Prove $cot^{-1}left(frac{sqrt{1+sin x}+sqrt{1-sin x}}{sqrt{1+sin x}-sqrt{1-sin x}}right)=frac x2$
I have the following question:
Prove that: $$ cot^{-1}Biggl(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}Biggl) = frac x2, x in biggl(0, frac pi4biggl) $$
The solution:
My question is about the second step (i.e., highlighted). We know that it has came from the following resolution: $$ sqrt{1pm sin x} = sqrt{sin^2frac x 2 + cos^2frac x 2 pm2sinfrac x 2cosfrac x 2} = pm biggl( cosfrac x 2 pm sinfrac x 2 biggl) $$
I've included the $pm$ symbol in accordance with the standard definition of square root. The above solution uses the positive resolution (i.e., $ cosfrac x 2 - sinfrac x 2 $) for $ sqrt{1- sin x} $. But, if I use the negative one, then the result changes. The sixth step changes to: $$ cot^{-1}biggl(tanfrac x 2biggl) = cot^{-1}Biggl(cotleft(frac pi 2 - frac x 2right)Biggl) $$
which yield the result: $$ frac pi 2 - frac x 2 $$
Mathematically, this result is different from that provided in the RHS of question.
Is the question statement wrong or I've been hacked up?
trigonometry proof-verification
asked Dec 1 '18 at 12:45
rv7
1339
add a comment |
I have the following question:
Prove that: $$ cot^{-1}Biggl(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}Biggl) = frac x2, x in biggl(0, frac pi4biggl) $$
The solution:
My question is about the second step (i.e., highlighted). We know that it has came from the following resolution: $$ sqrt{1pm sin x} = sqrt{sin^2frac x 2 + cos^2frac x 2 pm2sinfrac x 2cosfrac x 2} = pm biggl( cosfrac x 2 pm sinfrac x 2 biggl) $$
I've included the $pm$ symbol in accordance with the standard definition of square root. The above solution uses the positive resolution (i.e., $ cosfrac x 2 - sinfrac x 2 $) for $ sqrt{1- sin x} $. But, if I use the negative one, then the result changes. The sixth step changes to: $$ cot^{-1}biggl(tanfrac x 2biggl) = cot^{-1}Biggl(cotleft(frac pi 2 - frac x 2right)Biggl) $$
which yield the result: $$ frac pi 2 - frac x 2 $$
Mathematically, this result is different from that provided in the RHS of question.
Is the question statement wrong or I've been hacked up?
trigonometry proof-verification
asked Dec 1 '18 at 12:45
rv7
1339
@Winther Ok, I edited the question.
– rv7
Dec 1 '18 at 13:42
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
– StubbornAtom
Dec 1 '18 at 13:55
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
– Fawad
Dec 2 '18 at 5:17
add a comment |
I have the following question:
Prove that: $$ cot^{-1}Biggl(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}Biggl) = frac x2, x in biggl(0, frac pi4biggl) $$
The solution:
My question is about the second step (i.e., highlighted). We know that it has came from the following resolution: $$ sqrt{1pm sin x} = sqrt{sin^2frac x 2 + cos^2frac x 2 pm2sinfrac x 2cosfrac x 2} = pm biggl( cosfrac x 2 pm sinfrac x 2 biggl) $$
I've included the $pm$ symbol in accordance with the standard definition of square root. The above solution uses the positive resolution (i.e., $ cosfrac x 2 - sinfrac x 2 $) for $ sqrt{1- sin x} $. But, if I use the negative one, then the result changes. The sixth step changes to: $$ cot^{-1}biggl(tanfrac x 2biggl) = cot^{-1}Biggl(cotleft(frac pi 2 - frac x 2right)Biggl) $$
which yield the result: $$ frac pi 2 - frac x 2 $$
Mathematically, this result is different from that provided in the RHS of question.
Is the question statement wrong or I've been hacked up?
trigonometry proof-verification
asked Dec 1 '18 at 12:45
rv7
1339
I have the following question:
Prove that: $$ cot^{-1}Biggl(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}Biggl) = frac x2, x in biggl(0, frac pi4biggl) $$
The solution:
My question is about the second step (i.e., highlighted). We know that it has came from the following resolution: $$ sqrt{1pm sin x} = sqrt{sin^2frac x 2 + cos^2frac x 2 pm2sinfrac x 2cosfrac x 2} = pm biggl( cosfrac x 2 pm sinfrac x 2 biggl) $$
I've included the $pm$ symbol in accordance with the standard definition of square root. The above solution uses the positive resolution (i.e., $ cosfrac x 2 - sinfrac x 2 $) for $ sqrt{1- sin x} $. But, if I use the negative one, then the result changes. The sixth step changes to: $$ cot^{-1}biggl(tanfrac x 2biggl) = cot^{-1}Biggl(cotleft(frac pi 2 - frac x 2right)Biggl) $$
which yield the result: $$ frac pi 2 - frac x 2 $$
Mathematically, this result is different from that provided in the RHS of question.
Is the question statement wrong or I've been hacked up?
trigonometry proof-verification
trigonometry proof-verification
asked Dec 1 '18 at 12:45
rv7
1339
asked Dec 1 '18 at 12:45
rv7
1339
edited Dec 2 '18 at 4:53
asked Dec 1 '18 at 12:45
rv7
1339
asked Dec 1 '18 at 12:45
rv7
1339
asked Dec 1 '18 at 12:45
rv7
1339
1339
@Winther Ok, I edited the question.
– rv7
Dec 1 '18 at 13:42
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
– StubbornAtom
Dec 1 '18 at 13:55
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
– Fawad
Dec 2 '18 at 5:17
add a comment |
@Winther Ok, I edited the question.
– rv7
Dec 1 '18 at 13:42
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
– StubbornAtom
Dec 1 '18 at 13:55
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
– Fawad
Dec 2 '18 at 5:17
@Winther Ok, I edited the question.
– rv7
Dec 1 '18 at 13:42
@Winther Ok, I edited the question.
– rv7
Dec 1 '18 at 13:42
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
– StubbornAtom
Dec 1 '18 at 13:55
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
– StubbornAtom
Dec 1 '18 at 13:55
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
– Fawad
Dec 2 '18 at 5:17
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
– Fawad
Dec 2 '18 at 5:17
add a comment |
2 Answers
2
active
oldest
votes
Rationalize the numerator to find
$$f(x)=dfrac{1+|cos x|}{sin x}$$
Now if $cos xge0,$ $$f(x)=cotdfrac x2$$
$cot^{-1}f(x)=?$
Else $f(x)=dfrac{1-cos x}{sin x}=tandfrac x2$
Now $cot^{-1}f(x)=dfracpi2-tan^{-1}f(x)=?$
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
add a comment |
As StubbornAtom mentioned in comment, for $x in biggl(0, frac pi4biggl)$, the square root is:
$$sqrt{1-sin x}=cos frac x2-sin frac x2,$$
because:
$$cos frac x2-sin frac x2>0.$$
If, for example, $x in biggl(frac pi2, pibiggl)$, the square root would be:
$$sqrt{1-sin x}=sinfrac x2-cos frac x2,$$
because:
$$sin frac x2-cos frac x2>0.$$
Addendum. Note that $pm$ is placed to make the number positive and it does not imply one gets positive and negative values for the square root. Here is another way to write the same equality:
$$sqrt{1-sin x}=left|cos frac x2-sin frac x2right|=begin{cases}+left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2ge 0 \ -left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2< 0 \ end{cases}.$$
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
add a comment |
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2 Answers
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2 Answers
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Rationalize the numerator to find
$$f(x)=dfrac{1+|cos x|}{sin x}$$
Now if $cos xge0,$ $$f(x)=cotdfrac x2$$
$cot^{-1}f(x)=?$
Else $f(x)=dfrac{1-cos x}{sin x}=tandfrac x2$
Now $cot^{-1}f(x)=dfracpi2-tan^{-1}f(x)=?$
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
add a comment |
Rationalize the numerator to find
$$f(x)=dfrac{1+|cos x|}{sin x}$$
Now if $cos xge0,$ $$f(x)=cotdfrac x2$$
$cot^{-1}f(x)=?$
Else $f(x)=dfrac{1-cos x}{sin x}=tandfrac x2$
Now $cot^{-1}f(x)=dfracpi2-tan^{-1}f(x)=?$
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
add a comment |
Rationalize the numerator to find
$$f(x)=dfrac{1+|cos x|}{sin x}$$
Now if $cos xge0,$ $$f(x)=cotdfrac x2$$
$cot^{-1}f(x)=?$
Else $f(x)=dfrac{1-cos x}{sin x}=tandfrac x2$
Now $cot^{-1}f(x)=dfracpi2-tan^{-1}f(x)=?$
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
Rationalize the numerator to find
$$f(x)=dfrac{1+|cos x|}{sin x}$$
Now if $cos xge0,$ $$f(x)=cotdfrac x2$$
$cot^{-1}f(x)=?$
Else $f(x)=dfrac{1-cos x}{sin x}=tandfrac x2$
Now $cot^{-1}f(x)=dfracpi2-tan^{-1}f(x)=?$
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
answered Dec 1 '18 at 13:11
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
As StubbornAtom mentioned in comment, for $x in biggl(0, frac pi4biggl)$, the square root is:
$$sqrt{1-sin x}=cos frac x2-sin frac x2,$$
because:
$$cos frac x2-sin frac x2>0.$$
If, for example, $x in biggl(frac pi2, pibiggl)$, the square root would be:
$$sqrt{1-sin x}=sinfrac x2-cos frac x2,$$
because:
$$sin frac x2-cos frac x2>0.$$
Addendum. Note that $pm$ is placed to make the number positive and it does not imply one gets positive and negative values for the square root. Here is another way to write the same equality:
$$sqrt{1-sin x}=left|cos frac x2-sin frac x2right|=begin{cases}+left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2ge 0 \ -left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2< 0 \ end{cases}.$$
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
add a comment |
As StubbornAtom mentioned in comment, for $x in biggl(0, frac pi4biggl)$, the square root is:
$$sqrt{1-sin x}=cos frac x2-sin frac x2,$$
because:
$$cos frac x2-sin frac x2>0.$$
If, for example, $x in biggl(frac pi2, pibiggl)$, the square root would be:
$$sqrt{1-sin x}=sinfrac x2-cos frac x2,$$
because:
$$sin frac x2-cos frac x2>0.$$
Addendum. Note that $pm$ is placed to make the number positive and it does not imply one gets positive and negative values for the square root. Here is another way to write the same equality:
$$sqrt{1-sin x}=left|cos frac x2-sin frac x2right|=begin{cases}+left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2ge 0 \ -left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2< 0 \ end{cases}.$$
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
add a comment |
As StubbornAtom mentioned in comment, for $x in biggl(0, frac pi4biggl)$, the square root is:
$$sqrt{1-sin x}=cos frac x2-sin frac x2,$$
because:
$$cos frac x2-sin frac x2>0.$$
If, for example, $x in biggl(frac pi2, pibiggl)$, the square root would be:
$$sqrt{1-sin x}=sinfrac x2-cos frac x2,$$
because:
$$sin frac x2-cos frac x2>0.$$
Addendum. Note that $pm$ is placed to make the number positive and it does not imply one gets positive and negative values for the square root. Here is another way to write the same equality:
$$sqrt{1-sin x}=left|cos frac x2-sin frac x2right|=begin{cases}+left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2ge 0 \ -left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2< 0 \ end{cases}.$$
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
As StubbornAtom mentioned in comment, for $x in biggl(0, frac pi4biggl)$, the square root is:
$$sqrt{1-sin x}=cos frac x2-sin frac x2,$$
because:
$$cos frac x2-sin frac x2>0.$$
If, for example, $x in biggl(frac pi2, pibiggl)$, the square root would be:
$$sqrt{1-sin x}=sinfrac x2-cos frac x2,$$
because:
$$sin frac x2-cos frac x2>0.$$
Addendum. Note that $pm$ is placed to make the number positive and it does not imply one gets positive and negative values for the square root. Here is another way to write the same equality:
$$sqrt{1-sin x}=left|cos frac x2-sin frac x2right|=begin{cases}+left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2ge 0 \ -left(cos frac x2-sin frac x2right), cos frac x2-sin frac x2< 0 \ end{cases}.$$
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
edited Dec 2 '18 at 5:11
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
answered Dec 1 '18 at 16:16
farruhota
19.3k2736
19.3k2736
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
add a comment |
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
– rv7
Dec 2 '18 at 2:28
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
Firstly, $0le 1-sin xle 2<9$. Secondly, $sqrt{9}=3$ and $pm sqrt{9}=pm 3$. Thirdly, $0<x<pi/4$. For example, when $x=pi/6$, $sqrt{1-sin (pi/6)}=1/sqrt{2}=cos (pi/12)-sin (pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$. Fourthly, for $x=5pi/6>pi/2>pi/4$, we get what you are saying: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
– farruhota
Dec 2 '18 at 4:20
add a comment |
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@Winther Ok, I edited the question.
– rv7
Dec 1 '18 at 13:42
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
– StubbornAtom
Dec 1 '18 at 13:55
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
– Fawad
Dec 2 '18 at 5:17