Higher Engineering Mathematics, Sixth Edition

Compound angles 169

8. Solve the following equations for values of
   θ√between 0◦and 360◦:(a)6cosθ+sinθ=
   3(b)2sin3θ+8cos3θ=1.
   ⎡
   ⎣

(a) 82. 9 ◦, 296 ◦

(b) 32. 36 ◦, 97 ◦, 152. 36 ◦, 217 ◦,

272. 36 ◦and 337◦

⎤

⎦

9. The thirdharmonic of a wave motionis given
   by 4.3cos3θ− 6 .9sin3θ. Express this in the
   formRsin( 3 θ±α).[8.13sin( 3 θ+ 2. 584 )]


10. The displacementxmetres of a mass from
    a fixed point about which it is oscillating is
    given byx= 2 .4sinωt+ 3 .2cosωt,wheret
    is the time in seconds. Expressxin the form
    Rsin(ωt+α).[x= 4 .0sin(ωt+ 0. 927 )m]


11. Two voltages,v 1 =5cosωtand
    v 2 =−8sinωtare inputs to an analogue cir-
    cuit. Determine an expression for the output
    voltage if this is given by(v 1 +v 2 ).
    [9.434sin(ωt+ 2. 583 )]

17.3 Double angles

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

sin(A+B),weletB=Athen

sin2A=2sinAcosA

Also, for example,

sin4A=2sin2Acos 2A

and sin8A=2sin4Acos 4A,and so on.

(ii) If, in the compound-angle formula for
cos(A+B),weletB=Athen

cos2A=cos^2 A−sin^2 A

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

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

two further formula for cos2Acan be produced.

Thus cos2A=cos^2 A−sin^2 A

=( 1 −sin^2 A)−sin^2 A

i.e. cos2A= 1 −2sin^2 A

and cos2A=cos^2 A−sin^2 A

=cos^2 A−( 1 −cos^2 A)

i.e. cos2A=2cos^2 A− 1

Also, for example,

cos4A=cos^22 A−sin^22 Aor

1 −2sin^22 Aor

2cos^22 A− 1

and cos6A=cos^23 A−sin^23 Aor

1 −2sin^23 Aor

2cos^23 A− 1 ,

andsoon.

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

tan(A+B),weletB=Athen

tan2A=

2tanA

1 −tan^2 A

Also, for example,

tan4A=

2tan2A

1 −tan^22 A

and tan5A=

2tan^52 A

1 −tan^252 A

and so on.

Problem 11. I 3 sin3θis the third harmonic of a

waveform. Express the third harmonic in terms of

thefirstharmonicsinθ,whenI 3 =1.

WhenI 3 = 1 ,

I 3 sin3θ=sin3θ=sin( 2 θ+θ)

=sin2θcosθ+cos2θsinθ,

from the sin(A+B)formula

=(2sinθcosθ)cosθ+( 1 −2sin^2 θ)sinθ,

from the double angle expansions

=2sinθcos^2 θ+sinθ−2sin^3 θ

=2sinθ( 1 −sin^2 θ)+sinθ−2sin^3 θ,

(since cos^2 θ= 1 −sin^2 θ)

=2sinθ−2sin^3 θ+sinθ−2sin^3 θ

i.e.sin3θ=3sinθ−4sin^3 θ

Problem 12. Prove that

1 −cos2θ

sin2θ

=tanθ.