y 2 Well it is a cos graph with amplitude of 3 and only half a waveform between 0 and 2π because the cos argument is x /2: y =3cos π 2

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1 Complete Solutions Eamination Questions Complete Solutions to Eamination Questions 4. (a) How do we sketch the graph of y 3? Well it is a graph with amplitude of 3 and only half a waveform between 0 and π because the argument is /: 3 y y 3 π π π π 3π π (b) Using the waveform section in the tetbook with reference to (4.8), (4.9) and (4.30) we have the following: π The period T is the time taken to complete one cycle, that is T 4π. The frequency f / is defined as f T 4π Hz. The amplitude of the given function is 3.. Taking inverse of both sides gives 3 0. By formula R 360 n α α (4.7) The general solution of ( ) is ( ) ± where ( R) the general solution is given by n ± 0 ( n) ( n) 360 ± { ± ( 0n) ± Dividing by 3 3. How do we convert sin 3 into r( α )? Finding the angle α is difficult but to determine r is a straightforward application of Pythagoras, that is r Plotting the quadrant where sin 3 lies:

2 Complete Solutions Eamination Questions α 3-30 By using a calculator or TABLE we have tan 30 3 Clearly 30 is the wrong quadrant. Hence α Substituting r and α 50 we have sin 3 50 sin ( A+ B) sin A A 4. Cross-multiplying the Left Hand Side of + gives sin B B sin B sin A B+ Asin B sin ( A+ B) Because sin B B sin B B sin A B+ Asin B sin ( A+ B) sin ( A+ B) Because sin ( ) sin B B B sin B sin( A+ B) Because sin B / sin 75 How do we evaluate sin 60? We need to write the 75 and 60 in terms of well known trigonometric ratios such as those in TABLE. We can write and 60 ( 30 ) Substituting these into the above we have sin 75 sin ( ) sin 60 sin ( [ 30 ]) sin A A sin ( A+ B) We can evaluate this by using the above proven result +. sin B B sin B Dividing this by we have sin ( A+ B) sina A sin B + sin B B (*)

3 Complete Solutions Eamination Questions 3 Substituting A 45 and B 30 into (*) gives sin 75 sin ( ) sin { + sin 60 From sin 30 sin above ( [ ]) / / + Because sin ( 45 ) ( 45 ) / 3/ { / + { + / 3 3 Taking out Taking out common factor common factor / 4 5. What does arctan 7 mean? Remember arctan is tan 4 4, that is arctan tan. Let be the angle such that tan therefore tan ( ). We have By Pythagoras adj Therefore ( ) : hyp ( ) arctan tan 6. We need to find the values of t which satisfy ( t) ( t) sin 0. How? We use a trig identity but which one? Use sin t. Thus substituting this into the given ( t ) ( t) sin ( t) equation ( t) sin ( t) 0 yields: sin () t sin ( t) 4sin ( t) 0 Taking the square root gives sin 4 () t [ Transposing]

4 Complete Solutions Eamination Questions 4 sin () t, What are the values of t where sin () t? π Using our calculator or by the table of values in the tet we have t sin. The 6 π π domain of t is between 0 and π therefore t π. We can use CAST to find 6 6 the other angle: t S A T C π 6 sin () t (Negative) in the quadrants C and T. π 7π The other angle t is given by π +. Do we have any other values of t? 6 6 Yes where sin () t +. In this case t lies in the quadrants A and S because we have π π 5π positive value of sin. Hence t sin and π π 5π 7π π Collecting our t values we have t,, and We can write 3 π 3π π 3π as +. We have 4 3 3π 3π π ( A+ B) ( A) ( B) ( A) ( B) 3π π 3π π Using sin sin sin sin 3π 3π Because, sin, π π 3 and sin { Taking Out a Common Factor

5 Complete Solutions Eamination Questions 5 8. Since tan ( ) is in the fourth quadrant we have: The hypotenuse + 5. We have sin ( ) and ( ) 5 5 sin we have Using the double angle formula ( ) sin Re-arranging this gives sin ( ) sin ± Taking Square Root we have 5 Substituting the above result ( ) [ ] sin ± ( ) ± 5 We are given that is in the fourth quadrant which means that 3 π < < π. Dividing this by : 3π < < π 4 Therefore is in the second quadrant where the sin value is positive. Removing the negative square root we have sin 5

6 Complete Solutions Eamination Questions 6 9. Since is in the second quadrant such that sin 3, we have: By applying Pythagoras and noting that lies to the left of the origin we have Remember the negative only signifies length is to the left of the origin. Therefore using the definition of ine and tangent 4 3 ( ), tan ( ) 5 4 What other trigonometric ratios do we need to find? 5 ec( ) sin 3/ How do we verify Note that csc ec sec cot ( ) ( ) csc ( ) 5 4 / tan 3/ 4 3 ( ) sec? which means that sin Substituting this into the given equation we have sin. csc ec Because sin sin csc csc sec Because sec. In the tetbook we have only converted from a combination of sine and ine to

7 Complete Solutions Eamination Questions 7 a ine only waveform. In this case we will use the otherwise option, that is convert the Left Hand Side of sin 3.4 into R α. By + (4.75) a( ) bsin ( ) R( α) with a 3 and b we have + where R α tan 8.43 [By calculator] 3 Solving the given equation we have sin b R a + b and α tan a Dividing through by 0 and taking inverse ine we have.4 ( 8.43 ) Hence Are there any other angles? Yes because is positive in the fourth quadrant as well S A T C The angle shown is Using this angle we have The angles are 58.97, π. (a) How do we show sin Apply the identity sin ( A B) sin ( A) ( B) ( A) sin ( B)? with A and 3π 3π 3π sin sin sin π B :

8 Complete Solutions Eamination Questions 8 (b) Writing sec into the LHS of the given equation sec sin tan : sec sin sin + Because sin sin sin sin tan Because tan 3. We are given that C b a A c B Which rule do we use to find the length b? We can apply the sine rule because we have two angles and one length. a b (*) sin ( A) sin ( B) Substituting a 0, A 30 and B 45 into (*) we have 0 b sin ( 30 ) sin ( 45 ) 0 Transposing this gives b sin ( 45 ) 8.8. sin Using the identity ( A + B) ( A) ( B) sin ( A) sin ( B) B arc we have arctan+ arc A B sin A sin B with A arctan and ( arctan) ( arc ) sin ( arctan) sin ( arc ) ( ) Using our table of values we have arctan 45. Let be the angle such that ( ) then arc. Drawing a right-angled triangle with hypotenuse and adjacent because ( ) :

9 Complete Solutions Eamination Questions 9 Using the sine definition (4.) we have ( ) Substituting ar ctan 45 and sin arc yields:. into ( + ) ( 45 ) ( ) sin ( 45 ) sin ( ) arctan arc arctan arc sin arctan sin arc 3 3 Because ( 45 ) sin ( 45 ) Factorising Therefore the epression in is. 5. The given waveform is vt () 57 sin ( 67π t 0.6) tet that For. Remember we know from the main v R sin(ωt ± α) phase angle Time Ampiltude Angular velocity () vt 57 sin 67π t 0.6 we have R 57 and ω 67π. Using these values: π / π 3 (i) Amplitude 57, period timet ms. Using ω 67 / π 67 f T we have frequency f 33.5 Hz. The phase is 0. 6 rads. How do we convert this into degrees? 80 By applying (4.7) rads π : π The phase angle in degrees is 5. How do we find the phase time? By using (4.33) Rsin ( ωt α ) lags Rsin ( t) ω by α ω. Hence Phase time ms 67π Hence the phase time is.35ms. (ii) Substitute t 0 vt 57 sin 67πt 0.6 : into the given equation v π At t 0 the voltage is 4.65 volts. (iii) Substituting t 8 ms sin sin into 57 sin ( 67πt 0.6) vt gives

10 Complete Solutions Eamination Questions π v sin sin The voltage at t 8 ms is volts. (iv) How do we find the time when the voltage is first a maimum? This occurs when sin 67π t 0.6) because the maimum sine value is. Taking inverse sine gives ( t π () π 67π t / π 0.6 t π 67π 0.6 sin ( / π) The voltage is first a maimum at time t 8.70 ms. (v) When does v 40 volts? Substituting () 40 vt 57 sin 67π t 0.6 : Taking inverse sin gives vt into () 57sin ( 67π t 0.6) sin ( 67π t 0.6) πt 0.6 sin Transposing to make t the subject gives t π The waveform vt () reaches 40volts at t 4.9ms. We can sketch vt () over one cycle by using the above evaluations: v(t) v 57sin(67πt) 3 ( 8.70^(-3), 57 ) v 57sin(67πt 0.6) t