Mixing
About proof: $cot^{-1}left(frac{sqrt{1+sin x}+sqrt{1-sin x}}{sqrt{1+sin x}-sqrt{1-sin x}}right)=frac x2$
$begingroup$
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
rv7rv7
10311
$endgroup$
|
show 2 more comments
$begingroup$
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
rv7rv7
10311
$endgroup$
$begingroup$
@Winther Ok, I edited the question.
$endgroup$
– rv7
Dec 1 '18 at 13:42
$begingroup$
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
$endgroup$
– StubbornAtom
Dec 1 '18 at 13:55
$begingroup$
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
$endgroup$
– Fawad
Dec 2 '18 at 5:17
1
$begingroup$
The authors use a specific range to get positive numbers. They don't spell it out but they DO mean PRINCIPAL ROOTS (square root values are positive). They also imply real numbers in this problem. And no, the result doesn’t change if “standard” square roots are extracted: You just get two values. Yet nonsense may still arise: $sqrt{1}=1;;sqrt{1}=-1$. Hence, $-1=+1$. Non-principal square root values should be combined properly!
$endgroup$
– Ken Draco
Jan 21 at 7:44
$begingroup$
Note that Arccot may have infinite number of values since $cot$ is periodic. So we need to restrict its range to avoid dragging around infinite number of values. Hence, we define $cot^{-1}$ range from $0$ to $pi$ for convenience (to make $cot^{-1}$ continuous). Hope you can figure out everything checks out for negative values too if root values are combined properly! And the range of $cot^{-1}$ can also be defined –pi/2 to +pi/2 to deal with negative values (not needed here). +1 for figuring out roots aren’t principal roots and confusion is possible if a student is not a rote-learner.
$endgroup$
– Ken Draco
Jan 21 at 7:47
|
show 2 more comments
$begingroup$
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
rv7rv7
10311
$endgroup$
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
rv7rv7
10311
asked Dec 1 '18 at 12:45
rv7rv7
10311
edited Jan 19 at 16:56
rv7
asked Dec 1 '18 at 12:45
rv7rv7
10311
asked Dec 1 '18 at 12:45
rv7rv7
10311
asked Dec 1 '18 at 12:45
rv7rv7
10311
10311
$begingroup$
@Winther Ok, I edited the question.
$endgroup$
– rv7
Dec 1 '18 at 13:42
$begingroup$
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
$endgroup$
– StubbornAtom
Dec 1 '18 at 13:55
$begingroup$
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
$endgroup$
– Fawad
Dec 2 '18 at 5:17
1
$begingroup$
The authors use a specific range to get positive numbers. They don't spell it out but they DO mean PRINCIPAL ROOTS (square root values are positive). They also imply real numbers in this problem. And no, the result doesn’t change if “standard” square roots are extracted: You just get two values. Yet nonsense may still arise: $sqrt{1}=1;;sqrt{1}=-1$. Hence, $-1=+1$. Non-principal square root values should be combined properly!
$endgroup$
– Ken Draco
Jan 21 at 7:44
$begingroup$
Note that Arccot may have infinite number of values since $cot$ is periodic. So we need to restrict its range to avoid dragging around infinite number of values. Hence, we define $cot^{-1}$ range from $0$ to $pi$ for convenience (to make $cot^{-1}$ continuous). Hope you can figure out everything checks out for negative values too if root values are combined properly! And the range of $cot^{-1}$ can also be defined –pi/2 to +pi/2 to deal with negative values (not needed here). +1 for figuring out roots aren’t principal roots and confusion is possible if a student is not a rote-learner.
$endgroup$
– Ken Draco
Jan 21 at 7:47
|
show 2 more comments
$begingroup$
@Winther Ok, I edited the question.
$endgroup$
– rv7
Dec 1 '18 at 13:42
$begingroup$
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
$endgroup$
– StubbornAtom
Dec 1 '18 at 13:55
$begingroup$
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
$endgroup$
– Fawad
Dec 2 '18 at 5:17
1
$begingroup$
The authors use a specific range to get positive numbers. They don't spell it out but they DO mean PRINCIPAL ROOTS (square root values are positive). They also imply real numbers in this problem. And no, the result doesn’t change if “standard” square roots are extracted: You just get two values. Yet nonsense may still arise: $sqrt{1}=1;;sqrt{1}=-1$. Hence, $-1=+1$. Non-principal square root values should be combined properly!
$endgroup$
– Ken Draco
Jan 21 at 7:44
$begingroup$
Note that Arccot may have infinite number of values since $cot$ is periodic. So we need to restrict its range to avoid dragging around infinite number of values. Hence, we define $cot^{-1}$ range from $0$ to $pi$ for convenience (to make $cot^{-1}$ continuous). Hope you can figure out everything checks out for negative values too if root values are combined properly! And the range of $cot^{-1}$ can also be defined –pi/2 to +pi/2 to deal with negative values (not needed here). +1 for figuring out roots aren’t principal roots and confusion is possible if a student is not a rote-learner.
$endgroup$
– Ken Draco
Jan 21 at 7:47
$begingroup$
@Winther Ok, I edited the question.
$endgroup$
– rv7
Dec 1 '18 at 13:42
$begingroup$
@Winther Ok, I edited the question.
$endgroup$
– rv7
Dec 1 '18 at 13:42
$begingroup$
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
$endgroup$
– StubbornAtom
Dec 1 '18 at 13:55
$begingroup$
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
$endgroup$
– StubbornAtom
Dec 1 '18 at 13:55
$begingroup$
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
$endgroup$
– Fawad
Dec 2 '18 at 5:17
$begingroup$
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
$endgroup$
– Fawad
Dec 2 '18 at 5:17
1
1
$begingroup$
The authors use a specific range to get positive numbers. They don't spell it out but they DO mean PRINCIPAL ROOTS (square root values are positive). They also imply real numbers in this problem. And no, the result doesn’t change if “standard” square roots are extracted: You just get two values. Yet nonsense may still arise: $sqrt{1}=1;;sqrt{1}=-1$. Hence, $-1=+1$. Non-principal square root values should be combined properly!
$endgroup$
– Ken Draco
Jan 21 at 7:44
$begingroup$
The authors use a specific range to get positive numbers. They don't spell it out but they DO mean PRINCIPAL ROOTS (square root values are positive). They also imply real numbers in this problem. And no, the result doesn’t change if “standard” square roots are extracted: You just get two values. Yet nonsense may still arise: $sqrt{1}=1;;sqrt{1}=-1$. Hence, $-1=+1$. Non-principal square root values should be combined properly!
$endgroup$
– Ken Draco
Jan 21 at 7:44
$begingroup$
Note that Arccot may have infinite number of values since $cot$ is periodic. So we need to restrict its range to avoid dragging around infinite number of values. Hence, we define $cot^{-1}$ range from $0$ to $pi$ for convenience (to make $cot^{-1}$ continuous). Hope you can figure out everything checks out for negative values too if root values are combined properly! And the range of $cot^{-1}$ can also be defined –pi/2 to +pi/2 to deal with negative values (not needed here). +1 for figuring out roots aren’t principal roots and confusion is possible if a student is not a rote-learner.
$endgroup$
– Ken Draco
Jan 21 at 7:47
$begingroup$
Note that Arccot may have infinite number of values since $cot$ is periodic. So we need to restrict its range to avoid dragging around infinite number of values. Hence, we define $cot^{-1}$ range from $0$ to $pi$ for convenience (to make $cot^{-1}$ continuous). Hope you can figure out everything checks out for negative values too if root values are combined properly! And the range of $cot^{-1}$ can also be defined –pi/2 to +pi/2 to deal with negative values (not needed here). +1 for figuring out roots aren’t principal roots and confusion is possible if a student is not a rote-learner.
$endgroup$
– Ken Draco
Jan 21 at 7:47
|
show 2 more comments
4 Answers
4
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oldest
votes
$begingroup$
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 bhattacharjeelab bhattacharjee
226k15157275
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add a comment |
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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
farruhotafarruhota
20.6k2740
$endgroup$
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
add a comment |
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Note that for $xin (0,{pi over 4})$ we have $$sqrt{1-{sqrt 2over 2}}<sqrt{1-sin x}<1\sqrt{1-{sqrt 2over 2}}=cos {pi over 8}-sin {pi over 8}<cos {xover 2}-sin{xover 2}<1\-1<sin {x over 2}-cos {x over 2}<-sqrt{1-{sqrt 2over 2}}$$comparing the range for all $xin (0,{pi over 4})$ we obtain$$sqrt{1-sin x}=cos {xover 2}-sin{xover 2}$$and$$sqrt{1-sin x}ne sin {xover 2}-cos{xover 2}$$
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
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We see that
$$frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
$$= sqrt{(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})^2}$$
$$= sqrt{frac{1+sin x +1-sin x + 2cdotsqrt{1-sin^2 x}}{1+sin x +1-sin x -2cdotsqrt{1-sin^2 x}}}$$
$$= sqrt{frac{1+cos x}{1-cos x}}$$
$$= sqrt{frac{2cdotcos^2frac{x}{2}}{2cdotsin^2frac{x}{2}}}$$
$$= cot frac{x}{2}$$
And hence,
$$cot^{-1} (frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})$$
$$= cot^{-1} (cot frac{x}{2})$$
$$= frac{x}{2}$$
Note: as $xin [0,frac{pi}{4}]$, $sqrt{1-sin x} le sqrt{1+sin x}$.
And hence
$$0lefrac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
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$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
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– rv7
Jan 21 at 15:10
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
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 bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
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 bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
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 bhattacharjeelab bhattacharjee
226k15157275
answered Dec 1 '18 at 13:11
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 1 '18 at 13:11
lab bhattacharjeelab bhattacharjee
226k15157275
answered Dec 1 '18 at 13:11
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
$begingroup$
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
farruhotafarruhota
20.6k2740
$endgroup$
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
add a comment |
$begingroup$
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
farruhotafarruhota
20.6k2740
$endgroup$
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
add a comment |
$begingroup$
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
farruhotafarruhota
20.6k2740
$endgroup$
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
farruhotafarruhota
20.6k2740
edited Dec 2 '18 at 5:11
answered Dec 1 '18 at 16:16
farruhotafarruhota
20.6k2740
answered Dec 1 '18 at 16:16
farruhotafarruhota
20.6k2740
answered Dec 1 '18 at 16:16
farruhotafarruhota
20.6k2740
20.6k2740
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
add a comment |
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
Why not $sqrt{1-sin x}= - bigl( cos frac x2-sin frac x2 bigl)$? Because $sqrt9 = pm3$
$endgroup$
– rv7
Dec 2 '18 at 2:28
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
$begingroup$
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: $sqrt{1-sin (5pi/6)}=1/sqrt{2}=sin (5pi/12)-cos (5pi/12)=frac{1+sqrt{3}}{2sqrt{2}}-frac{sqrt{3}-1}{2sqrt{2}}$.
$endgroup$
– farruhota
Dec 2 '18 at 4:20
add a comment |
$begingroup$
Note that for $xin (0,{pi over 4})$ we have $$sqrt{1-{sqrt 2over 2}}<sqrt{1-sin x}<1\sqrt{1-{sqrt 2over 2}}=cos {pi over 8}-sin {pi over 8}<cos {xover 2}-sin{xover 2}<1\-1<sin {x over 2}-cos {x over 2}<-sqrt{1-{sqrt 2over 2}}$$comparing the range for all $xin (0,{pi over 4})$ we obtain$$sqrt{1-sin x}=cos {xover 2}-sin{xover 2}$$and$$sqrt{1-sin x}ne sin {xover 2}-cos{xover 2}$$
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Note that for $xin (0,{pi over 4})$ we have $$sqrt{1-{sqrt 2over 2}}<sqrt{1-sin x}<1\sqrt{1-{sqrt 2over 2}}=cos {pi over 8}-sin {pi over 8}<cos {xover 2}-sin{xover 2}<1\-1<sin {x over 2}-cos {x over 2}<-sqrt{1-{sqrt 2over 2}}$$comparing the range for all $xin (0,{pi over 4})$ we obtain$$sqrt{1-sin x}=cos {xover 2}-sin{xover 2}$$and$$sqrt{1-sin x}ne sin {xover 2}-cos{xover 2}$$
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Note that for $xin (0,{pi over 4})$ we have $$sqrt{1-{sqrt 2over 2}}<sqrt{1-sin x}<1\sqrt{1-{sqrt 2over 2}}=cos {pi over 8}-sin {pi over 8}<cos {xover 2}-sin{xover 2}<1\-1<sin {x over 2}-cos {x over 2}<-sqrt{1-{sqrt 2over 2}}$$comparing the range for all $xin (0,{pi over 4})$ we obtain$$sqrt{1-sin x}=cos {xover 2}-sin{xover 2}$$and$$sqrt{1-sin x}ne sin {xover 2}-cos{xover 2}$$
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Note that for $xin (0,{pi over 4})$ we have $$sqrt{1-{sqrt 2over 2}}<sqrt{1-sin x}<1\sqrt{1-{sqrt 2over 2}}=cos {pi over 8}-sin {pi over 8}<cos {xover 2}-sin{xover 2}<1\-1<sin {x over 2}-cos {x over 2}<-sqrt{1-{sqrt 2over 2}}$$comparing the range for all $xin (0,{pi over 4})$ we obtain$$sqrt{1-sin x}=cos {xover 2}-sin{xover 2}$$and$$sqrt{1-sin x}ne sin {xover 2}-cos{xover 2}$$
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 19 at 19:16
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
$begingroup$
We see that
$$frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
$$= sqrt{(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})^2}$$
$$= sqrt{frac{1+sin x +1-sin x + 2cdotsqrt{1-sin^2 x}}{1+sin x +1-sin x -2cdotsqrt{1-sin^2 x}}}$$
$$= sqrt{frac{1+cos x}{1-cos x}}$$
$$= sqrt{frac{2cdotcos^2frac{x}{2}}{2cdotsin^2frac{x}{2}}}$$
$$= cot frac{x}{2}$$
And hence,
$$cot^{-1} (frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})$$
$$= cot^{-1} (cot frac{x}{2})$$
$$= frac{x}{2}$$
Note: as $xin [0,frac{pi}{4}]$, $sqrt{1-sin x} le sqrt{1+sin x}$.
And hence
$$0lefrac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
$endgroup$
$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
$endgroup$
– rv7
Jan 21 at 15:10
add a comment |
$begingroup$
We see that
$$frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
$$= sqrt{(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})^2}$$
$$= sqrt{frac{1+sin x +1-sin x + 2cdotsqrt{1-sin^2 x}}{1+sin x +1-sin x -2cdotsqrt{1-sin^2 x}}}$$
$$= sqrt{frac{1+cos x}{1-cos x}}$$
$$= sqrt{frac{2cdotcos^2frac{x}{2}}{2cdotsin^2frac{x}{2}}}$$
$$= cot frac{x}{2}$$
And hence,
$$cot^{-1} (frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})$$
$$= cot^{-1} (cot frac{x}{2})$$
$$= frac{x}{2}$$
Note: as $xin [0,frac{pi}{4}]$, $sqrt{1-sin x} le sqrt{1+sin x}$.
And hence
$$0lefrac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
$endgroup$
$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
$endgroup$
– rv7
Jan 21 at 15:10
add a comment |
$begingroup$
We see that
$$frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
$$= sqrt{(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})^2}$$
$$= sqrt{frac{1+sin x +1-sin x + 2cdotsqrt{1-sin^2 x}}{1+sin x +1-sin x -2cdotsqrt{1-sin^2 x}}}$$
$$= sqrt{frac{1+cos x}{1-cos x}}$$
$$= sqrt{frac{2cdotcos^2frac{x}{2}}{2cdotsin^2frac{x}{2}}}$$
$$= cot frac{x}{2}$$
And hence,
$$cot^{-1} (frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})$$
$$= cot^{-1} (cot frac{x}{2})$$
$$= frac{x}{2}$$
Note: as $xin [0,frac{pi}{4}]$, $sqrt{1-sin x} le sqrt{1+sin x}$.
And hence
$$0lefrac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
$endgroup$
We see that
$$frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
$$= sqrt{(frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})^2}$$
$$= sqrt{frac{1+sin x +1-sin x + 2cdotsqrt{1-sin^2 x}}{1+sin x +1-sin x -2cdotsqrt{1-sin^2 x}}}$$
$$= sqrt{frac{1+cos x}{1-cos x}}$$
$$= sqrt{frac{2cdotcos^2frac{x}{2}}{2cdotsin^2frac{x}{2}}}$$
$$= cot frac{x}{2}$$
And hence,
$$cot^{-1} (frac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}})$$
$$= cot^{-1} (cot frac{x}{2})$$
$$= frac{x}{2}$$
Note: as $xin [0,frac{pi}{4}]$, $sqrt{1-sin x} le sqrt{1+sin x}$.
And hence
$$0lefrac{sqrt{1+sin x} + sqrt{1-sin x}}{sqrt{1+sin x} - sqrt{1-sin x}}$$
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
edited Jan 21 at 15:26
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
answered Jan 19 at 20:36
Jonas De SchouwerJonas De Schouwer
3518
3518
$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
$endgroup$
– rv7
Jan 21 at 15:10
add a comment |
$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
$endgroup$
– rv7
Jan 21 at 15:10
$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
$endgroup$
– rv7
Jan 21 at 15:10
$begingroup$
Yet another way of solving the problem. But, you forgot to mention $pm$ in sixth line before cot x/2
$endgroup$
– rv7
Jan 21 at 15:10
add a comment |
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$begingroup$
@Winther Ok, I edited the question.
$endgroup$
– rv7
Dec 1 '18 at 13:42
$begingroup$
The range $xin(0,pi/4)$ has to be kept in mind while removing the square root.
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– StubbornAtom
Dec 1 '18 at 13:55
$begingroup$
It’s already given x belong to 0 to pi/4. That means x is positive. Which also means sinx + cos is positive
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– Fawad
Dec 2 '18 at 5:17
1
$begingroup$
The authors use a specific range to get positive numbers. They don't spell it out but they DO mean PRINCIPAL ROOTS (square root values are positive). They also imply real numbers in this problem. And no, the result doesn’t change if “standard” square roots are extracted: You just get two values. Yet nonsense may still arise: $sqrt{1}=1;;sqrt{1}=-1$. Hence, $-1=+1$. Non-principal square root values should be combined properly!
$endgroup$
– Ken Draco
Jan 21 at 7:44
$begingroup$
Note that Arccot may have infinite number of values since $cot$ is periodic. So we need to restrict its range to avoid dragging around infinite number of values. Hence, we define $cot^{-1}$ range from $0$ to $pi$ for convenience (to make $cot^{-1}$ continuous). Hope you can figure out everything checks out for negative values too if root values are combined properly! And the range of $cot^{-1}$ can also be defined –pi/2 to +pi/2 to deal with negative values (not needed here). +1 for figuring out roots aren’t principal roots and confusion is possible if a student is not a rote-learner.
$endgroup$
– Ken Draco
Jan 21 at 7:47