If $z=costheta + isin theta$ prove $frac{z^2-1}{z^2+1}=itantheta$
$begingroup$
If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$
I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.
trigonometry complex-numbers
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
$endgroup$
add a comment |
$begingroup$
If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$
I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.
trigonometry complex-numbers
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
$endgroup$
$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58
add a comment |
$begingroup$
If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$
I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.
trigonometry complex-numbers
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
$endgroup$
If $z=costheta + isin theta$ prove $$frac{z^2-1}{z^2+1}=itantheta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
$$frac{(cos^2theta + 2isin theta costheta - sin^2theta)-1}{(cos^2theta + 2isin theta costheta - sin^2theta)+1}$$
$$frac{(cos^2theta - sin^2theta)+( 2isin theta costheta) -1}{(cos^2theta - sin^2theta)+( 2isin theta costheta)+1}$$
$$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}$$
I understand how I can do it with using $z=e^{i theta}$, however I want to solve it using double angle identities.
trigonometry complex-numbers
trigonometry complex-numbers
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
edited Jan 14 at 21:48
H.Linkhorn
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
asked Jan 14 at 21:29
H.LinkhornH.Linkhorn
383113
383113
$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58
add a comment |
$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58
$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58
$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
$endgroup$
add a comment |
$begingroup$
$z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$
answered Jan 14 at 21:38
MahranMahran
634
$endgroup$
add a comment |
$begingroup$
$cos theta +isin theta =e ^{i theta}$. Then
$$begin{aligned}
frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
&=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
&=frac{2 i sin theta}{2 cos theta}\
&=itantheta
end{aligned}$$
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
$endgroup$
add a comment |
$begingroup$
If you divide the numerator and denominator of
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain
$$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$
which simplifies to
$$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$
Then factor the numerator to get
$$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$
Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
$endgroup$
add a comment |
$begingroup$
Now where you've left off,
use $cos2y=2cos^2y-1=1-2sin^2y (1)$
to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
$endgroup$
add a comment |
$begingroup$
Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
$endgroup$
add a comment |
$begingroup$
Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
$endgroup$
Your approach will work with double angle formulae, but this is quicker: since $z=exp itheta$, $frac{z-1/z}{z+1/z}=frac{2isintheta}{2costheta}$.
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
answered Jan 14 at 21:33
J.G.J.G.
25.5k22539
25.5k22539
add a comment |
add a comment |
$begingroup$
$z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$
answered Jan 14 at 21:38
MahranMahran
634
$endgroup$
add a comment |
$begingroup$
$z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$
answered Jan 14 at 21:38
MahranMahran
634
$endgroup$
add a comment |
$begingroup$
$z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$
answered Jan 14 at 21:38
MahranMahran
634
$endgroup$
$z=e^{itheta}$, $frac{z^2-1}{z^2+1}=frac{e^{2 itheta}-1}{e^{2itheta}+1}=frac{e^{ itheta}-e^{ -itheta}}{e^{itheta}+e^{ itheta}}=frac{2 i, sin(theta)}{2,cos(theta)}=itan(theta)$
answered Jan 14 at 21:38
MahranMahran
634
answered Jan 14 at 21:38
MahranMahran
634
answered Jan 14 at 21:38
MahranMahran
634
answered Jan 14 at 21:38
MahranMahran
634
634
add a comment |
add a comment |
$begingroup$
$cos theta +isin theta =e ^{i theta}$. Then
$$begin{aligned}
frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
&=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
&=frac{2 i sin theta}{2 cos theta}\
&=itantheta
end{aligned}$$
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
$endgroup$
add a comment |
$begingroup$
$cos theta +isin theta =e ^{i theta}$. Then
$$begin{aligned}
frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
&=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
&=frac{2 i sin theta}{2 cos theta}\
&=itantheta
end{aligned}$$
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
$endgroup$
add a comment |
$begingroup$
$cos theta +isin theta =e ^{i theta}$. Then
$$begin{aligned}
frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
&=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
&=frac{2 i sin theta}{2 cos theta}\
&=itantheta
end{aligned}$$
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
$endgroup$
$cos theta +isin theta =e ^{i theta}$. Then
$$begin{aligned}
frac{z^2-1}{z^2+1}&=frac{z-1/z}{z+1/z}\
&=frac{e^{itheta}-e^{-itheta}}{e^{itheta}+e^{-itheta}}\
&=frac{2 i sin theta}{2 cos theta}\
&=itantheta
end{aligned}$$
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
answered Jan 14 at 21:37
mathcounterexamples.netmathcounterexamples.net
26.5k22157
26.5k22157
add a comment |
add a comment |
$begingroup$
If you divide the numerator and denominator of
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain
$$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$
which simplifies to
$$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$
Then factor the numerator to get
$$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$
Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
$endgroup$
add a comment |
$begingroup$
If you divide the numerator and denominator of
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain
$$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$
which simplifies to
$$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$
Then factor the numerator to get
$$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$
Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
$endgroup$
add a comment |
$begingroup$
If you divide the numerator and denominator of
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain
$$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$
which simplifies to
$$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$
Then factor the numerator to get
$$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$
Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
$endgroup$
If you divide the numerator and denominator of
$$frac{(costheta + isin theta)^2-1}{(costheta + isin theta)^2+1}$$
by $cos^2theta$ and then use the identity $sec^2theta=1+tan^2$ on the result you obtain
$$ frac{(1+i,tantheta)^2-1-tan^2theta}{(1+i,tantheta)^2+1+tan^2theta} $$
which simplifies to
$$ frac{i,tantheta-tan^2theta}{1+i,tantheta} $$
Then factor the numerator to get
$$ frac{i,tantheta(1+i,tantheta)}{1+i,tantheta}=i,tantheta $$
Note: This is an approach which can be taken if the student does not yet know that $e^{i,theta}=costheta+i,sintheta$
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
answered Jan 14 at 22:07
John Wayland BalesJohn Wayland Bales
14k21238
14k21238
add a comment |
add a comment |
$begingroup$
Now where you've left off,
use $cos2y=2cos^2y-1=1-2sin^2y (1)$
to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Now where you've left off,
use $cos2y=2cos^2y-1=1-2sin^2y (1)$
to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Now where you've left off,
use $cos2y=2cos^2y-1=1-2sin^2y (1)$
to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
Now where you've left off,
use $cos2y=2cos^2y-1=1-2sin^2y (1)$
to find $$frac{cos2theta + isin 2theta -1}{cos2theta + isin 2theta +1}=dfrac{-2sin^2theta+ isin 2theta}{2cos^2theta+ isin 2theta}=dfrac{2isintheta(costheta+isintheta)}{2costheta(costheta+isintheta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(costheta+isintheta)^n=cos ntheta+isin ntheta $$
use $(1)$ to establish $$dfrac{z^n-1}{z^n+1}=2itandfrac{ntheta}2$$
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 15 at 4:25
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
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$begingroup$
Just using trig identities multiply top and bottom by $(cos theta -isintheta).$ You could also use $(cos 2theta -isin2 theta + 1)$
$endgroup$
– Doug M
Jan 14 at 21:58