180 GEOMETRY AND TRIGONOMETRY
Hence
4 .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)cosωt
Equating coefficients gives:
− 2. 7 =Rcosα, from which, cosα=
− 2. 7
R
and − 4. 1 =Rsinα, from which, sinα=
− 4. 1
R
There 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. 909
and θ=tan−^1
4. 1
2. 7
= 56. 63 ◦
Figure 18.5
Henceα= 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)cosθ
Equating coefficients gives:
3 =Rcosα, from which, cosα=
3
R
and 5=Rsinα, from which, sinα=
5
R
Since both sinαand cosαare positive,Rlies in the
first quadrant, as shown in Fig. 18.6.
Figure 18.6
From 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 ′