If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and...
$begingroup$
If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?
I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?
calculus algebra-precalculus trigonometry
edited Dec 14 '18 at 12:19
amWhy
192k28225439
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
$endgroup$
add a comment |
$begingroup$
If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?
I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?
calculus algebra-precalculus trigonometry
edited Dec 14 '18 at 12:19
amWhy
192k28225439
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
$endgroup$
$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17
$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19
add a comment |
$begingroup$
If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?
I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?
calculus algebra-precalculus trigonometry
edited Dec 14 '18 at 12:19
amWhy
192k28225439
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
$endgroup$
If $cosbeta = −cos alpha$ and $sin beta =−sin alpha $, what must be the relation between $alpha$ and $beta$?
I know that then $alpha = beta + k pi$, but I cannot understand why. Would anyone explain this for me please?
calculus algebra-precalculus trigonometry
calculus algebra-precalculus trigonometry
edited Dec 14 '18 at 12:19
amWhy
192k28225439
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
edited Dec 14 '18 at 12:19
amWhy
192k28225439
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
edited Dec 14 '18 at 12:19
amWhy
192k28225439
edited Dec 14 '18 at 12:19
amWhy
192k28225439
edited Dec 14 '18 at 12:19
amWhy
192k28225439
192k28225439
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
asked Dec 14 '18 at 12:14
hopefullyhopefully
166112
166112
$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17
$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19
add a comment |
$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17
$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19
$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17
$begingroup$
I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17
$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19
$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Use Prosthaphaeresis Formulas,
$$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
$$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
What if $cosdfrac{alpha-beta}2$$ne0$
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
1
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
1
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
1
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
|
show 1 more comment
$begingroup$
$(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.
The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.
This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
$endgroup$
add a comment |
$begingroup$
Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
$$
sinalphacosbeta-cosalphasinbeta=0
$$
that is, $sin(alpha-beta)=0$. This entails
$$
alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
$$
($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.
For the second set of solutions, we have
begin{align}
cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
sinalpha&=sin(beta+pi+2kpi)=-sinbeta
end{align}
for every $beta$. So the conclusion is that your conditions are equivalent to
$$
alpha=beta+pi+2kpi qquadtext{($k$ integer)}
$$
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
By the definition we have
$$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$
$$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$
then we have
$beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible
$beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$
and therefore finally we have
- $alpha =hpi$
- $beta=pi-hpi+2kpi=(1-h)pi+2kpi$
or
- $alpha+beta=pi + 2kpi$
answered Dec 14 '18 at 13:18
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
$$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$
$impliesalpha=npi+beta$ where $n$ is any integer
Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use Prosthaphaeresis Formulas,
$$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
$$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
What if $cosdfrac{alpha-beta}2$$ne0$
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
1
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
1
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
1
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
|
show 1 more comment
$begingroup$
Use Prosthaphaeresis Formulas,
$$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
$$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
What if $cosdfrac{alpha-beta}2$$ne0$
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
1
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
1
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
1
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
|
show 1 more comment
$begingroup$
Use Prosthaphaeresis Formulas,
$$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
$$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
What if $cosdfrac{alpha-beta}2$$ne0$
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
Use Prosthaphaeresis Formulas,
$$0=cosalpha+cosbeta=2cosdfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
$$0=sinalpha+sinbeta=2sindfrac{alpha+beta}2cosdfrac{alpha-beta}2$$
What if $cosdfrac{alpha-beta}2$$ne0$
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 12:19
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
1
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
1
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
1
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
|
show 1 more comment
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
1
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
1
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
1
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
$begingroup$
Could not understand you in case of $cos frac {alpha - beta}{2} ne 0$ ..... could you provide more details please?
$endgroup$
– hopefully
Dec 14 '18 at 12:28
1
1
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
$begingroup$
cosine and sine functions never equal 0 at the same time
$endgroup$
– hopefully
Dec 14 '18 at 12:29
1
1
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, If $cosdfrac{alpha-beta}2$$ne0,$ we have $$cosdfrac{alpha+beta}2=sindfrac{alpha+beta}2=0$$ But $cos^2dfrac{alpha+beta}2+sin^2dfrac{alpha+beta}2=?$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:29
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
$begingroup$
@hopefully, We must have $cosdfrac{alpha-beta}2=0$
$endgroup$
– lab bhattacharjee
Dec 14 '18 at 12:30
1
1
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
$begingroup$
equals one, so we have a contradiction ...... correct?
$endgroup$
– hopefully
Dec 14 '18 at 12:31
|
show 1 more comment
$begingroup$
$(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.
The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.
This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
$endgroup$
add a comment |
$begingroup$
$(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.
The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.
This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
$endgroup$
add a comment |
$begingroup$
$(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.
The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.
This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
$endgroup$
$(cos beta,sin beta)$ are coordinates of a point $M$ on unit circle.
The point $P(cos alpha, sin alpha)$ is symmetrical to $M$ through $0$.
This gives $$alpha=beta + (2k+1)pi, kin mathbb{Z}.$$
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
edited Dec 14 '18 at 13:27
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
answered Dec 14 '18 at 12:20
user376343user376343
3,2982825
3,2982825
add a comment |
add a comment |
$begingroup$
Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
$$
sinalphacosbeta-cosalphasinbeta=0
$$
that is, $sin(alpha-beta)=0$. This entails
$$
alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
$$
($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.
For the second set of solutions, we have
begin{align}
cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
sinalpha&=sin(beta+pi+2kpi)=-sinbeta
end{align}
for every $beta$. So the conclusion is that your conditions are equivalent to
$$
alpha=beta+pi+2kpi qquadtext{($k$ integer)}
$$
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
$$
sinalphacosbeta-cosalphasinbeta=0
$$
that is, $sin(alpha-beta)=0$. This entails
$$
alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
$$
($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.
For the second set of solutions, we have
begin{align}
cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
sinalpha&=sin(beta+pi+2kpi)=-sinbeta
end{align}
for every $beta$. So the conclusion is that your conditions are equivalent to
$$
alpha=beta+pi+2kpi qquadtext{($k$ integer)}
$$
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
$endgroup$
add a comment |
$begingroup$
Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
$$
sinalphacosbeta-cosalphasinbeta=0
$$
that is, $sin(alpha-beta)=0$. This entails
$$
alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
$$
($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.
For the second set of solutions, we have
begin{align}
cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
sinalpha&=sin(beta+pi+2kpi)=-sinbeta
end{align}
for every $beta$. So the conclusion is that your conditions are equivalent to
$$
alpha=beta+pi+2kpi qquadtext{($k$ integer)}
$$
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
$endgroup$
Multiply the first equation by $sinbeta$ and the second equation by $cosbeta$ and subtract:
$$
sinalphacosbeta-cosalphasinbeta=0
$$
that is, $sin(alpha-beta)=0$. This entails
$$
alpha=beta+2kpi qquadtext{or}qquad alpha=beta+pi+2kpi
$$
($k$ an integer). Now let's test the first set of solutions: if $alpha=beta+2kpi$, then $cosalpha=cosbeta$ and $sinalpha=sinbeta$: it's impossible that $cosbeta=-cosbeta$ and $sinbeta=-sinbeta$.
For the second set of solutions, we have
begin{align}
cosalpha&=cos(beta+pi+2kpi)=-cosbeta \
sinalpha&=sin(beta+pi+2kpi)=-sinbeta
end{align}
for every $beta$. So the conclusion is that your conditions are equivalent to
$$
alpha=beta+pi+2kpi qquadtext{($k$ integer)}
$$
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
answered Dec 14 '18 at 13:30
egregegreg
180k1485202
180k1485202
add a comment |
add a comment |
$begingroup$
By the definition we have
$$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$
$$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$
then we have
$beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible
$beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$
and therefore finally we have
- $alpha =hpi$
- $beta=pi-hpi+2kpi=(1-h)pi+2kpi$
or
- $alpha+beta=pi + 2kpi$
answered Dec 14 '18 at 13:18
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
By the definition we have
$$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$
$$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$
then we have
$beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible
$beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$
and therefore finally we have
- $alpha =hpi$
- $beta=pi-hpi+2kpi=(1-h)pi+2kpi$
or
- $alpha+beta=pi + 2kpi$
answered Dec 14 '18 at 13:18
gimusigimusi
1
$endgroup$
add a comment |
$begingroup$
By the definition we have
$$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$
$$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$
then we have
$beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible
$beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$
and therefore finally we have
- $alpha =hpi$
- $beta=pi-hpi+2kpi=(1-h)pi+2kpi$
or
- $alpha+beta=pi + 2kpi$
answered Dec 14 '18 at 13:18
gimusigimusi
1
$endgroup$
By the definition we have
$$cosbeta = −cos alpha=cos (pi-alpha) iff beta=pi-alpha+2k_1pi quad lor quad -(pi-alpha)+2k_2pi$$
$$sinbeta = −sin alpha=sin (-alpha) iff beta=-alpha+2k_3pi quad lor quad pi-(-alpha)+2k_4pi$$
then we have
$beta=pi-alpha+2k_1pi=-alpha+2k_3piiff pi =2hpi$ which is impossible
$beta=pi-alpha+2k_1pi=pi-(-alpha)+2k_4piiff 2alpha =2hpi iff alpha =hpi$
and therefore finally we have
- $alpha =hpi$
- $beta=pi-hpi+2kpi=(1-h)pi+2kpi$
or
- $alpha+beta=pi + 2kpi$
answered Dec 14 '18 at 13:18
gimusigimusi
1
edited Dec 14 '18 at 13:32
answered Dec 14 '18 at 13:18
gimusigimusi
1
answered Dec 14 '18 at 13:18
gimusigimusi
1
answered Dec 14 '18 at 13:18
gimusigimusi
1
1
add a comment |
add a comment |
$begingroup$
$$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$
$impliesalpha=npi+beta$ where $n$ is any integer
Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$
$impliesalpha=npi+beta$ where $n$ is any integer
Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
add a comment |
$begingroup$
$$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$
$impliesalpha=npi+beta$ where $n$ is any integer
Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
$endgroup$
$$tanalpha=dfrac{-sinbeta}{-cosbeta}=?$$
$impliesalpha=npi+beta$ where $n$ is any integer
Now check the given condition for even $=2m$(say) and odd $=2m+1$(say) values of $n$
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
answered Dec 14 '18 at 17:46
lab bhattacharjeelab bhattacharjee
224k15156274
224k15156274
add a comment |
add a comment |
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I would recommend solving the two equations separately, and then considering in which cases both of the equations are fulfilled at the same time.
$endgroup$
– Matti P.
Dec 14 '18 at 12:17
$begingroup$
The answer $alpha = beta + k pi$ is incomplete. You have to say what kind of number $k$ is. Usually $k$ is any integer, in which case the answer would be wrong. (Consider $k = 2$ or even $k = 0$). However, $k = 1,$ $k = 3,$ and so forth are good values for $k$ in this formula.
$endgroup$
– David K
Dec 14 '18 at 12:19