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
)