Writing linear combination of exponentials as cosine
Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.
So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$
Am I on the right track? I can't seem to figure out where to go next.
differential-equations complex-numbers
add a comment |
Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.
So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$
Am I on the right track? I can't seem to figure out where to go next.
differential-equations complex-numbers
Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58
The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19
add a comment |
Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.
So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$
Am I on the right track? I can't seem to figure out where to go next.
differential-equations complex-numbers
Show that $$y = A_1e^{ix} + A_2e^{-ix}$$
can be written as $$y = Acos(x - delta)$$ where A and $delta$ are real.
So far I have done the following:
$$y = (A+B)cos(x) + (A-B) isin(x)$$
Am I on the right track? I can't seem to figure out where to go next.
differential-equations complex-numbers
differential-equations complex-numbers
asked Dec 5 '18 at 2:51
51n84d51n84d
383
383
Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58
The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19
add a comment |
Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58
The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19
Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58
Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58
The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19
The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19
add a comment |
1 Answer
1
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oldest
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Hint
Since
$$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
$$A_1+A_2=A cos (delta )$$
$$i(A_1-A_2)=A sin (delta )$$
Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint
Since
$$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
$$A_1+A_2=A cos (delta )$$
$$i(A_1-A_2)=A sin (delta )$$
Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.
add a comment |
Hint
Since
$$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
$$A_1+A_2=A cos (delta )$$
$$i(A_1-A_2)=A sin (delta )$$
Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.
add a comment |
Hint
Since
$$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
$$A_1+A_2=A cos (delta )$$
$$i(A_1-A_2)=A sin (delta )$$
Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.
Hint
Since
$$ A_1e^{ix} + A_2e^{-ix}= Acos(x - delta)$$
must hold for any $x$, write it for $x=0$ and $x=frac pi 2$. This gives
$$A_1+A_2=A cos (delta )$$
$$i(A_1-A_2)=A sin (delta )$$
Start squaring both equations and add them; this could give $A$. Make the ratio to get $tan(delta )$.
answered Dec 5 '18 at 3:54
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
add a comment |
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Should the coefficients be conjugate? Otherwise $y$ might not be real.
– xbh
Dec 5 '18 at 2:58
The question didn't specify that the coefficients are conjugate...
– 51n84d
Dec 5 '18 at 3:19