It is really However, we shall outline the procedure in detail. First, we can, if we are clever with mathematics and know enough about cosines and sines, simply work it out. The easiest such case is the one where A I and A2 are equal, let us say they are both equal to A. In those circumstances, for example (we could call this the ric method of solving the R=A[cos(at+φ1)+cos(at+φ) Once, in our trigonometry class, we may have learned the rule that cos A cos B= 2 cos(A B)cos I(A-B) (29.10) If we know that, then we can immediately write R as R=2Acos(中1一中)cos(at+妙1+她2) (29.11) So we find that we have an oscillatory wave with a new phase and a new amplitude In general, the result will be an oscillatory wave with a new amplitude A R, which we may call the resultant amplitude, oscillating at the same frequency but with phase difference R, called the resultant phase. In view of this, our particular case has the following result: that the resultant amplitude is AR=2Acos如(中1-中2) (29.12) and the resultant phase is the average of the two phases, and we have completely solved our problem Now suppose that we cannot remember that the sum of two cosines is twice the cosine of half the sum times the cosine of half the difference. Then we may use another method of analysis which is more geometrical. Any cosine function of at can be considered as the horizontal projection of a rotating vector. Suppose there were a vector A 1 of length A 1 rotating with time, so that its angle with the horizontal axis is wt pI. ( We shall leave out the at in a minute and see that it makes no difference. Suppose that we take a snapshot at the time t=0, although in fact, the picture is rotating with angular velocity w( Fig. 29-9). The projection Fig. 29-9. a geometrical method of A 1 along the horizontal axis is precisely A1 cos(ot p1). Now at t=0 the combining two cosine waves. The entire second wave could be represented by another vector, A2, of length A2 and at an diagram is thought of as rotating counter- angle p2, and also rotating. They are both rotating with the same angular velocity clockwise with angular frequency w w, and therefore the relative positions of the two are fixed. The system goes around like a rigid body. The horizontal projection of a2 is A2 cos(t + 2). But we know from the theory of vectors that if we add the two vectors in the ordinary way, by the parallelogram rule, and draw the resultant vector AR, the x-componen of the resultant is the sum of the x-components of the other two vectors. That solves our problem. It is easy to check that this gives the correct result for the special case we treated above, where A,= A2= A. In this case we see 29-9 that AR lies midway between A1 and A 2 and makes an angle 2(o with each. Therefore we see that AR 2A cos 2( 2-1), as before. Al an of A1 and A2 when the two amplitudes are equal. Clearly, we can also solve for the case where the amplitudes are not equal, just as easily. We can call that the geometrical way of solving the proble There is still another way of solving the problem, and that is the analytical way. That is, instead of having actually to draw a picture like Fig. 29-9, we can write something down which says the same thing as the picture: instead of drawing the vectors, we write a complex number to represent each of the vectors. The real parts of the complex numbers are the actual physical quantities. So in our par ticular case the waves could be written in this way: Ale at+op[the real part of this is A, cos(ot u] and A2e of+ea. Now we can add the two R