Compound angles 169
8. Solve the following equations for values of
θ between 0
◦ and 360
◦ : (a) 6 cosθ + sin θ =
√
3 (b) 2 sin3θ + 8 cos 3θ = 1.
⎡
⎣
(a) 82.9 ◦ , 296 ◦
(b) 32.36 ◦ , 97 ◦ , 152.36 ◦ , 217 ◦ ,
272.36 ◦ and 337 ◦
⎤
⎦
9. The third harmonic of a wave motion is given
by 4.3 cos 3θ − 6.9 sin3θ. Express this in the
form R sin(3θ ± α). [8.13 sin(3θ + 2.584)]
10. The displacement x metres of a mass from
a fixed point about which it is oscillating is
given by x = 2.4 sin ωt + 3.2 cosωt , where t
is the time in seconds. Express x in the form
R sin(ωt + α). [x = 4.0 sin(ωt + 0.927)m]
11. Two voltages, v 1 = 5 cos ωt and
v 2 = −8 sinωt are inputs to an analogue circuit. Determine an expression for the output
voltage if this is given by (v 1 + v 2 ).
[9.434 sin(ωt + 2.583)]
17.3 Double angles
(i) If, in the compound-angle formula for
sin(A + B), we let B = A then
sin 2A = 2 sin A cos A
Also, for example,
sin 4 A = 2 sin2A cos 2 A
and sin 8 A = 2 sin4A cos 4 A, and so on.
(ii) If, in the compound-angle formula for
cos(A + B), we let B = A then
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 2 A can be produced.
Thus cos 2 A = cos
2 A − sin
2 A
= (1 − sin
2 A) − sin
2 A
i.e.
cos 2A = 1 − 2 sin
2 A
and
cos 2 A = cos
2 A − sin
2 A
= cos
2 A − (1 − cos
2 A)
i.e.
cos 2 A = 2cos
2 A − 1
Also, for example,
cos 4 A = cos
2 2 A − sin
2 2 A or
1 − 2 sin
2 2 A or
2 cos
2 2 A − 1
and cos 6 A = cos
2 3 A − sin
2 3 A or
1 − 2 sin
2 3 A or
2 cos
2 3 A − 1,
and so on.
(iii) If, in the compound-angle formula for
tan(A + B), we let B = A then
tan 2A =
2 tan A
1 − tan 2 A
Also, for example,
tan 4 A =
2 tan 2A
1 − tan 2 2 A
and tan 5 A =
2 tan
5
2 A
1 − tan 2 5
2 A
and so on.
Problem 11. I 3 sin 3θ is the third harmonic of a
waveform. Express the third harmonic in terms of
the first harmonic sin θ, when I 3 = 1.
When I 3 = 1,
I 3 sin 3θ = sin 3θ = sin(2θ + θ)
= sin 2θ cos θ + cos 2θ sin θ,
from the sin(A + B) formula
= (2 sinθ cos θ) cos θ + (1 − 2 sin
2
θ) sin θ,
from the double angle expansions
= 2 sinθ cos
2
θ + sin θ − 2 sin
3
θ
= 2 sinθ(1 − sin
2
θ) + sin θ − 2 sin
3
θ,
(since cos
2
θ = 1 − sin
2
θ)
= 2 sinθ − 2 sin
3
θ + sin θ − 2 sin
3
θ
i.e. sin 3θ = 3 sinθ − 4 sin 3 θ
Problem 12. Prove that
1 − cos 2θ
sin 2θ
= tan θ.