COMPOUND ANGLES 185
B
From equation (7),
cos 6x+cos 2x=2 cos 4xcos 2x
From equation (5),
sin 6x+sin 2x=2 sin 4xcos 2x
Hence
cos 6x+cos 2x
sin 6x+sin 2x
=
2 cos 4xcos 2x
2 sin 4xcos 2x
=
cos 4x
sin 4x
=cot 4x
Now try the following exercise.
Exercise 84 Further problems on changing
sums or differences of sines and cosines into
products
In Problems 1 to 5, express as products:
- sin 3x+sinx [2 sin 2xcosx]
2.^12 (sin 9θ−sin 7θ) [cos 8θsinθ] - cos 5t+cos 3t [2 cos 4tcost]
4.^18 (cos 5t−cost)
[
−^14 sin 3tsin 2t
]
5.^12
(
cos
π
3
+cos
π
4
) [
cos
7 π
24
cos
π
24
]
- Show that:
(a)
sin 4x−sin 2x
cos 4x+cos 2x
=tanx
(b)^12 {sin(5x−α)−sin(x+α)}
=cos 3xsin(2x−α)
18.6 Power waveforms in a.c. circuits
(a) Purely resistive a.c. circuits
Let a voltagev=Vmsinωtbe applied to a circuit
comprising resistance only. The resulting current is
i=Imsinωt, and the corresponding instantaneous
power,p, is given by:
p=vi=(Vmsinωt)(Imsinωt)
i.e., p=VmImsin^2 ωt
From double angle formulae of Section 18.3,
cos 2A= 1 −2 sin^2 A, from which,
sin^2 A=^12 (1−cos 2A) thus
sin^2 ωt=^12 (1−cos 2ωt)
Then power p=VmIm[^12 (l−cos 2ωt)]
i.e. p=^12 VmIm(1−cos 2ωt)
The waveforms ofv,iandpare shown in Fig. 18.8.
The waveform of power repeats itself afterπ/ω
seconds and hence the power has a frequency twice
that of voltage and current. The power is always pos-
itive, having a maximum value ofVmIm. The average
or mean value of the power is^12 VmIm.
Figure 18.8
The rms value of voltageV= 0. 707 Vm, i.e.V=
Vm
√
2
,
from which,Vm=
√
2 V.
Similarly, the rms value of current,I=
Im
√
2
, from
which, Im=
√
2 I. Hence the average power, P,
developed in a purely resistive a.c. circuit is given
byP= 21 VmIm=^12 (
√
2 V)(
√
2 I)=VIwatts.
Also, powerP=I^2 RorV^2 /Ras for a d.c. circuit,
sinceV=IR.
Summarizing, the average powerPin a purely
resistive a.c. circuit given by
P=VI=I^2 R=
V^2
R
whereVandIare rms values.
(b) Purely inductive a.c. circuits
Let a voltagev=Vmsinωtbe applied to a circuit
containing pure inductance (theoretical case). The