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Making Phase-Angle Calculations Easier Nicholas R. Kirchner February 28, 2012 1 Getting Started When we have a function whose graph looks precisely like a sine wave, it is useful to express it in the form ) = sin( ωt ) or ) = cos( ωt ) so that certain operations become simpler. (Note: we choose a “+ ” in the sine and ” in the cosine solely because of the formats of the sum formulas for sine and cosine.) It happens that a function of the form ) = cos ωt sin ωt is a sine wave. 2 Pattern Matching We would like to write ) = cos ωt sin ωt as sin( ωt The sum formula for sin reads sin( ) = sin cos + cos sin Applying this formula yields sin( ωt ) = sin cos ωt + cos sin ωt Noting the similarity of the right hand side with ), we would be done if we could ﬁnd a such that sin and cos . However, this is only possible if = 1, since sin + cos = 1 is a well-known identity. Therefore, we let so that we can factor this out: ) = cos ωt sin ωt cos ωt sin ωt

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Now the coeﬃcients in front of the sine and cosine do satisfy the required identity: = 1 Thus, we want sin and cos , and once we ﬁnd this , we’re done! What about the cosine case? If we want to rewrite ) in the form cos( ωt ), take advantage of the identity cos( ) = cos cos + sin sin to get cos( ωt ) = cos cos ωt + sin sin ωt The only change from the sine case is that we want cos and sin In other words, reverse the roles of and 3 Getting the Angle Right We’re still trying to write ) = cos ωt sin ωt as sin( ωt We have seen that and are fairly easy to come by, and must satisfy sin cos Dividing these two equations yields tan (and notice in particular how this equation does not need ). It is very tempting to take an arctangent here, but that will only be correct half the time. The problem is that the tangent has a period of , which means there are two positons on the unit circle which can give the desired tangent. For example, tan π/ 4 = 1 and tan 5 π/ 4 = 1. The arctangent will work ﬁne if our angle is in quadrants 1 or 4, since the range of the arctangent function is π/ 4 to π/ 4. If we are in quadrants 2 or 3, we will add to the value of the arctangent.

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Notice that in quadrants 1 and 4, the -coordinate (cosine) is positive. Since cos we are in quadrants 1 or 4 precisely when 0. So, if 0, take to be the arctangent. If 0, take to be the arctangent plus Hello? Talk about cosine too? For the cos( ωt ) case, we have sin cos Dividing, tan This time, take to be the arctangent if 0. If 0, take to be the arctangent plus Yup, all we do is reverse the roles that and played in the sine case. Note for programmers: Many languages have two arctangent functions. For example, C has atan and atan2 atan takes one argument and computes its arctangent. atan2 takes two arguments: atan2(y,x) and computes arctan . Why bother having this atan2 function? Because the atan2 function can determine in which quadrant the point (x,y) lies and return an angle from that quadrant. The atan function can only output angles in quadrants 1 and 4. 4 Formulas and A Calculator Program The moment you’ve all been waiting for: the easy to use formulas. We can write ) = cos ωt sin ωt sin( ωt where arctan + arctan What if = 0 If is zero then we are on the -axis, so our angle is either π/ 2 or 3 π/ 2. You can tell which by the sign of

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The Cosines are feeling left out! To write ) = cos ωt sin ωt cos( ωt use the same formulas as above for and , but change arctan + arctan The program This program should work on a TI-83 to handle the sine case. Writing a program for the cosine case is an exercise for the reader. Prompt B Prompt C Disp "A=" Disp sqrt(B^2+C^2) Disp "D=" if C > 0 Disp atan(B/C) if C < 0 Disp atan(B/C) + pi if C == 0 and B > 0 Disp pi/2 if C == 0 and B < 0 Disp 3*pi/2 5 Example and Checking Your Answer Example Consider ) = 3 cos 2 4 sin 2 To write it in the form sin( ωt ) we take + ( 4) = 5 = 2 and for we note that we require the -coordinate to be negative (since cos ). In fact, this angle must be in quadrant 2. Thus, + arctan 46685 Therefore, ) = 5 sin(2 + 2 46685) The way to check this answer is to plot both this result and the original on the same axes. If the graphs coincide, the answer is correct.

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Now, let’s rewrite in the form cos( ωt ). We want = 5 and = 2 as before. This time, we require to be in the fourth quadrant (since cos and sin Thus = arctan 92730 Note: We are going to want to be positive on homework and exams. If you compute a negative , which will happen precisely when is in the fourth quadrant, add 2 So, we really want 92730 + 2 35589 and the answer is ) = 5 cos(2 35589) Checking: To convince that the results are correct, the graphs of each of our three expressions follow. If these were graphed on the same set of axes, it would be impossible to tell them apart. -5 -4 -3 -2 -1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 Graph of y(t) = 3 cos 2t - 4 sin 2t 6 A Final Thought Converting cos ωt sin ωt to sin( ωt ) is precisely the process of con- verting the rectangular point ( x,y ) = ( ,C ) into polar form. The and thus obtained become and , respectively, in the converted form. Converting to cos( ωt ) is the same thing, but the rectangular point is x,y ) = ( ,C ).

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-5 -4 -3 -2 -1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 Graph of y(t) = 5 sin(2t + 2.46685) -5 -4 -3 -2 -1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 Graph of y(t) = 5 cos(2t - 5.35589)

Kirchner February 28 2012 1 Getting Started When we have a function whose graph looks precisely like a sine wave it is useful to express it in the form sin 969t or cos 969t so that certain operations become simpler Note we choose a in t ID: 22906

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Page 1

Making Phase-Angle Calculations Easier Nicholas R. Kirchner February 28, 2012 1 Getting Started When we have a function whose graph looks precisely like a sine wave, it is useful to express it in the form ) = sin( ωt ) or ) = cos( ωt ) so that certain operations become simpler. (Note: we choose a “+ ” in the sine and ” in the cosine solely because of the formats of the sum formulas for sine and cosine.) It happens that a function of the form ) = cos ωt sin ωt is a sine wave. 2 Pattern Matching We would like to write ) = cos ωt sin ωt as sin( ωt The sum formula for sin reads sin( ) = sin cos + cos sin Applying this formula yields sin( ωt ) = sin cos ωt + cos sin ωt Noting the similarity of the right hand side with ), we would be done if we could ﬁnd a such that sin and cos . However, this is only possible if = 1, since sin + cos = 1 is a well-known identity. Therefore, we let so that we can factor this out: ) = cos ωt sin ωt cos ωt sin ωt

Page 2

Now the coeﬃcients in front of the sine and cosine do satisfy the required identity: = 1 Thus, we want sin and cos , and once we ﬁnd this , we’re done! What about the cosine case? If we want to rewrite ) in the form cos( ωt ), take advantage of the identity cos( ) = cos cos + sin sin to get cos( ωt ) = cos cos ωt + sin sin ωt The only change from the sine case is that we want cos and sin In other words, reverse the roles of and 3 Getting the Angle Right We’re still trying to write ) = cos ωt sin ωt as sin( ωt We have seen that and are fairly easy to come by, and must satisfy sin cos Dividing these two equations yields tan (and notice in particular how this equation does not need ). It is very tempting to take an arctangent here, but that will only be correct half the time. The problem is that the tangent has a period of , which means there are two positons on the unit circle which can give the desired tangent. For example, tan π/ 4 = 1 and tan 5 π/ 4 = 1. The arctangent will work ﬁne if our angle is in quadrants 1 or 4, since the range of the arctangent function is π/ 4 to π/ 4. If we are in quadrants 2 or 3, we will add to the value of the arctangent.

Page 3

Notice that in quadrants 1 and 4, the -coordinate (cosine) is positive. Since cos we are in quadrants 1 or 4 precisely when 0. So, if 0, take to be the arctangent. If 0, take to be the arctangent plus Hello? Talk about cosine too? For the cos( ωt ) case, we have sin cos Dividing, tan This time, take to be the arctangent if 0. If 0, take to be the arctangent plus Yup, all we do is reverse the roles that and played in the sine case. Note for programmers: Many languages have two arctangent functions. For example, C has atan and atan2 atan takes one argument and computes its arctangent. atan2 takes two arguments: atan2(y,x) and computes arctan . Why bother having this atan2 function? Because the atan2 function can determine in which quadrant the point (x,y) lies and return an angle from that quadrant. The atan function can only output angles in quadrants 1 and 4. 4 Formulas and A Calculator Program The moment you’ve all been waiting for: the easy to use formulas. We can write ) = cos ωt sin ωt sin( ωt where arctan + arctan What if = 0 If is zero then we are on the -axis, so our angle is either π/ 2 or 3 π/ 2. You can tell which by the sign of

Page 4

The Cosines are feeling left out! To write ) = cos ωt sin ωt cos( ωt use the same formulas as above for and , but change arctan + arctan The program This program should work on a TI-83 to handle the sine case. Writing a program for the cosine case is an exercise for the reader. Prompt B Prompt C Disp "A=" Disp sqrt(B^2+C^2) Disp "D=" if C > 0 Disp atan(B/C) if C < 0 Disp atan(B/C) + pi if C == 0 and B > 0 Disp pi/2 if C == 0 and B < 0 Disp 3*pi/2 5 Example and Checking Your Answer Example Consider ) = 3 cos 2 4 sin 2 To write it in the form sin( ωt ) we take + ( 4) = 5 = 2 and for we note that we require the -coordinate to be negative (since cos ). In fact, this angle must be in quadrant 2. Thus, + arctan 46685 Therefore, ) = 5 sin(2 + 2 46685) The way to check this answer is to plot both this result and the original on the same axes. If the graphs coincide, the answer is correct.

Page 5

Now, let’s rewrite in the form cos( ωt ). We want = 5 and = 2 as before. This time, we require to be in the fourth quadrant (since cos and sin Thus = arctan 92730 Note: We are going to want to be positive on homework and exams. If you compute a negative , which will happen precisely when is in the fourth quadrant, add 2 So, we really want 92730 + 2 35589 and the answer is ) = 5 cos(2 35589) Checking: To convince that the results are correct, the graphs of each of our three expressions follow. If these were graphed on the same set of axes, it would be impossible to tell them apart. -5 -4 -3 -2 -1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 Graph of y(t) = 3 cos 2t - 4 sin 2t 6 A Final Thought Converting cos ωt sin ωt to sin( ωt ) is precisely the process of con- verting the rectangular point ( x,y ) = ( ,C ) into polar form. The and thus obtained become and , respectively, in the converted form. Converting to cos( ωt ) is the same thing, but the rectangular point is x,y ) = ( ,C ).

Page 6

-5 -4 -3 -2 -1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 Graph of y(t) = 5 sin(2t + 2.46685) -5 -4 -3 -2 -1 0 1 2 3 4 5 -3 -2 -1 0 1 2 3 Graph of y(t) = 5 cos(2t - 5.35589)

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