082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

3.7 Modelling waves using sint and cost 137
From (3.3) and (3.4), both sinφ and cosφ are negative and so φ lies in the
third quadrant. Division of (3.3)by (3.4) gives
3
−2 = −Rsin φ
Rcosφ =−tanφ
tanφ = 1.5
Usingacalculatorandnotingthat φ liesinthethirdquadrantwefind φ = 4.124.
Finally
3sin4t−2cos4t = √ 13cos(4t +4.124)
Soi 3
(t) = √ 13cos(4t +4.124).Thereforei 3
(t)hasanamplitudeof √ 13amps
and a phase of4.124 radians.
Example3.11 Express0.5cos3t +sin3t asasingle cosinewave.
Solution Let
0.5cos3t +sin3t =Rcos(3t + φ)
Hence
=R(cos3tcosφ −sin3tsinφ)
= (Rcosφ)cos3t −(Rsinφ)sin3t
0.5=Rcosφ (3.5)
1=−Rsinφ (3.6)
Bysquaring and adding weobtain
1.25 =R 2
R = √ 1.25 = 1.1180
Division of (3.6)by (3.5) yields
2=−tanφ
(4 d.p.)
From (3.5), cosφ is positive; from (3.6), sin φ is negative; and so φ lies in the fourth
quadrant. Hence, using a calculator, φ = 5.1760. So
0.5cos3t +sin3t = 1.1180cos(3t +5.1760)
Example3.12 Ifacos ωt +bsinωt is expressed in the formRcos(ωt − θ) show thatR = √ a 2 +b 2
and tanθ = b a .
Solution Let
acosωt+bsinωt =Rcos(ωt−θ)
Then, using the trigonometric identity forcos(A −B), we can write
acosωt+bsinωt =Rcos(ωt−θ)
=R(cosωtcosθ +sinωtsinθ)
= (Rcosθ)cosωt+(Rsinθ)sinωt