180 GEOMETRY AND TRIGONOMETRYHence4 .6 sinωt− 7 .3 cosωt= 8 .628 sin(ωt− 1. 008 ).Problem 8. Express −2.7 sinωt− 4 .1 cosωt
in the formRsin(ωt+α).Let−2.7 sinωt− 4 .1 cosωt=Rsin(ωt+α)=R[sinωtcosα+cosωtsinα]
=(Rcosα)sinωt+(Rsinα)cosωtEquating coefficients gives:− 2. 7 =Rcosα, from which, cosα=− 2. 7
Rand − 4. 1 =Rsinα, from which, sinα=− 4. 1
RThere is only one quadrant in which both cosineand
sine are negative, i.e. the third quadrant, as shown in
Fig. 18.5. From Fig. 18.5,
R=√
[(− 2 .7)^2 +(− 4 .1)^2 ]= 4. 909and θ=tan−^14. 1
2. 7= 56. 63 ◦Figure 18.5Henceα= 180 ◦+ 56. 63 ◦= 236. 63 ◦or 4.130 radi-
ans.Thus,− 2 .7 sinωt− 4 .1 cosωt= 4 .909 sin(ωt+ 4. 130 ).An angle of 236. 63 ◦is the same as−123.37◦or
−2.153 radians.
Hence− 2 .7 sinωt− 4 .1 cosωtmay be expressed
also as4.909 sin(ωt− 2. 153 ), which is preferred
since it is theprincipal value(i.e.−π≤α≤π).Problem 9. Express 3 sinθ+5 cosθ in the
formRsin(θ+α), and hence solve the equation
3 sinθ+5 cosθ=4, for values ofθbetween 0◦
and 360◦.Let 3 sinθ+5 cosθ=Rsin(θ+α)=R[sinθcosα+cosθsinα]=(Rcosα)sinθ+(Rsinα)cosθEquating coefficients gives:3 =Rcosα, from which, cosα=3
Rand 5=Rsinα, from which, sinα=5
R
Since both sinαand cosαare positive,Rlies in the
first quadrant, as shown in Fig. 18.6.Figure 18.6From Fig. 18.6,R=√
(3^2 + 52 )= 5 .831 and
α=tan−^153 = 59 ◦ 2 ′.
Hence 3 sinθ+5 cosθ= 5 .831 sin(θ+ 59 ◦ 2 ′)However 3 sinθ+5 cosθ= 4
Thus 5.831 sin(θ+ 59 ◦ 2 ′)=4, from which(θ+ 59 ◦ 2 ′)=sin−^1(
4
5. 831)i.e. θ+ 59 ◦ 2 ′= 43 ◦ 19 ′or 136◦ 41 ′
Hence θ= 43 ◦ 19 ′− 59 ◦ 2 ′=− 15 ◦ 43 ′
or θ= 136 ◦ 41 ′− 59 ◦ 2 ′= 77 ◦ 39 ′