Mixing
Expression for $1-sin x$
$begingroup$
Do we take $1-sin x$ as $(sin frac{x}{2}-cos frac{x}{2})^2$ or $(cos frac{x}{2}-sin frac{x}{2})^2$. My teacher says that in most of the cases we take the latter. In one problem the interval was specified as something like $(frac{pi}{2},pi)$ and my teacher told me to take the former. I am facing problem in understanding this. Can someone shed some light on this? Also what about $1+sin x$.
Both the expressions may be the same but they yield different results (in differentiation and integration problems) and I have myself seen this.
trigonometry
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
$endgroup$
|
show 3 more comments
$begingroup$
Do we take $1-sin x$ as $(sin frac{x}{2}-cos frac{x}{2})^2$ or $(cos frac{x}{2}-sin frac{x}{2})^2$. My teacher says that in most of the cases we take the latter. In one problem the interval was specified as something like $(frac{pi}{2},pi)$ and my teacher told me to take the former. I am facing problem in understanding this. Can someone shed some light on this? Also what about $1+sin x$.
Both the expressions may be the same but they yield different results (in differentiation and integration problems) and I have myself seen this.
trigonometry
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
$endgroup$
1
$begingroup$
$(a-b)^{2} = (b-a)^{2}$, hence I don't understand why it is even bothering you.
$endgroup$
– Anik Bhowmick
Jan 18 at 13:58
$begingroup$
@AnikBhowmick, Both may be the same but they yield different results.
$endgroup$
– MrAP
Jan 18 at 14:00
$begingroup$
What is the context of this question? If I were to see "$1-sin x$" in the wild, I'd leave it as-is; re-writing in either suggested form just adds visual and mental clutter. However, if I had $sqrt{1-sin x}$, then it would certainly be convenient to identify the expression as a square in order to cancel the radical. In this case, you could still use either suggested form, since you need to wrap it in absolute value bars anyway; however, the latter has a sliver of an advantage, in that you can drop the absolute value bars when you know $x$ happens to be acute.
$endgroup$
– Blue
Jan 18 at 14:07
$begingroup$
Is this question related to taking square roots?
$endgroup$
– Thomas Shelby
Jan 18 at 14:08
$begingroup$
One of the problems did involve square roots.
$endgroup$
– MrAP
Jan 18 at 14:09
|
show 3 more comments
$begingroup$
Do we take $1-sin x$ as $(sin frac{x}{2}-cos frac{x}{2})^2$ or $(cos frac{x}{2}-sin frac{x}{2})^2$. My teacher says that in most of the cases we take the latter. In one problem the interval was specified as something like $(frac{pi}{2},pi)$ and my teacher told me to take the former. I am facing problem in understanding this. Can someone shed some light on this? Also what about $1+sin x$.
Both the expressions may be the same but they yield different results (in differentiation and integration problems) and I have myself seen this.
trigonometry
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
$endgroup$
Do we take $1-sin x$ as $(sin frac{x}{2}-cos frac{x}{2})^2$ or $(cos frac{x}{2}-sin frac{x}{2})^2$. My teacher says that in most of the cases we take the latter. In one problem the interval was specified as something like $(frac{pi}{2},pi)$ and my teacher told me to take the former. I am facing problem in understanding this. Can someone shed some light on this? Also what about $1+sin x$.
Both the expressions may be the same but they yield different results (in differentiation and integration problems) and I have myself seen this.
trigonometry
trigonometry
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
edited Jan 18 at 14:04
MrAP
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
asked Jan 18 at 13:54
MrAPMrAP
1,15021432
1,15021432
1
$begingroup$
$(a-b)^{2} = (b-a)^{2}$, hence I don't understand why it is even bothering you.
$endgroup$
– Anik Bhowmick
Jan 18 at 13:58
$begingroup$
@AnikBhowmick, Both may be the same but they yield different results.
$endgroup$
– MrAP
Jan 18 at 14:00
$begingroup$
What is the context of this question? If I were to see "$1-sin x$" in the wild, I'd leave it as-is; re-writing in either suggested form just adds visual and mental clutter. However, if I had $sqrt{1-sin x}$, then it would certainly be convenient to identify the expression as a square in order to cancel the radical. In this case, you could still use either suggested form, since you need to wrap it in absolute value bars anyway; however, the latter has a sliver of an advantage, in that you can drop the absolute value bars when you know $x$ happens to be acute.
$endgroup$
– Blue
Jan 18 at 14:07
$begingroup$
Is this question related to taking square roots?
$endgroup$
– Thomas Shelby
Jan 18 at 14:08
$begingroup$
One of the problems did involve square roots.
$endgroup$
– MrAP
Jan 18 at 14:09
|
show 3 more comments
1
$begingroup$
$(a-b)^{2} = (b-a)^{2}$, hence I don't understand why it is even bothering you.
$endgroup$
– Anik Bhowmick
Jan 18 at 13:58
$begingroup$
@AnikBhowmick, Both may be the same but they yield different results.
$endgroup$
– MrAP
Jan 18 at 14:00
$begingroup$
What is the context of this question? If I were to see "$1-sin x$" in the wild, I'd leave it as-is; re-writing in either suggested form just adds visual and mental clutter. However, if I had $sqrt{1-sin x}$, then it would certainly be convenient to identify the expression as a square in order to cancel the radical. In this case, you could still use either suggested form, since you need to wrap it in absolute value bars anyway; however, the latter has a sliver of an advantage, in that you can drop the absolute value bars when you know $x$ happens to be acute.
$endgroup$
– Blue
Jan 18 at 14:07
$begingroup$
Is this question related to taking square roots?
$endgroup$
– Thomas Shelby
Jan 18 at 14:08
$begingroup$
One of the problems did involve square roots.
$endgroup$
– MrAP
Jan 18 at 14:09
1
1
$begingroup$
$(a-b)^{2} = (b-a)^{2}$, hence I don't understand why it is even bothering you.
$endgroup$
– Anik Bhowmick
Jan 18 at 13:58
$begingroup$
$(a-b)^{2} = (b-a)^{2}$, hence I don't understand why it is even bothering you.
$endgroup$
– Anik Bhowmick
Jan 18 at 13:58
$begingroup$
@AnikBhowmick, Both may be the same but they yield different results.
$endgroup$
– MrAP
Jan 18 at 14:00
$begingroup$
@AnikBhowmick, Both may be the same but they yield different results.
$endgroup$
– MrAP
Jan 18 at 14:00
$begingroup$
What is the context of this question? If I were to see "$1-sin x$" in the wild, I'd leave it as-is; re-writing in either suggested form just adds visual and mental clutter. However, if I had $sqrt{1-sin x}$, then it would certainly be convenient to identify the expression as a square in order to cancel the radical. In this case, you could still use either suggested form, since you need to wrap it in absolute value bars anyway; however, the latter has a sliver of an advantage, in that you can drop the absolute value bars when you know $x$ happens to be acute.
$endgroup$
– Blue
Jan 18 at 14:07
$begingroup$
What is the context of this question? If I were to see "$1-sin x$" in the wild, I'd leave it as-is; re-writing in either suggested form just adds visual and mental clutter. However, if I had $sqrt{1-sin x}$, then it would certainly be convenient to identify the expression as a square in order to cancel the radical. In this case, you could still use either suggested form, since you need to wrap it in absolute value bars anyway; however, the latter has a sliver of an advantage, in that you can drop the absolute value bars when you know $x$ happens to be acute.
$endgroup$
– Blue
Jan 18 at 14:07
$begingroup$
Is this question related to taking square roots?
$endgroup$
– Thomas Shelby
Jan 18 at 14:08
$begingroup$
Is this question related to taking square roots?
$endgroup$
– Thomas Shelby
Jan 18 at 14:08
$begingroup$
One of the problems did involve square roots.
$endgroup$
– MrAP
Jan 18 at 14:09
$begingroup$
One of the problems did involve square roots.
$endgroup$
– MrAP
Jan 18 at 14:09
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Those expressions are equal so... Both. Regarding the last question, plug $x mapsto -x$ and you'll have an almost equal expression.
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
Use that $$sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)=1-sin(x)$$ and $$(sin(x/2)+cos(x/2))^2=1+sin(x)$$
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
$endgroup$
add a comment |
$begingroup$
For any real number $x$, you have that
$$x^2 = (-x)^2 $$
Since
$$sinfrac x2 - cos frac x2 = -left(cos frac x2 - sinfrac x 2right)$$
The two expressions, (1):
$$left(sinfrac x2 - cos frac x2right)^2 $$
and (2):
$$left(-left(cos frac x2 - sinfrac x 2right)right)^2 = left(cosfrac x2 - sinfrac x 2right)^2 $$
are equal.
Since they are the same, they cannot yield different results. An error must have been made. Unless I'm misunderstanding what exactly is being asked.
answered Jan 18 at 14:01
EffEff
11.6k21638
$endgroup$
add a comment |
$begingroup$
Hint:
As for real $a,sqrt{a^2}=|a|$ which will be $=+a$ if $age0$
Now $1-sin2y=(cos y-sin y)^2=2sin^2left(dfracpi4-yright)$
Now $sin uge0$ if $u$ lies in the first & second quadrant
Similarly, $(sin y+cos y)^2=?$
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
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
$begingroup$
Those expressions are equal so... Both. Regarding the last question, plug $x mapsto -x$ and you'll have an almost equal expression.
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
Those expressions are equal so... Both. Regarding the last question, plug $x mapsto -x$ and you'll have an almost equal expression.
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
add a comment |
$begingroup$
Those expressions are equal so... Both. Regarding the last question, plug $x mapsto -x$ and you'll have an almost equal expression.
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
Those expressions are equal so... Both. Regarding the last question, plug $x mapsto -x$ and you'll have an almost equal expression.
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
answered Jan 18 at 13:57
Lucas HenriqueLucas Henrique
1,026414
1,026414
add a comment |
add a comment |
$begingroup$
Use that $$sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)=1-sin(x)$$ and $$(sin(x/2)+cos(x/2))^2=1+sin(x)$$
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
$endgroup$
add a comment |
$begingroup$
Use that $$sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)=1-sin(x)$$ and $$(sin(x/2)+cos(x/2))^2=1+sin(x)$$
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
$endgroup$
add a comment |
$begingroup$
Use that $$sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)=1-sin(x)$$ and $$(sin(x/2)+cos(x/2))^2=1+sin(x)$$
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
$endgroup$
Use that $$sin^2(x/2)+cos^2(x/2)-2sin(x/2)cos(x/2)=1-sin(x)$$ and $$(sin(x/2)+cos(x/2))^2=1+sin(x)$$
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
answered Jan 18 at 13:57
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.5k42866
76.5k42866
add a comment |
add a comment |
$begingroup$
For any real number $x$, you have that
$$x^2 = (-x)^2 $$
Since
$$sinfrac x2 - cos frac x2 = -left(cos frac x2 - sinfrac x 2right)$$
The two expressions, (1):
$$left(sinfrac x2 - cos frac x2right)^2 $$
and (2):
$$left(-left(cos frac x2 - sinfrac x 2right)right)^2 = left(cosfrac x2 - sinfrac x 2right)^2 $$
are equal.
Since they are the same, they cannot yield different results. An error must have been made. Unless I'm misunderstanding what exactly is being asked.
answered Jan 18 at 14:01
EffEff
11.6k21638
$endgroup$
add a comment |
$begingroup$
For any real number $x$, you have that
$$x^2 = (-x)^2 $$
Since
$$sinfrac x2 - cos frac x2 = -left(cos frac x2 - sinfrac x 2right)$$
The two expressions, (1):
$$left(sinfrac x2 - cos frac x2right)^2 $$
and (2):
$$left(-left(cos frac x2 - sinfrac x 2right)right)^2 = left(cosfrac x2 - sinfrac x 2right)^2 $$
are equal.
Since they are the same, they cannot yield different results. An error must have been made. Unless I'm misunderstanding what exactly is being asked.
answered Jan 18 at 14:01
EffEff
11.6k21638
$endgroup$
add a comment |
$begingroup$
For any real number $x$, you have that
$$x^2 = (-x)^2 $$
Since
$$sinfrac x2 - cos frac x2 = -left(cos frac x2 - sinfrac x 2right)$$
The two expressions, (1):
$$left(sinfrac x2 - cos frac x2right)^2 $$
and (2):
$$left(-left(cos frac x2 - sinfrac x 2right)right)^2 = left(cosfrac x2 - sinfrac x 2right)^2 $$
are equal.
Since they are the same, they cannot yield different results. An error must have been made. Unless I'm misunderstanding what exactly is being asked.
answered Jan 18 at 14:01
EffEff
11.6k21638
$endgroup$
For any real number $x$, you have that
$$x^2 = (-x)^2 $$
Since
$$sinfrac x2 - cos frac x2 = -left(cos frac x2 - sinfrac x 2right)$$
The two expressions, (1):
$$left(sinfrac x2 - cos frac x2right)^2 $$
and (2):
$$left(-left(cos frac x2 - sinfrac x 2right)right)^2 = left(cosfrac x2 - sinfrac x 2right)^2 $$
are equal.
Since they are the same, they cannot yield different results. An error must have been made. Unless I'm misunderstanding what exactly is being asked.
answered Jan 18 at 14:01
EffEff
11.6k21638
answered Jan 18 at 14:01
EffEff
11.6k21638
answered Jan 18 at 14:01
EffEff
11.6k21638
answered Jan 18 at 14:01
EffEff
11.6k21638
11.6k21638
add a comment |
add a comment |
$begingroup$
Hint:
As for real $a,sqrt{a^2}=|a|$ which will be $=+a$ if $age0$
Now $1-sin2y=(cos y-sin y)^2=2sin^2left(dfracpi4-yright)$
Now $sin uge0$ if $u$ lies in the first & second quadrant
Similarly, $(sin y+cos y)^2=?$
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
As for real $a,sqrt{a^2}=|a|$ which will be $=+a$ if $age0$
Now $1-sin2y=(cos y-sin y)^2=2sin^2left(dfracpi4-yright)$
Now $sin uge0$ if $u$ lies in the first & second quadrant
Similarly, $(sin y+cos y)^2=?$
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
As for real $a,sqrt{a^2}=|a|$ which will be $=+a$ if $age0$
Now $1-sin2y=(cos y-sin y)^2=2sin^2left(dfracpi4-yright)$
Now $sin uge0$ if $u$ lies in the first & second quadrant
Similarly, $(sin y+cos y)^2=?$
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
As for real $a,sqrt{a^2}=|a|$ which will be $=+a$ if $age0$
Now $1-sin2y=(cos y-sin y)^2=2sin^2left(dfracpi4-yright)$
Now $sin uge0$ if $u$ lies in the first & second quadrant
Similarly, $(sin y+cos y)^2=?$
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 18 at 14:02
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
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1
$begingroup$
$(a-b)^{2} = (b-a)^{2}$, hence I don't understand why it is even bothering you.
$endgroup$
– Anik Bhowmick
Jan 18 at 13:58
$begingroup$
@AnikBhowmick, Both may be the same but they yield different results.
$endgroup$
– MrAP
Jan 18 at 14:00
$begingroup$
What is the context of this question? If I were to see "$1-sin x$" in the wild, I'd leave it as-is; re-writing in either suggested form just adds visual and mental clutter. However, if I had $sqrt{1-sin x}$, then it would certainly be convenient to identify the expression as a square in order to cancel the radical. In this case, you could still use either suggested form, since you need to wrap it in absolute value bars anyway; however, the latter has a sliver of an advantage, in that you can drop the absolute value bars when you know $x$ happens to be acute.
$endgroup$
– Blue
Jan 18 at 14:07
$begingroup$
Is this question related to taking square roots?
$endgroup$
– Thomas Shelby
Jan 18 at 14:08
$begingroup$
One of the problems did involve square roots.
$endgroup$
– MrAP
Jan 18 at 14:09