where does the $cos(theta)=1-2sin^{2}(theta/2)$ come from?
$begingroup$
I have seen people using $cos(theta)=1-2sin^{2}(theta/2)$ but how do we get it?
trigonometry
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
asked Dec 6 '18 at 10:54
Student123Student123
536
$endgroup$
add a comment |
$begingroup$
I have seen people using $cos(theta)=1-2sin^{2}(theta/2)$ but how do we get it?
trigonometry
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
asked Dec 6 '18 at 10:54
Student123Student123
536
$endgroup$
add a comment |
$begingroup$
I have seen people using $cos(theta)=1-2sin^{2}(theta/2)$ but how do we get it?
trigonometry
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
asked Dec 6 '18 at 10:54
Student123Student123
536
$endgroup$
I have seen people using $cos(theta)=1-2sin^{2}(theta/2)$ but how do we get it?
trigonometry
trigonometry
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
asked Dec 6 '18 at 10:54
Student123Student123
536
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
asked Dec 6 '18 at 10:54
Student123Student123
536
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
edited Dec 6 '18 at 10:59
Tianlalu
3,08621038
3,08621038
asked Dec 6 '18 at 10:54
Student123Student123
536
asked Dec 6 '18 at 10:54
Student123Student123
536
asked Dec 6 '18 at 10:54
Student123Student123
536
536
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
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It's a special case of the compound-angle formula $cos (A+B)=cos Acos B-sin Asin B$. Take $A=B=frac{theta}{2}$ so $costheta=cos^2frac{theta}{2}-sin^2frac{theta}{2}$. This can be written in two equivalent forms using $cos^2frac{theta}{2}+sin^2frac{theta}{2}=1$, one being $1-2sin^2frac{theta}{2}$ (the other is $2cos^2frac{theta}{2}-1$).
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
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add a comment |
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$$
cos(2theta)=1-2sin^2(theta).
$$
Change $thetarightarrowtheta/2$.
answered Dec 6 '18 at 11:00
JonJon
4,42511122
$endgroup$
add a comment |
$begingroup$
Hint: What happens if you use the double-angle identity for cosine and use the substitution $theta to frac{theta}{2}$?
Also, note that
$$cos(apm b) = cos acos bmpsin asin b$$
which allows you to derive the double-angle identity.
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
$endgroup$
add a comment |
$begingroup$
Hint: Use the definition of the half angle formula $$sin frac {theta} 2 = pm sqrt {frac {1 + cos theta}{2}}$$ Square both sides and manipulate as needed.
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
$endgroup$
add a comment |
$begingroup$
De Moivre:$$cos^2alpha-sin^2alpha+2icosalphasinalpha=(cosalpha+isinalpha)^2=cos(2alpha)+2isin(2alpha)$$
So that: $$1-2sin^2alpha=cos^2alpha+sin^2alpha-2sin^2alpha=cos^2alpha-sin^2alpha=cos2alpha$$
Now substitute: $alpha=frac12theta$.
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's a special case of the compound-angle formula $cos (A+B)=cos Acos B-sin Asin B$. Take $A=B=frac{theta}{2}$ so $costheta=cos^2frac{theta}{2}-sin^2frac{theta}{2}$. This can be written in two equivalent forms using $cos^2frac{theta}{2}+sin^2frac{theta}{2}=1$, one being $1-2sin^2frac{theta}{2}$ (the other is $2cos^2frac{theta}{2}-1$).
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
$endgroup$
add a comment |
$begingroup$
It's a special case of the compound-angle formula $cos (A+B)=cos Acos B-sin Asin B$. Take $A=B=frac{theta}{2}$ so $costheta=cos^2frac{theta}{2}-sin^2frac{theta}{2}$. This can be written in two equivalent forms using $cos^2frac{theta}{2}+sin^2frac{theta}{2}=1$, one being $1-2sin^2frac{theta}{2}$ (the other is $2cos^2frac{theta}{2}-1$).
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
$endgroup$
add a comment |
$begingroup$
It's a special case of the compound-angle formula $cos (A+B)=cos Acos B-sin Asin B$. Take $A=B=frac{theta}{2}$ so $costheta=cos^2frac{theta}{2}-sin^2frac{theta}{2}$. This can be written in two equivalent forms using $cos^2frac{theta}{2}+sin^2frac{theta}{2}=1$, one being $1-2sin^2frac{theta}{2}$ (the other is $2cos^2frac{theta}{2}-1$).
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
$endgroup$
It's a special case of the compound-angle formula $cos (A+B)=cos Acos B-sin Asin B$. Take $A=B=frac{theta}{2}$ so $costheta=cos^2frac{theta}{2}-sin^2frac{theta}{2}$. This can be written in two equivalent forms using $cos^2frac{theta}{2}+sin^2frac{theta}{2}=1$, one being $1-2sin^2frac{theta}{2}$ (the other is $2cos^2frac{theta}{2}-1$).
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
answered Dec 6 '18 at 11:00
J.G.J.G.
25.7k22540
25.7k22540
add a comment |
add a comment |
$begingroup$
$$
cos(2theta)=1-2sin^2(theta).
$$
Change $thetarightarrowtheta/2$.
answered Dec 6 '18 at 11:00
JonJon
4,42511122
$endgroup$
add a comment |
$begingroup$
$$
cos(2theta)=1-2sin^2(theta).
$$
Change $thetarightarrowtheta/2$.
answered Dec 6 '18 at 11:00
JonJon
4,42511122
$endgroup$
add a comment |
$begingroup$
$$
cos(2theta)=1-2sin^2(theta).
$$
Change $thetarightarrowtheta/2$.
answered Dec 6 '18 at 11:00
JonJon
4,42511122
$endgroup$
$$
cos(2theta)=1-2sin^2(theta).
$$
Change $thetarightarrowtheta/2$.
answered Dec 6 '18 at 11:00
JonJon
4,42511122
answered Dec 6 '18 at 11:00
JonJon
4,42511122
answered Dec 6 '18 at 11:00
JonJon
4,42511122
answered Dec 6 '18 at 11:00
JonJon
4,42511122
4,42511122
add a comment |
add a comment |
$begingroup$
Hint: What happens if you use the double-angle identity for cosine and use the substitution $theta to frac{theta}{2}$?
Also, note that
$$cos(apm b) = cos acos bmpsin asin b$$
which allows you to derive the double-angle identity.
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
$endgroup$
add a comment |
$begingroup$
Hint: What happens if you use the double-angle identity for cosine and use the substitution $theta to frac{theta}{2}$?
Also, note that
$$cos(apm b) = cos acos bmpsin asin b$$
which allows you to derive the double-angle identity.
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
$endgroup$
add a comment |
$begingroup$
Hint: What happens if you use the double-angle identity for cosine and use the substitution $theta to frac{theta}{2}$?
Also, note that
$$cos(apm b) = cos acos bmpsin asin b$$
which allows you to derive the double-angle identity.
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
$endgroup$
Hint: What happens if you use the double-angle identity for cosine and use the substitution $theta to frac{theta}{2}$?
Also, note that
$$cos(apm b) = cos acos bmpsin asin b$$
which allows you to derive the double-angle identity.
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
edited Dec 6 '18 at 11:11
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
answered Dec 6 '18 at 11:00
KM101KM101
5,9251524
5,9251524
add a comment |
add a comment |
$begingroup$
Hint: Use the definition of the half angle formula $$sin frac {theta} 2 = pm sqrt {frac {1 + cos theta}{2}}$$ Square both sides and manipulate as needed.
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
$endgroup$
add a comment |
$begingroup$
Hint: Use the definition of the half angle formula $$sin frac {theta} 2 = pm sqrt {frac {1 + cos theta}{2}}$$ Square both sides and manipulate as needed.
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
$endgroup$
add a comment |
$begingroup$
Hint: Use the definition of the half angle formula $$sin frac {theta} 2 = pm sqrt {frac {1 + cos theta}{2}}$$ Square both sides and manipulate as needed.
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
$endgroup$
Hint: Use the definition of the half angle formula $$sin frac {theta} 2 = pm sqrt {frac {1 + cos theta}{2}}$$ Square both sides and manipulate as needed.
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
answered Dec 6 '18 at 11:06
bjcolby15bjcolby15
1,29711016
1,29711016
add a comment |
add a comment |
$begingroup$
De Moivre:$$cos^2alpha-sin^2alpha+2icosalphasinalpha=(cosalpha+isinalpha)^2=cos(2alpha)+2isin(2alpha)$$
So that: $$1-2sin^2alpha=cos^2alpha+sin^2alpha-2sin^2alpha=cos^2alpha-sin^2alpha=cos2alpha$$
Now substitute: $alpha=frac12theta$.
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
De Moivre:$$cos^2alpha-sin^2alpha+2icosalphasinalpha=(cosalpha+isinalpha)^2=cos(2alpha)+2isin(2alpha)$$
So that: $$1-2sin^2alpha=cos^2alpha+sin^2alpha-2sin^2alpha=cos^2alpha-sin^2alpha=cos2alpha$$
Now substitute: $alpha=frac12theta$.
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
$endgroup$
add a comment |
$begingroup$
De Moivre:$$cos^2alpha-sin^2alpha+2icosalphasinalpha=(cosalpha+isinalpha)^2=cos(2alpha)+2isin(2alpha)$$
So that: $$1-2sin^2alpha=cos^2alpha+sin^2alpha-2sin^2alpha=cos^2alpha-sin^2alpha=cos2alpha$$
Now substitute: $alpha=frac12theta$.
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
$endgroup$
De Moivre:$$cos^2alpha-sin^2alpha+2icosalphasinalpha=(cosalpha+isinalpha)^2=cos(2alpha)+2isin(2alpha)$$
So that: $$1-2sin^2alpha=cos^2alpha+sin^2alpha-2sin^2alpha=cos^2alpha-sin^2alpha=cos2alpha$$
Now substitute: $alpha=frac12theta$.
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
answered Dec 6 '18 at 11:07
drhabdrhab
101k544130
101k544130
add a comment |
add a comment |
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