Higher Engineering Mathematics

182 GEOMETRY AND TRIGONOMETRY

18.3 Double angles

(i) If, in the compound-angle formula for

sin(A+B), we letB=Athen

sin 2A=2 sinAcosA

Also, for example,

sin 4A=2 sin 2Acos 2A

and sin 8A=2 sin 4Acos 4A, and so on.

(ii) If, in the compound-angle formula for

cos(A+B), we letB=Athen

cos 2A=cos^2 A−sin^2 A

Since cos^2 A+sin^2 A=1, then

cos^2 A= 1 −sin^2 A, and sin^2 A= 1 −cos^2 A,

and two further formula for cos 2A can be

produced.

Thus cos 2A=cos^2 A−sin^2 A

=(1−sin^2 A)−sin^2 A

i.e. cos 2A= 1 −2 sin^2 A

and cos 2A=cos^2 A−sin^2 A

=cos^2 A−(1−cos^2 A)

i.e. cos 2A=2cos^2 A− 1

Also, for example,

cos 4A=cos^22 A−sin^22 A or

1 −2 sin^22 Aor

2 cos^22 A− 1

and cos 6A=cos^23 A−sin^23 A or

1 −2 sin^23 Aor

2 cos^23 A−1,

and so on.

(iii) If, in the compound-angle formula for
tan(A+B), we letB=Athen

tan 2A=

2 tanA

1 −tan^2 A

Also, for example,

tan 4A=

2 tan 2A

1 −tan^22 A

and tan 5A=

2 tan^52 A

1 −tan^252 A

and so on.

Problem 11. I 3 sin 3θis the third harmonic of a

waveform. Express the third harmonic in terms

of the first harmonic sinθ, whenI 3 =1.

WhenI 3 =1,

I 3 sin 3θ=sin 3θ=sin (2θ+θ)

=sin 2θcosθ+cos 2θsinθ,

from the sin (A+B) formula

=(2 sinθcosθ) cosθ+(1−2 sin^2 θ) sinθ,

from the double angle expansions

=2 sinθcos^2 θ+sinθ−2 sin^3 θ

=2 sinθ(1−sin^2 θ)+sinθ−2 sin^3 θ,

(since cos^2 θ= 1 −sin^2 θ)

=2 sinθ−2 sin^3 θ+sinθ−2 sin^3 θ

i.e.sin 3θ=3 sinθ−4 sin^3 θ

Problem 12. Prove that

1 −cos 2θ

sin 2θ

=tanθ.

LHS=

1 −cos 2θ

sin 2θ

=

1 −(1−2 sin^2 θ)

2 sinθcosθ

=

2 sin^2 θ

2 sinθcosθ

=

sinθ

cosθ

=tanθ=RHS

Problem 13. Prove that

cot 2x+cosec 2x=cotx.

LHS=cot 2x+cosec 2x=

cos 2x

sin 2x

+

1

sin 2x

=

cos 2x+ 1

sin 2x

=

(2 cos^2 x−1)+ 1

sin 2x

=

2 cos^2 x

sin 2x

=

2 cos^2 x

2 sinxcosx

=

cosx

sinx

=cotx=RHS