Understanding Engineering Mathematics

wherem,nare integers.

So the general solution is

θ=kπ,or

2 π

3

+ 2 mπ,or−

2 π

3

+ 2 nπ k,m,nare integers

NB – it is a common error to forget the sinθ=0 part of the solution.

(ii) cos 3θ=cosθis an example of cosA=cosBfor which we know

3 θ= 2 kπ±θ wherek=an integer

So 4θ= 2 kπ or 2θ= 2 kπ

We therefore obtain

θ=

2 kπ

4

=

kπ

2

or

2 kπ

2

=kπ

and so the general solution is

kπ

2

orkπ, withk=an integer.

6.2.9 TheacosqYbsinqform

➤

173 197➤

The compound angle identities may be used, with a little help from Pythagoras (154

➤
),
to convert an expression of the formacosθ+bsinθ to one of the more manageable
formsrcos(θ±α)orrsin(θ±α). The latter forms, for example, tell us immediately by
inspection the maximum and minimum values of such expressions, and where they occur.
This section uses a lot of what we have done, in particular the ‘minimal set’ of results to
remember – cos^2 θ+sin^2 θ=1, and the expansions of sin(A+B)and cos(A+B).To
illustrate conversion of this form, we consider the example of

acosθ+bsinθ≡rsin(θ+α)

Using the compound angle formula this implies

acosθ+bsinθ≡rcosαsinθ+rsinαcosθ

This has got to be true for all possible values ofθ, and to ensure this we equate the
coefficients of cosθ,sinθon each side to get

rcosα=a

rsinα=b

where, by squaring and adding (185
➤
),r=

√

a^2 +b^2 and

cosα=

a

r

sinα=

b

r

from whichαmay be determined. Care is needed here if eitheraorbis negative – we
have to look carefully at the quadrant thatαlies in.