Higher Engineering Mathematics, Sixth Edition

Compound angles 165

3. Show that:
   (a) sin

(

x+

π

3

)

+sin

(

x+

2 π

3

)

=

√

3cosx

and

(b)−sin

(

3 π

2

−φ

)

=cosφ

4. Prove that:
   (a) sin

(

θ+

π

4

)

−sin

(

θ−

3 π

4

)

=

√

2 (sinθ+cosθ)

(b)

cos( 270 ◦+θ)

cos( 360 ◦−θ)

=tanθ

5. Given cosA= 0 .42 and sinB= 0 .73 evaluate
   (a) sin(A−B),(b)cos(A−B),(c)tan(A+B),
   correct to 4 decimal places.
   [(a) 0.3136 (b) 0.9495 (c)−2.4687]
   In Problems 6 and 7, solve the equations for
   values ofθbetween 0◦and 360◦.


6. 3sin(θ+ 30 ◦)=7cosθ
   [64. 72 ◦or 244. 72 ◦]


7. 4sin(θ− 40 ◦)=2sinθ
   [67. 52 ◦or 247. 52 ◦]

17.2 Conversion ofasinωt+bcosωt

intoRsin(ωt+α)

(i) Rsin(ωt+α)represents a sine wave of maxi-

mum valueR, periodic time 2π/ω, frequency

ω/ 2 π and leading Rsinωt by angleα.(See

Chapter 14).

(ii) Rsin(ωt+α) may be expanded using the

compound-angle formula for sin(A+B),where

A=ωtandB=α. Hence,

Rsin(ωt+α)

=R[sinωtcosα+cosωtsinα]

=Rsinωtcosα+Rcosωtsinα

=(Rcosα)sinωt+(Rsinα)cosωt

(iii) If a=Rcosα and b=Rsinα,wherea and
bare constants, thenRsin(ωt+α)=asinωt+
bcosωt,i.e.asineandcosinefunctionofthesame
frequency when added produce a sine wave of the

same frequency (which isfurtherdemonstrated in

Chapter 25).

(iv) Sincea=Rcosα, then cosα=a/R, and since

b=Rsinα,thensinα=b/R.

R b

a



Figure 17.1

If the values ofaandbare known then the values

ofRandαmay be calculated. The relationship between

constantsa,b,Randαare shown in Fig. 17.1.

From Fig. 17.1, by Pythagoras’ theorem:

R=

√

a^2 +b^2

and from trigonometric ratios:

α=tan−^1 b/a

Problem 6. Find an expression for 3sinωt+ 4

cosωtin the formRsin(ωt+α)and sketch graphs

of 3sinωt,4cosωtandRsin(ωt+α)on the

same axes.

Let 3sinωt+4cosωt=Rsin(ωt+α)

then 3sinωt+4cosωt

=R[sinωtcosα+cosωtsinα]

=(Rcosα)sinωt+(Rsinα)cosωt

Equating coefficients of sinωtgives:

3 =Rcosα,from which,cosα=

3

R

Equating coefficients of cosωtgives:

4 =Rsinα,from which,sinα=

4

R