Higher Engineering Mathematics, Sixth Edition

Compound angles 167

R 2 7.3

4.6



Figure 17.4

Problem 8. Express−2.7sinωt− 4 .1cosωtin

the formRsin(ωt+α).

Let−2.7sinωt− 4 .1cosωt=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. 17.5. From Fig. 17.5,

R=

√

[(− 2. 7 )^2 +(− 4. 1 )^2 ]= 4. 909

and θ=tan−^1

4. 1

2. 7

= 56. 63 ◦

2708

908

(^18083608)
08

u
2 4.1 R
2 2.7
Figure 17.5
Henceα= 180 ◦+ 56. 63 ◦= 236. 63 ◦or 4.130 radians.
Thus,
− 2 .7sinωt− 4 .1cosωt= 4 .909sin(ω t+ 4. 130 ).
An angle of 236. 63 ◦is the same as−123.37◦or−2.153
radians.
Hence− 2 .7sinωt− 4 .1cosωtmay be expressed also
as4.909sin(ω t− 2. 153 ), which is preferred since it is
theprincipal value(i.e.−π≤α≤π).
Problem 9. Express 3sinθ+5cosθin the form
Rsin(θ+α), and hence solve the equation
3sinθ+5cosθ=4, for values ofθbetween 0◦and
360 ◦.
Let 3sinθ+5cosθ=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. 17.6.
R 5
3

Figure 17.6
From Fig. 17.6,R=
√
( 32 + 52 )= 5 .831 and
α=tan−^153 = 59. 03 ◦.
Hence 3sinθ+5cosθ= 5 .831sin(θ+ 59. 03 ◦)
However 3sinθ+5cosθ= 4
Thus 5.831sin(θ+ 59. 03 ◦)= 4 ,from which
(θ+ 59. 03 ◦)=sin−^1
(
4
5. 831
)