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Fig. 11-A
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Now, with the above facts in mind, let x and y be two adjacent angles, as in Fig. 13-A.
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Fig. 13-A
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Now consider Fig. 14-A, in which it is given that lines CA and DB are perpendicular to line OB, line ED is perpendicular to line CA, line CD is perpendicular to line OD. In Fig. 14-A, consider the two triangles OBD and CED. Close inspection will show that the corresponding sides of these two triangles are perpendicular and thus that they are similar triangles, with angle x as shown. Hence, from direct inspection of Fig. 14-A we have that cos x y OA OB AB OB ED since AB ED OC OC OC OC
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Then, since OB OD cos x and ED CD sin x, we have cos x y OD CD cos x sin x OC OC
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But note that OD=OC cos y and CD=OC sin y; thus the preceding equation can be written in the standard form cos x y cos x cos y sin x sin y which, for the special case of y x, becomes cos 2x cos2 x sin2 x Now make use of the identity sin2 x cos2 x 1 (from problem 64). Doing this, then writing  in place of x, gives the required eq. (98).
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From Fig. 14-A: sin x y
Identity for sin(x y)
AC CE AE CE BD since BD AE OC OC OC OC
Then, since, CE CD cos x and BD OD sin x, we have sin x y CD OD cos x sin x OC OC
Appendix
But note that CD=OC sin y and OD=OC cos y; thus the preceding equation can be written in the standard form sin x y sin x cos y cos x sin y
Note 8.
Often-Used Greek Letters
The following are Greek letters most often used in engineering work. a alpha b beta BAY tah  epsilon d delta small delta capital o omega small  omega capital y theta THAY tah m mu p pi f phi fee
Note 9.
Sinusoidal Waves of the Same Frequency
Applying the identity found in note 7 to the left-hand side of eq. (124) gives the equality A sin !t B sin !t a A B cos a sin !t B sin a cos !t Or, letting E and F denote the CONSTANT values A B cos a and B sin a, the above becomes A sin !t B sin !t a E sin !t F cos !t which can also be written in the form p E F A sin !t B sin !t a E 2 F 2 p sin !t p cos !t 2 F2 2 F2 E E ! 1-A
Now, letting  be a constant angle, construct the right triangle shown in Fig. 15-A.
Fig. 15-A
Appendix
E F Note that p cos  and p sin , and thus eq. (1-A) becomes E2 F 2 E2 F 2 p A sin !t B sin !t a E 2 F 2 sin !t cos  cos !t sin  Now apply the identity found in note 7 to the right-hand side above (setting x !t and y  to get the nal result that p 2-A A sin !t B sin !t a E 2 F 2 sin !t  thus proving that the sum of two sinusoidal waves of the same frequency is equivalent to a single sinusoidal wave of the same frequency.
Note 10.
Sinusoidal Waves as Vectors
Let A and B denote the PEAK VALUES of two sine waves of the same frequency, with the rst wave lagging the second wave by a degrees. In deriving eq. (2-A) above for the same two waves we algebraically showed that p peak value of the SUM of the two waves E 2 F 2 p A2 B2 2AB cos a (making use of the identity sin2 a cos2 a 1), and also, from inspection of Fig. 15-A, that PHASE ANGLE of resultant wave  arctan F B sin a arctan E A B cos a
We now wish to show that the same results can be obtained graphically by using the phasor representation of sine waves, and by assuming that phasors can be treated as vector quantities; that is, that phasors ADD together in accordance with the PARALLELOGRAM LAW of addition of vectors. To do this, let us begin with a phasor diagram of the two waves, such as shown in Fig. 16-A. Now let R be the VECTOR sum of A and B, as shown in Fig. 17-A. Then, upon applying the Pythagorean theorem to the large right triangle, we have R2 A B cos a 2 B2 sin2 a hence, after applying the identity sin2 a cos2 a 1, we have that the MAGNITUDE of the resultant vector R is p R A2 B2 2AB cos a
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