Higher Engineering Mathematics

Geometry and trigonometry

18

Compound angles

18.1 Compound angle formulae

An electric currentimay be expressed as

i = 5 sin(ωt− 0 .33) amperes. Similarly, the dis-

placementx of a body from a fixed point can

be expressed asx=10 sin(2t+ 0 .67) metres. The

angles (ωt− 0 .33) and (2t+ 0 .67) are calledcom-

pound anglesbecause they are the sum or difference

of two angles. Thecompound angle formulaefor

sines and cosines of the sum and difference of two

anglesAandBare:

sin(A+B)=sinAcosB+cosAsinB

sin(A−B)=sinAcosB−cosAsinB

cos(A+B)=cosAcosB−sinAsinB

cos(A−B)=cosAcosB+sinAsinB

(Note, sin(A+B)isnotequal to (sinA+sinB), and

so on.)

The formulae stated above may be used to derive two

further compound angle formulae:

tan(A+B)=

tanA+tanB

1 −tanAtanB

tan(A−B)=

tanA−tanB

1 +tanAtanB

The compound-angle formulae are true for all values

ofAandB, and by substituting values ofAandBinto

the formulae they may be shown to be true.

Problem 1. Expand and simplify the following

expressions:

(a) sin(π+α) (b)−cos(90◦+β)

(c) sin(A−B)−sin(A+B)

(a) sin(π+α)=sinπcosα+cosπsinα(from

the formula for sin (A+B))

=(0)(cosα)+(−1) sinα=−sinα

(b)−cos (90◦+β)

=−[cos 90◦cosβ−sin 90◦sinβ]

=−[(0)(cosβ)−(1) sinβ]=sinβ

(c) sin(A−B)−sin(A+B)

=[sinAcosB−cosAsinB]

−[sinAcosB+cosAsinB]

=−2cos A sin B

Problem 2. Prove that

cos(y−π)+sin

(

y+

π

2

)

= 0.

cos (y−π)=cosycosπ+sinysinπ

=(cosy)(−1)+(siny)(0)

=−cosy

sin

(

y+

π

2

)

=sinycos

π

2

+cosysin

π

2

=(siny)(0)+(cosy)(1)=cosy

Hence cos(y−π)+sin

(

y+

π

2

)

=(−cosy)+(cosy)= 0

Problem 3. Show that

tan

(

x+

π

4

)

tan

(

x−

π

4

)

=− 1.

tan

(

x+

π

4

)

=

tanx+tanπ 4

1 −tanxtanπ 4

from the formula for tan(A+B)

=

tanx+ 1

1 −(tanx)(1)

=

(

1 +tanx

1 −tanx

)

since tan

π

4

= 1

tan

(

x−

π

4

)

=

tanx−tan

π

4

1 +tanxtan

π

4

=

(

tanx− 1

1 +tanx

)