COMPOUND ANGLES 185B
From equation (7),
cos 6x+cos 2x=2 cos 4xcos 2xFrom equation (5),
sin 6x+sin 2x=2 sin 4xcos 2xHence
cos 6x+cos 2x
sin 6x+sin 2x=2 cos 4xcos 2x
2 sin 4xcos 2x=cos 4x
sin 4x=cot 4xNow try the following exercise.
Exercise 84 Further problems on changing
sums or differences of sines and cosines into
productsIn 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
(
cosπ
3+cosπ
4) [
cos7 π
24cosπ
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) thussin^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 18.8The 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, fromwhich, 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 byP=VI=I^2 R=V^2
RwhereVandIare rms values.(b) Purely inductive a.c. circuitsLet a voltagev=Vmsinωtbe applied to a circuit
containing pure inductance (theoretical case). The