University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2022
Views:

103

Pages:

2
Author:

Brooks Hill

Some Trigonometry and Complexification The following is a little more explanation of the trig that comes up in Chapter 4. Amplitude, Period , Phase Shift — Given the function Acos( u )t S ) , the amplitude of the function is A, the period is 2ir / u , the frequency is ui / ( 2n ) , the circular frequency is w , and the phase shift is 5 / UJ . So, for example, if we were to graph the function ( ^ 3 COS TTt + , the amplitude is 3, the period is 2, and the phase shift is to the left by 1/ 4 of the period ” . The result: —|, which would translate as “shift . j! Cosine Sum Formula You might recall the cosine sum formula: cos(rr — y ) — cos( r) cos( y ) + sin(:r) sin (y) : Which in fact is the real part of the complex number: cos( x — y ) + isin (x — y ) = = etxe ~ty — (cos(x) + i sin ( r ) ) (cos( — y ) + ism( -y ) ) : Continuing the RHS, we get: el ( x ~- y ) = (cos (a:) cos(y ) — + sin (a:) sin ( y )) + i ( sin( x ) cos( y ) sin ( y ) cos (.x ) ) The real part gives us the cosine difference formula, and the imaginary part gives us the sine difference formula. Writing a Trig Sum Using this , we see that: Rcos( u )t — S ) = ( Rcos( 6 ) ) cos(ut ) + (i? sin(5) ) sin(wf ) This implies that we can write the following sum as a single cosine function: A cos ( u>t ) + Bsin( ajt ) — Rcos( u>t — §) 1 <=> A = Rcos ( S ) B = R sin ( tf ) or , given A , B then: - B A ( the inverse tangent is the four quadrant version ) . As a shortcut to remembering this, let z — A + iB R V -42 + B2 8 = Tan and - i then R , 5 come from the magnitude and argument of 2. Side Remark: Careful with the parentheses... If we write the periodic function as Rcos( u>( t - 5 ) ) , then 5 itself is the phase shift , but then we don’t have the form for the cosine sum formula. Worked Examples 1. Rewrite the expression as Rcos ( u )t — 8 ) : cos ( v/31) + 2 sin ( v/31 ) , SOLUTION : Thinking of the complex number z and — 1 + 2i , then R = \/ l 2 + 22 = \/5 , 8 — tan * ( 2) « 1.1 rad Therefore, cos ( v/3 f ) + 2 sin ( \/31 ) = v/5 cos ( v/3 t — 1.1) Now it is easy to analyze the sum as a single periodic function- It has an amplitude of y / E , a period of 2n /\/ 3, and a phase shift of 1.1/ >/3 radians. 2 . Rewrite the expression as a single cosine: — cos(3f ) + sin (3t ) SOLUTION: Thinking of z = —1 + *, the magnitude is \/ 2 and the inverse tangent gives — 7T / 4. BUT, we should notice that our point is actually in the second quadrant , so the argument is actually 7r 7r / 4 '’,>n / 4 ( we should use the “four quadrant ” inverse tangent ) . . Therefore, the sum is equal to: \pl cos ( 3t — — The amplitude is \ /2, the period is 27T / 3, and the phase shift is 7r / 4. Exercises 1. Write the expression 3 cos(51 ) - 2 sin (5f ) in the form Rcosizit - 8 ) . — cos(2t ) + sin(2t ) in the form Rcos( uit — 5 ) . 3. Write the expression — cos( t ) 3 sin ( t ) in the form Rcos ( ut — 5 ) . 2. Write the expression - 2

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