Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

PRELIMINARY ALGEBRA


and use made of equation (1.15), the following results are obtained:


sin 2θ=2sinθcosθ, (1.29)

cos 2θ=cos^2 θ−sin^2 θ

=2cos^2 θ− 1

=1−2sin^2 θ, (1.30)

tan 2θ=

2tanθ
1 −tan^2 θ

. (1.31)


A further set of identities enables sinusoidal functions ofθto be expressed in

terms of polynomial functions of a variablet= tan(θ/2). They are not used in


their primary role until the next chapter, but we give a derivation of them here


for reference.


Ift= tan(θ/2), then it follows from (1.16) that 1+t^2 =sec^2 (θ/2) and cos(θ/2) =

(1 +t^2 )−^1 /^2 , whilst sin(θ/2) =t(1 +t^2 )−^1 /^2. Now, using (1.29) and (1.30), we may


write:


sinθ=2sin

θ
2

cos

θ
2

=

2 t
1+t^2

, (1.32)

cosθ=cos^2

θ
2

−sin^2

θ
2

=

1 −t^2
1+t^2

, (1.33)

tanθ=

2 t
1 −t^2

. (1.34)


It can be further shown that the derivative ofθwith respect tottakes the


algebraic form 2/(1 +t^2 ). This completes a package of results that enables


expressions involving sinusoids, particularly when they appear as integrands, to


be cast in more convenient algebraic forms. The proof of the derivative property


and examples of use of the above results are given in subsection (2.2.7).


We conclude this section with a worked example which is of such a commonly

occurring form that it might be considered a standard procedure.


Solve forθthe equation
asinθ+bcosθ=k,
wherea, bandkare given real quantities.

To solve this equation we make use of result (1.18) by settinga=Kcosφandb=Ksinφ
for suitable values ofKandφ. We then have


k=Kcosφsinθ+Ksinφcosθ=Ksin(θ+φ),

with


K^2 =a^2 +b^2 and φ=tan−^1

b
a

.


Whetherφlies in 0≤φ≤πor in−π<φ<0 has to be determined by the individual
signs ofaandb. The solution is thus


θ=sin−^1

(


k
K

)


−φ,
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