Mixing
Double Angle Formulas: Finding $tan 2theta$
I am trying to find $tan 2theta$ where $sin theta = frac{5}{13}$ and $theta$ is in Quadrant One.
According to my textbook, $tan 2theta = frac{120}{119}$, but I get $frac{-10}{13}$ instead.
The Identity I am using:
$$tan 2theta = frac{2 tan theta}{1 - tan^{2}theta}$$
My Process:
Since $y = 5;$ and $r = 13,; x = 12.$
Apply Tangent Double Angle Formula:
$$frac{2(frac{5}{12})}{1 - (frac{5}{12})^2}$$
$$frac{frac{10}{12}}{1 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{10}{12-25}$$
$$frac{-10}{13}$$
What am I doing wrong?
algebra-precalculus trigonometry
edited Nov 22 '18 at 9:34
user376343
2,9132823
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
add a comment |
I am trying to find $tan 2theta$ where $sin theta = frac{5}{13}$ and $theta$ is in Quadrant One.
According to my textbook, $tan 2theta = frac{120}{119}$, but I get $frac{-10}{13}$ instead.
The Identity I am using:
$$tan 2theta = frac{2 tan theta}{1 - tan^{2}theta}$$
My Process:
Since $y = 5;$ and $r = 13,; x = 12.$
Apply Tangent Double Angle Formula:
$$frac{2(frac{5}{12})}{1 - (frac{5}{12})^2}$$
$$frac{frac{10}{12}}{1 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{10}{12-25}$$
$$frac{-10}{13}$$
What am I doing wrong?
algebra-precalculus trigonometry
edited Nov 22 '18 at 9:34
user376343
2,9132823
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
4
$(5/12)^2$ is not $25/12$. It is $25/144$.
– Nick
Nov 22 '18 at 0:12
Erm, that's probably my issue then.
– LuminousNutria
Nov 22 '18 at 0:12
add a comment |
I am trying to find $tan 2theta$ where $sin theta = frac{5}{13}$ and $theta$ is in Quadrant One.
According to my textbook, $tan 2theta = frac{120}{119}$, but I get $frac{-10}{13}$ instead.
The Identity I am using:
$$tan 2theta = frac{2 tan theta}{1 - tan^{2}theta}$$
My Process:
Since $y = 5;$ and $r = 13,; x = 12.$
Apply Tangent Double Angle Formula:
$$frac{2(frac{5}{12})}{1 - (frac{5}{12})^2}$$
$$frac{frac{10}{12}}{1 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{10}{12-25}$$
$$frac{-10}{13}$$
What am I doing wrong?
algebra-precalculus trigonometry
edited Nov 22 '18 at 9:34
user376343
2,9132823
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
I am trying to find $tan 2theta$ where $sin theta = frac{5}{13}$ and $theta$ is in Quadrant One.
According to my textbook, $tan 2theta = frac{120}{119}$, but I get $frac{-10}{13}$ instead.
The Identity I am using:
$$tan 2theta = frac{2 tan theta}{1 - tan^{2}theta}$$
My Process:
Since $y = 5;$ and $r = 13,; x = 12.$
Apply Tangent Double Angle Formula:
$$frac{2(frac{5}{12})}{1 - (frac{5}{12})^2}$$
$$frac{frac{10}{12}}{1 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{10}{12-25}$$
$$frac{-10}{13}$$
What am I doing wrong?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Nov 22 '18 at 9:34
user376343
2,9132823
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
edited Nov 22 '18 at 9:34
user376343
2,9132823
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
edited Nov 22 '18 at 9:34
user376343
2,9132823
edited Nov 22 '18 at 9:34
user376343
2,9132823
edited Nov 22 '18 at 9:34
user376343
2,9132823
2,9132823
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
asked Nov 22 '18 at 0:09
LuminousNutriaLuminousNutria
1709
1709
4
$(5/12)^2$ is not $25/12$. It is $25/144$.
– Nick
Nov 22 '18 at 0:12
Erm, that's probably my issue then.
– LuminousNutria
Nov 22 '18 at 0:12
add a comment |
4
$(5/12)^2$ is not $25/12$. It is $25/144$.
– Nick
Nov 22 '18 at 0:12
Erm, that's probably my issue then.
– LuminousNutria
Nov 22 '18 at 0:12
4
4
$(5/12)^2$ is not $25/12$. It is $25/144$.
– Nick
Nov 22 '18 at 0:12
$(5/12)^2$ is not $25/12$. It is $25/144$.
– Nick
Nov 22 '18 at 0:12
Erm, that's probably my issue then.
– LuminousNutria
Nov 22 '18 at 0:12
Erm, that's probably my issue then.
– LuminousNutria
Nov 22 '18 at 0:12
add a comment |
2 Answers
2
active
oldest
votes
Alternatively:
$sintheta = 5/13$, implies $costheta = sqrt{1-(5/13)^2} = 12/13$.
$sin 2theta = 2sin theta cos theta = 120/169$.
$cos 2theta = 2 cos^2 theta-1 = 1 - 2 sin^2 theta = 119/169$.
This gives $tan 2theta = sin 2theta/ cos 2 theta = 120/119$.
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
add a comment |
I made a mistake solving the problem. $(frac{5}{12})^2 neq frac{25}{12}$.
Actually, $(frac{5}{12})^2 = frac{25}{144}$.
Taking that into account:
$$frac{frac{10}{12}}{1 - frac{25}{144}} cdot frac{12}{12}$$
$$frac{10}{12-frac{300}{144}}$$
$$frac{10}{12 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{120}{144-25}$$
$$frac{120}{119}$$
Therefore, $tan 2theta = frac{120}{119}$
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
Alternatively:
$sintheta = 5/13$, implies $costheta = sqrt{1-(5/13)^2} = 12/13$.
$sin 2theta = 2sin theta cos theta = 120/169$.
$cos 2theta = 2 cos^2 theta-1 = 1 - 2 sin^2 theta = 119/169$.
This gives $tan 2theta = sin 2theta/ cos 2 theta = 120/119$.
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
add a comment |
Alternatively:
$sintheta = 5/13$, implies $costheta = sqrt{1-(5/13)^2} = 12/13$.
$sin 2theta = 2sin theta cos theta = 120/169$.
$cos 2theta = 2 cos^2 theta-1 = 1 - 2 sin^2 theta = 119/169$.
This gives $tan 2theta = sin 2theta/ cos 2 theta = 120/119$.
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
add a comment |
Alternatively:
$sintheta = 5/13$, implies $costheta = sqrt{1-(5/13)^2} = 12/13$.
$sin 2theta = 2sin theta cos theta = 120/169$.
$cos 2theta = 2 cos^2 theta-1 = 1 - 2 sin^2 theta = 119/169$.
This gives $tan 2theta = sin 2theta/ cos 2 theta = 120/119$.
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
Alternatively:
$sintheta = 5/13$, implies $costheta = sqrt{1-(5/13)^2} = 12/13$.
$sin 2theta = 2sin theta cos theta = 120/169$.
$cos 2theta = 2 cos^2 theta-1 = 1 - 2 sin^2 theta = 119/169$.
This gives $tan 2theta = sin 2theta/ cos 2 theta = 120/119$.
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
answered Nov 22 '18 at 8:23
Aditya DuaAditya Dua
86418
86418
add a comment |
add a comment |
I made a mistake solving the problem. $(frac{5}{12})^2 neq frac{25}{12}$.
Actually, $(frac{5}{12})^2 = frac{25}{144}$.
Taking that into account:
$$frac{frac{10}{12}}{1 - frac{25}{144}} cdot frac{12}{12}$$
$$frac{10}{12-frac{300}{144}}$$
$$frac{10}{12 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{120}{144-25}$$
$$frac{120}{119}$$
Therefore, $tan 2theta = frac{120}{119}$
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
add a comment |
I made a mistake solving the problem. $(frac{5}{12})^2 neq frac{25}{12}$.
Actually, $(frac{5}{12})^2 = frac{25}{144}$.
Taking that into account:
$$frac{frac{10}{12}}{1 - frac{25}{144}} cdot frac{12}{12}$$
$$frac{10}{12-frac{300}{144}}$$
$$frac{10}{12 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{120}{144-25}$$
$$frac{120}{119}$$
Therefore, $tan 2theta = frac{120}{119}$
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
add a comment |
I made a mistake solving the problem. $(frac{5}{12})^2 neq frac{25}{12}$.
Actually, $(frac{5}{12})^2 = frac{25}{144}$.
Taking that into account:
$$frac{frac{10}{12}}{1 - frac{25}{144}} cdot frac{12}{12}$$
$$frac{10}{12-frac{300}{144}}$$
$$frac{10}{12 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{120}{144-25}$$
$$frac{120}{119}$$
Therefore, $tan 2theta = frac{120}{119}$
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
I made a mistake solving the problem. $(frac{5}{12})^2 neq frac{25}{12}$.
Actually, $(frac{5}{12})^2 = frac{25}{144}$.
Taking that into account:
$$frac{frac{10}{12}}{1 - frac{25}{144}} cdot frac{12}{12}$$
$$frac{10}{12-frac{300}{144}}$$
$$frac{10}{12 - frac{25}{12}} cdot frac{12}{12}$$
$$frac{120}{144-25}$$
$$frac{120}{119}$$
Therefore, $tan 2theta = frac{120}{119}$
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
answered Nov 22 '18 at 0:29
LuminousNutriaLuminousNutria
1709
1709
add a comment |
add a comment |
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4
$(5/12)^2$ is not $25/12$. It is $25/144$.
– Nick
Nov 22 '18 at 0:12
Erm, that's probably my issue then.
– LuminousNutria
Nov 22 '18 at 0:12