosecq -) (cosecq -)(cosecq +) f cosecq cosec q - cosecq cot q (+ cot q cosec q) sinq sin q cos q cosq sinq cosq secq tanq RHS (secq - tanq)(secq + tanq) LHS + tan q sec q - tan q sec q (+ tan q) - tan q sec q sec q cos q RHS 7 a LHS sin + cos + + cos sin sin + (+ cos ) (+ cos )sin sin ++ cos + cos (+ cos )sin + cos (+ cos )sin ( sin + cos ) (+ cos ) (+ cos )sin sin cosec RHS b Solve cosec = - 4, - p p Þ cosec = - Þ sin = - Calculator value is - p Solutions in - p p are - p, - p + p, p + p, p - p i.e. - p,- p, 4p, 5p 8 RHS (cosecq + cotq) sinq + cosq ö sinq ø (+ cosq) sin q (+ cosq) - cos q (+ cosq)(+ cosq) (+ cosq)(- cosq) + cosq - cosq LHS Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 9 a sec A = -, p < A < p, i.e. A is in nd quadrant. As + tan A = sec A + tan A = 9 tan A = 8 tan A = ± 8 = ± As A is in nd quadrant, tan A is negative. So tan A = - b sec A = -, so cos A = - As tan A = sin A cos A sin A = cos A tan A = - - = So cosec A = = = 4 An alternative approach is to use the fact that cosec q + cot q cosec A = + cot A = + 8 = 9 8 Þ cosec A = ± = ± 4 As p < A < p, cosec A is positive, so cosec A = 4 0 secq = k, k ³ q is in the nd quadrant Þ k is negative a cosq = secq = k b Using + tan q = sec q tan q = k - c tanq = ± k - In the nd quadrant, tanq is negative So tanq = - k - cotq = tanq = = - - k - d Using + cot q = cosec q k - cosec q = + k - = k -+ k - = k k - k So cosecq = ± k - In the nd quadrant, cosecq is positive As k is negative, cosecq = - sec + p ö 4 ø =, 0 p Þ cos + p ö 4 ø =, 0 p Þ + p 4 = cos-, p - cos- = p, p - p So = p - p 4, 5p - p 4 4p - p =, = p, 7p 0p - p k k - arcsin ö ø is the angle a in the interval - p a p whose sine is So arcsin ö ø = p 6 Similarly, arcsin - ö ø = - p 6 So arcsin ö ø - arcsin - ö ø = p p ö 6 ø = p Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 sec - tan - = 0, 0 p Þ (+ tan ) - tan - = 0 Þ tan - tan -= 0 (This does factorise!) tan + Þ tan = - ( ) = 0 ö tan - ø or tan = Calculator values are - p 6 and p 5 arctan ( - ) = - p Þ - = tan - p ö ø Þ - = - Þ = - 6 Solution set: p, 5p 6, 4p, p 6 4 a seccosec - sec - cosec + = sec (cosec - ) - (cosec - ) = (cosec - )(sec -) b So seccosec - sec - cosec + = 0 Þ (cosec - )(sec -) = 0 Þ cosec = or sec = Þ sin = or cos = sin =, 0 60 Þ = 0, (80-0) cos =, 0 60, Þ = 0, 60 (from the graph) As cosec is not defined for = 0, 60, the equation is not defined for these values, so = 0, 60 are not solutions So the solution set is: 0, 50 7 a As + tan sec sec - tan Þ (sec - tan )(sec + tan ) (difference of two squares) As tan + sec = - is given, so - (sec - tan ) = Þ sec - tan = - Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 7 b sec + tan = - and sec - tan = - i Add the equations Þ sec = - 0 Þ sec = - 5 ii Subtract the equations Þ tan = -- - ö ø = - 8 Þ tan = - 4 c As sec and tan are both negative, cos and tan are both negative. So must be in the nd quadrant. 9 a LHS sec 4 q - tan 4 q (sec q - tan q)(sec q + tan q) (sec q + tan q) (as sec q + tan q Þ sec q - tan q ) sec q + tan q RHS b sec 4 q = tan 4 q + tanq Þ sec 4 q - tan 4 q = tanq Þ sec q + tan q = tanq ( using part (a)) Þ (+ tan q) + tan q = tanq Þ tan q - tanq + = 0 Þ (tanq -)(tanq -) = 0 Þ tanq = or tanq = Solving tan = - 4, where is in the nd quadrant, gives = ( ) = 6.9 ( d.p.) 8 p = secq - tanq, q = secq + tanq Multiply together: pq = (secq - tanq)(secq + tanq) = sec q - tan q = (since + tan q = sec q) In the interval -80 q 80 tanq = Þ q = tan-, tan- = 6.6, -5.4 ( d.p.) tanq = Þq = tan -, tan - Solution set is: = 45, , -5, 6.6, 45 Þ p = q (There are several ways of solving this problem.) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 0 a y = sin c The curves are the same with the aes interchanged. The shaded area in (b) could be added to the graph in (a) to form a rectangle with sides and p, as in the diagram. p ò sin d represents the area between 0 y = sin, the - ais and = p b y = arcsin, - ò arcsin d represents the area between 0 y = arcsin, the -ais and = Area of rectangle = p = p p So ò sin d + arcsin d 0 ò = p 0 cot60 sec60 = tan60 cos60 = = = a The graph of y sec is y sec stretched by a scale factor in the y direction, then reflected in the -ais 0 and then translated by the vector. b -< k < 5 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 a The graph of y = arcsin - p is y arcsin stretched by a scale factor in the y direction and then translated by 0 the vector b Let y = arccos, - < 0 Þ cos y = Þ sin y = - p As < 0, < y p, so sin y is positive Þ tan y = sin y cos y = for < 0, - p < y p Note that as y > p, it is not in the range of y = arccos b Curve meets the -ais when y = 0 Þ arcsin - p = 0 Þ arcsin = p 6 Þ sin p 6 = Þ = Curve meets the -ais at, 0 ö ø However, from the tan curve, we know that tan( y - p) = tan y So tan( y - p) = - for < 0, - p < y - p 0 We can now use the inverse function y - p = arctan So y = p + arctan for < a Let y = arccos, 0 < Þ cos y = Þ sin y = - Note that as 0 <, 0 y < p, so sin y is positive - Thus tan y =, which is valid for 0 < Thus arccos = p + arctan for < 0 - Þ y = arctan - So arccos = arctan - for 0 < Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9