Question 9.5: If y1= 20 cos (ω t -30^°) and y2= 40 cos (ω t +60^°)express ...

a) First we expand both y_{1} and y_{2} using the cosine of the sum of two angles, to get

y_{1}= 20 \cos \omega t \cos 30^{\circ}+20 \sin \omega t \sin 30^{\circ};

Adding y_{1}andy_{2}, we obtain

=37.32 \cos \omega t-24.64 \sin \omega t .

To combine these two terms we treat the co-efficients of the cosine and sine as sides of a right triangle (Fig. 9.6) and then multiply and divide the right-hand side by the hypotenuse. Our expression for y becomes

= 44.72\left( cos 33.43^{\circ} \cos \omega t-sin33.43^{\circ}\sin \omega t\right).

Again, we invoke the identity involving the cosine of the sum of two angles and write

y= 44.72\cos \left(\omega t+33.43^{\circ}\right).

b) We can solve the problem by using phasors as follows: Because

y=y_{1}+y_{2},

then, from Eq. 9.24 V=V_{1}+V_{2}+…+V_{n},

\pmb{Y}=\pmb{Y}_{1}+\pmb{Y}_{2},

= 44.72\angle 33.43^{\circ}.

Once we know the phasor Y, we can write the corresponding trigonometric function for y by taking the inverse phasor transform:

= 44.72\cos \left(\omega t+33.43^{\circ}\right).

The superiority of the phasor approach for adding sinusoidal functions should be apparent. Note that it requires the ability to move back and forth between the polar and rectangular forms of complex numbers.