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Derivation of the tangent half angle identity
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I'm having trouble proceeding from
$$fracsin(theta)1+cos(theta)$$
to
$$tanleft(fractheta2right)$$
Context:
Consider the function $f$ defined for all $(x,y)$ such that $y neq 0$, with the rule $$f(x,y) = fracysqrtx^2+y^2+x$$
Show that $$f(rcos(theta),rsin(theta)) = tanleft(fractheta2right)$$
So far I've done:
$$f(rcos(theta),rsin(theta)) = fracrsin(theta)sqrtr^2cos^2(theta) + r^2sin^2(theta)+rcos(theta) = fracrsin(theta)sqrtr^2+rcos(theta)\=fracsin(theta)1+cos(theta)$$
Using $$cos(theta) = 2cos^2left(fractheta2right)-1 implies 2cos^2left(fractheta2right)=cos(theta)+1$$
We get $$f(rcos(theta),rsin(theta)) = fracsin(theta)2cos^2left(fractheta2right)$$
But I can't see how to proceed from here to the required result. Thanks in advance for any help!
algebra-precalculus multivariable-calculus trigonometry
asked 32 mins ago
Patrick Jankowski
281113
add a comment |Â
up vote
1
down vote
favorite
I'm having trouble proceeding from
$$fracsin(theta)1+cos(theta)$$
to
$$tanleft(fractheta2right)$$
Context:
Consider the function $f$ defined for all $(x,y)$ such that $y neq 0$, with the rule $$f(x,y) = fracysqrtx^2+y^2+x$$
Show that $$f(rcos(theta),rsin(theta)) = tanleft(fractheta2right)$$
So far I've done:
$$f(rcos(theta),rsin(theta)) = fracrsin(theta)sqrtr^2cos^2(theta) + r^2sin^2(theta)+rcos(theta) = fracrsin(theta)sqrtr^2+rcos(theta)\=fracsin(theta)1+cos(theta)$$
Using $$cos(theta) = 2cos^2left(fractheta2right)-1 implies 2cos^2left(fractheta2right)=cos(theta)+1$$
We get $$f(rcos(theta),rsin(theta)) = fracsin(theta)2cos^2left(fractheta2right)$$
But I can't see how to proceed from here to the required result. Thanks in advance for any help!
algebra-precalculus multivariable-calculus trigonometry
asked 32 mins ago
Patrick Jankowski
281113
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having trouble proceeding from
$$fracsin(theta)1+cos(theta)$$
to
$$tanleft(fractheta2right)$$
Context:
Consider the function $f$ defined for all $(x,y)$ such that $y neq 0$, with the rule $$f(x,y) = fracysqrtx^2+y^2+x$$
Show that $$f(rcos(theta),rsin(theta)) = tanleft(fractheta2right)$$
So far I've done:
$$f(rcos(theta),rsin(theta)) = fracrsin(theta)sqrtr^2cos^2(theta) + r^2sin^2(theta)+rcos(theta) = fracrsin(theta)sqrtr^2+rcos(theta)\=fracsin(theta)1+cos(theta)$$
Using $$cos(theta) = 2cos^2left(fractheta2right)-1 implies 2cos^2left(fractheta2right)=cos(theta)+1$$
We get $$f(rcos(theta),rsin(theta)) = fracsin(theta)2cos^2left(fractheta2right)$$
But I can't see how to proceed from here to the required result. Thanks in advance for any help!
algebra-precalculus multivariable-calculus trigonometry
asked 32 mins ago
Patrick Jankowski
281113
I'm having trouble proceeding from
$$fracsin(theta)1+cos(theta)$$
to
$$tanleft(fractheta2right)$$
Context:
Consider the function $f$ defined for all $(x,y)$ such that $y neq 0$, with the rule $$f(x,y) = fracysqrtx^2+y^2+x$$
Show that $$f(rcos(theta),rsin(theta)) = tanleft(fractheta2right)$$
So far I've done:
$$f(rcos(theta),rsin(theta)) = fracrsin(theta)sqrtr^2cos^2(theta) + r^2sin^2(theta)+rcos(theta) = fracrsin(theta)sqrtr^2+rcos(theta)\=fracsin(theta)1+cos(theta)$$
Using $$cos(theta) = 2cos^2left(fractheta2right)-1 implies 2cos^2left(fractheta2right)=cos(theta)+1$$
We get $$f(rcos(theta),rsin(theta)) = fracsin(theta)2cos^2left(fractheta2right)$$
But I can't see how to proceed from here to the required result. Thanks in advance for any help!
algebra-precalculus multivariable-calculus trigonometry
algebra-precalculus multivariable-calculus trigonometry
asked 32 mins ago
Patrick Jankowski
281113
asked 32 mins ago
Patrick Jankowski
281113
asked 32 mins ago
Patrick Jankowski
281113
asked 32 mins ago
Patrick Jankowski
281113
asked 32 mins ago
Patrick Jankowski
281113
281113
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
Hint: The numerator can be written as
$$ sintheta = sin left(2 cdot fractheta2right) = 2sinfractheta2cosfractheta2. $$
answered 30 mins ago
Sobi
3,6021520
2
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
1
Thank you for that, have a nice day!
– Sobi
27 mins ago
add a comment |Â
up vote
4
down vote
It is a lot easier to use some trig. identities: $sin (theta) =2 sin (theta /2) cos (theta /2)$ and $1+cos theta =2cos ^2 (theta /2)$. You will immediately get the result from these two formulas.
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
add a comment |Â
up vote
0
down vote
$ tan (theta/2)= dfrac sin(theta/2)cos(theta/2)=$
$dfrac 2sin (theta/2)(cos(theta/2) 2cos^2(theta/2)=$
$dfracsin (theta)2cos^2 (theta/2)= dfracsin theta1+cos (theta)$
Identities:
1) $sin (theta) =2 sin (theta/2)cos (theta/2)$
2) $cos theta = cos^2 (theta/2)-sin^2 (theta/2)$
answered 14 mins ago
Peter Szilas
9,6362720
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint: The numerator can be written as
$$ sintheta = sin left(2 cdot fractheta2right) = 2sinfractheta2cosfractheta2. $$
answered 30 mins ago
Sobi
3,6021520
2
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
1
Thank you for that, have a nice day!
– Sobi
27 mins ago
add a comment |Â
up vote
4
down vote
accepted
Hint: The numerator can be written as
$$ sintheta = sin left(2 cdot fractheta2right) = 2sinfractheta2cosfractheta2. $$
answered 30 mins ago
Sobi
3,6021520
2
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
1
Thank you for that, have a nice day!
– Sobi
27 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint: The numerator can be written as
$$ sintheta = sin left(2 cdot fractheta2right) = 2sinfractheta2cosfractheta2. $$
answered 30 mins ago
Sobi
3,6021520
Hint: The numerator can be written as
$$ sintheta = sin left(2 cdot fractheta2right) = 2sinfractheta2cosfractheta2. $$
answered 30 mins ago
Sobi
3,6021520
answered 30 mins ago
Sobi
3,6021520
answered 30 mins ago
Sobi
3,6021520
answered 30 mins ago
Sobi
3,6021520
3,6021520
2
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
1
Thank you for that, have a nice day!
– Sobi
27 mins ago
add a comment |Â
2
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
1
Thank you for that, have a nice day!
– Sobi
27 mins ago
2
2
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
beautiful, thanks man! so quick too
– Patrick Jankowski
29 mins ago
1
1
Thank you for that, have a nice day!
– Sobi
27 mins ago
Thank you for that, have a nice day!
– Sobi
27 mins ago
add a comment |Â
up vote
4
down vote
It is a lot easier to use some trig. identities: $sin (theta) =2 sin (theta /2) cos (theta /2)$ and $1+cos theta =2cos ^2 (theta /2)$. You will immediately get the result from these two formulas.
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
add a comment |Â
up vote
4
down vote
It is a lot easier to use some trig. identities: $sin (theta) =2 sin (theta /2) cos (theta /2)$ and $1+cos theta =2cos ^2 (theta /2)$. You will immediately get the result from these two formulas.
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It is a lot easier to use some trig. identities: $sin (theta) =2 sin (theta /2) cos (theta /2)$ and $1+cos theta =2cos ^2 (theta /2)$. You will immediately get the result from these two formulas.
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
It is a lot easier to use some trig. identities: $sin (theta) =2 sin (theta /2) cos (theta /2)$ and $1+cos theta =2cos ^2 (theta /2)$. You will immediately get the result from these two formulas.
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
answered 29 mins ago
Kavi Rama Murthy
34.4k31745
34.4k31745
add a comment |Â
add a comment |Â
up vote
0
down vote
$ tan (theta/2)= dfrac sin(theta/2)cos(theta/2)=$
$dfrac 2sin (theta/2)(cos(theta/2) 2cos^2(theta/2)=$
$dfracsin (theta)2cos^2 (theta/2)= dfracsin theta1+cos (theta)$
Identities:
1) $sin (theta) =2 sin (theta/2)cos (theta/2)$
2) $cos theta = cos^2 (theta/2)-sin^2 (theta/2)$
answered 14 mins ago
Peter Szilas
9,6362720
add a comment |Â
up vote
0
down vote
$ tan (theta/2)= dfrac sin(theta/2)cos(theta/2)=$
$dfrac 2sin (theta/2)(cos(theta/2) 2cos^2(theta/2)=$
$dfracsin (theta)2cos^2 (theta/2)= dfracsin theta1+cos (theta)$
Identities:
1) $sin (theta) =2 sin (theta/2)cos (theta/2)$
2) $cos theta = cos^2 (theta/2)-sin^2 (theta/2)$
answered 14 mins ago
Peter Szilas
9,6362720
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$ tan (theta/2)= dfrac sin(theta/2)cos(theta/2)=$
$dfrac 2sin (theta/2)(cos(theta/2) 2cos^2(theta/2)=$
$dfracsin (theta)2cos^2 (theta/2)= dfracsin theta1+cos (theta)$
Identities:
1) $sin (theta) =2 sin (theta/2)cos (theta/2)$
2) $cos theta = cos^2 (theta/2)-sin^2 (theta/2)$
answered 14 mins ago
Peter Szilas
9,6362720
$ tan (theta/2)= dfrac sin(theta/2)cos(theta/2)=$
$dfrac 2sin (theta/2)(cos(theta/2) 2cos^2(theta/2)=$
$dfracsin (theta)2cos^2 (theta/2)= dfracsin theta1+cos (theta)$
Identities:
1) $sin (theta) =2 sin (theta/2)cos (theta/2)$
2) $cos theta = cos^2 (theta/2)-sin^2 (theta/2)$
answered 14 mins ago
Peter Szilas
9,6362720
edited 6 mins ago
answered 14 mins ago
Peter Szilas
9,6362720
answered 14 mins ago
Peter Szilas
9,6362720
answered 14 mins ago
Peter Szilas
9,6362720
9,6362720
add a comment |Â
add a comment |Â
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