Higher Engineering Mathematics

COMPOUND ANGLES 181

B

Since − 15 ◦ 43 ′ is the same as− 15 ◦ 43 ′+ 360 ◦,
i.e. 344◦ 17 ′, then the solutions areθ= 77 ◦ 39 ′or
344 ◦ 17 ′, which may be checked by substituting into
the original equation.

Problem 10. Solve the equation

3 .5 cosA− 5 .8 sinA= 6 .5 for 0◦≤A≤ 360 ◦.

Let 3.5 cosA− 5 .8 sinA=Rsin(A+α)

=R[sinAcosα+cosAsinα]

=(Rcosα) sinA+(Rsinα) cosA

Equating coefficients gives:

3. 5 =Rsinα, from which, sinα=

3. 5

R

and − 5. 8 =Rcosα, from which, cosα=

− 5. 8

R

There is only one quadrant in which both sine is
positiveandcosine is negative, i.e. the second, as
shown in Fig. 18.7.

Figure 18.7

From Fig. 18.7, R=

√

[(3.5)^2 +(− 5 .8)^2 ]= 6. 774

andθ=tan−^1

3. 5

5. 8

= 31 ◦ 7 ′.

Henceα= 180 ◦− 31 ◦ 7 ′= 148 ◦ 53 ′.

Thus

3 .5cosA− 5 .8sinA= 6 .774 sin (A+ 148 ◦ 53 ′)= 6. 5

Hence sin(A+ 148 ◦ 53 ′)=

6. 5

6. 774

, from which,

(A+ 148 ◦ 53 ′)=sin−^1

6. 5

6. 774

= 73 ◦ 39 ′or 106◦ 21 ′

Thus A= 73 ◦ 39 ′− 148 ◦ 53 ′=− 75 ◦ 14 ′

≡(− 75 ◦ 14 ′+ 360 ◦)= 284 ◦ 46 ′

or A= 106 ◦ 21 ′− 148 ◦ 53 ′=− 42 ◦ 32 ′

≡(− 42 ◦ 32 ′+ 360 ◦)= 317 ◦ 28 ′

The solutions are thus A= 284 ◦ 46 ′or 317◦ 28 ′,

which may be checked in the original equation.

Now try the following exercise.

Exercise 81 Further problems on the

conversion ofasinωt+bcosωtinto

Rsin(ωt+α)

In Problems 1 to 4, change the functions into the

formRsin(ωt±α).

1. 5 sinωt+8 cosωt [9.434 sin(ωt+ 1 .012)]


2. 4 sinωt−3 cosωt [5 sin(ωt− 0 .644)]

3.−7 sinωt+4 cosωt

[8.062 sin(ωt+ 2 .622)]

4.−3 sinωt−6 cosωt

[6.708 sin(ωt− 2 .034)]

5. Solve the following equations for values ofθ
   between 0◦and 360◦: (a) 2 sinθ+4 cosθ= 3
   (b) 12 sinθ−9 cosθ[=7.
   (a) 74◦ 26 ′or 338◦ 42 ′
   (b) 64◦ 41 ′or 189◦ 3 ′

]

6. Solve the following equations for
   0 ◦<A< 360 ◦: (a) 3 cosA+2 sinA= 2. 8
   (b) 12 cosA−4 sin[A= 11
   (a) 72◦ 44 ′or 354◦ 38 ′
   (b) 11◦ 9 ′or 311◦ 59 ′

]

7. The third harmonic of a wave motion is given
   by 4.3 cos 3θ− 6 .9 sin 3θ. Express this in the
   formRsin(3θ±α). [8.13 sin(3θ+ 2 .584)]


8. The displacementxmetres of a mass from
   a fixed point about which it is oscillating is
   given byx= 2 .4 sinωt+ 3 .2 cosωt, wheret
   is the time in seconds. Expressxin the form
   Rsin(ωt+α). [x= 4 .0 sin(ωt+ 0 .927)m]


9. Two voltages,v 1 =5 cosωtand
   v 2 =−8 sinωtare inputs to an analogue cir-
   cuit. Determine an expression for the output
   voltage if this is given by (v 1 +v 2 ).
   [9.434 sin(ωt+ 2 .583)]