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Find $sin frac{theta}{2}$ when $sin theta = frac{3}{5}$, and $0^{circ} < theta < 90^{circ}$
I need to find $sin frac{theta}{2}$ when $sin theta = frac{3}{5}$, and $0^{circ} < theta < 90^{circ}$
The answer I'm currently getting is $sqrt{frac{1}{10}}$, but the answer must be either $frac{sqrt{10}}{10}$, $frac{3 sqrt{10}}{10}$, or $frac{1}{3}$.
My Process:
since, $sin frac{theta}{2} = pm sqrt{frac{1 - cos theta}{2}}$
$sqrt{ frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}} = sqrt{frac{1}{10}}$
What am I doing wrong?
algebra-precalculus trigonometry
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
add a comment |
I need to find $sin frac{theta}{2}$ when $sin theta = frac{3}{5}$, and $0^{circ} < theta < 90^{circ}$
The answer I'm currently getting is $sqrt{frac{1}{10}}$, but the answer must be either $frac{sqrt{10}}{10}$, $frac{3 sqrt{10}}{10}$, or $frac{1}{3}$.
My Process:
since, $sin frac{theta}{2} = pm sqrt{frac{1 - cos theta}{2}}$
$sqrt{ frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}} = sqrt{frac{1}{10}}$
What am I doing wrong?
algebra-precalculus trigonometry
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
1
Notational error: You should have $sqrt{frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}}$.
– N. F. Taussig
Nov 22 '18 at 1:18
@N.F.Taussig Thanks, I have fixed it.
– LuminousNutria
Nov 22 '18 at 1:25
add a comment |
I need to find $sin frac{theta}{2}$ when $sin theta = frac{3}{5}$, and $0^{circ} < theta < 90^{circ}$
The answer I'm currently getting is $sqrt{frac{1}{10}}$, but the answer must be either $frac{sqrt{10}}{10}$, $frac{3 sqrt{10}}{10}$, or $frac{1}{3}$.
My Process:
since, $sin frac{theta}{2} = pm sqrt{frac{1 - cos theta}{2}}$
$sqrt{ frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}} = sqrt{frac{1}{10}}$
What am I doing wrong?
algebra-precalculus trigonometry
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
I need to find $sin frac{theta}{2}$ when $sin theta = frac{3}{5}$, and $0^{circ} < theta < 90^{circ}$
The answer I'm currently getting is $sqrt{frac{1}{10}}$, but the answer must be either $frac{sqrt{10}}{10}$, $frac{3 sqrt{10}}{10}$, or $frac{1}{3}$.
My Process:
since, $sin frac{theta}{2} = pm sqrt{frac{1 - cos theta}{2}}$
$sqrt{ frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}} = sqrt{frac{1}{10}}$
What am I doing wrong?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
edited Nov 22 '18 at 1:24
LuminousNutria
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
asked Nov 22 '18 at 1:11
LuminousNutriaLuminousNutria
1709
1709
1
Notational error: You should have $sqrt{frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}}$.
– N. F. Taussig
Nov 22 '18 at 1:18
@N.F.Taussig Thanks, I have fixed it.
– LuminousNutria
Nov 22 '18 at 1:25
add a comment |
1
Notational error: You should have $sqrt{frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}}$.
– N. F. Taussig
Nov 22 '18 at 1:18
@N.F.Taussig Thanks, I have fixed it.
– LuminousNutria
Nov 22 '18 at 1:25
1
1
Notational error: You should have $sqrt{frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}}$.
– N. F. Taussig
Nov 22 '18 at 1:18
Notational error: You should have $sqrt{frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}}$.
– N. F. Taussig
Nov 22 '18 at 1:18
@N.F.Taussig Thanks, I have fixed it.
– LuminousNutria
Nov 22 '18 at 1:25
@N.F.Taussig Thanks, I have fixed it.
– LuminousNutria
Nov 22 '18 at 1:25
add a comment |
2 Answers
2
active
oldest
votes
...well
$$ frac{sqrt{10}}{10} = frac{sqrt{10}}{sqrt{10} times sqrt{10}} color{red}{=} frac 1{sqrt{10}}$$
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
add a comment |
I had forgotten that, $sqrt{frac{1}{10}} = frac{sqrt{1}}{sqrt{10}} = frac{1}{sqrt{10}} cdot (frac{sqrt{10}}{sqrt{10}}) = frac{sqrt{10}}{10}$
So, there is my answer.
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
add a comment |
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2 Answers
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active
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2 Answers
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active
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...well
$$ frac{sqrt{10}}{10} = frac{sqrt{10}}{sqrt{10} times sqrt{10}} color{red}{=} frac 1{sqrt{10}}$$
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
add a comment |
...well
$$ frac{sqrt{10}}{10} = frac{sqrt{10}}{sqrt{10} times sqrt{10}} color{red}{=} frac 1{sqrt{10}}$$
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
add a comment |
...well
$$ frac{sqrt{10}}{10} = frac{sqrt{10}}{sqrt{10} times sqrt{10}} color{red}{=} frac 1{sqrt{10}}$$
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
...well
$$ frac{sqrt{10}}{10} = frac{sqrt{10}}{sqrt{10} times sqrt{10}} color{red}{=} frac 1{sqrt{10}}$$
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
answered Nov 22 '18 at 1:15
Jaideep KhareJaideep Khare
17.7k32568
17.7k32568
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
add a comment |
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
Thank you, I realized what I should do about a second after posting.
– LuminousNutria
Nov 22 '18 at 1:18
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
@LuminousNutria Oh Great! Often happens with me too :P
– Jaideep Khare
Nov 22 '18 at 23:35
add a comment |
I had forgotten that, $sqrt{frac{1}{10}} = frac{sqrt{1}}{sqrt{10}} = frac{1}{sqrt{10}} cdot (frac{sqrt{10}}{sqrt{10}}) = frac{sqrt{10}}{10}$
So, there is my answer.
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
add a comment |
I had forgotten that, $sqrt{frac{1}{10}} = frac{sqrt{1}}{sqrt{10}} = frac{1}{sqrt{10}} cdot (frac{sqrt{10}}{sqrt{10}}) = frac{sqrt{10}}{10}$
So, there is my answer.
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
add a comment |
I had forgotten that, $sqrt{frac{1}{10}} = frac{sqrt{1}}{sqrt{10}} = frac{1}{sqrt{10}} cdot (frac{sqrt{10}}{sqrt{10}}) = frac{sqrt{10}}{10}$
So, there is my answer.
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
I had forgotten that, $sqrt{frac{1}{10}} = frac{sqrt{1}}{sqrt{10}} = frac{1}{sqrt{10}} cdot (frac{sqrt{10}}{sqrt{10}}) = frac{sqrt{10}}{10}$
So, there is my answer.
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
answered Nov 22 '18 at 1:18
LuminousNutriaLuminousNutria
1709
1709
add a comment |
add a comment |
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1
Notational error: You should have $sqrt{frac{1 - frac{4}{5}}{2}} cdot sqrt{frac{5}{5}} = sqrt{frac{5 - 4}{10}}$.
– N. F. Taussig
Nov 22 '18 at 1:18
@N.F.Taussig Thanks, I have fixed it.
– LuminousNutria
Nov 22 '18 at 1:25