Prove that $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)=tanfracX2tanfracY2$ [on hold]

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Prove that $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)=tanfracX2tanfracY2$

I tried expanding $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)$ and expanding $tanfracX2tanfracY2$ and then cross multiplying , but I can't seem to find the answer.

algebra-precalculus trigonometry

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mampuuu

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Prove that $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)=tanfracX2tanfracY2$

I tried expanding $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)$ and expanding $tanfracX2tanfracY2$ and then cross multiplying , but I can't seem to find the answer.

algebra-precalculus trigonometry

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edited 1 hour ago

user21820

36.2k440140

asked 10 hours ago

mampuuu

654

put on hold as off-topic by user21820, Xander Henderson, amWhy, alans, rschwieb 2 mins ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, amWhy, alans, rschwieb

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up vote
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up vote
1
down vote

favorite

Prove that $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)=tanfracX2tanfracY2$

I tried expanding $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)$ and expanding $tanfracX2tanfracY2$ and then cross multiplying , but I can't seem to find the answer.

algebra-precalculus trigonometry

share|cite|improve this question

edited 1 hour ago

user21820

36.2k440140

asked 10 hours ago

mampuuu

654

Prove that $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)=tanfracX2tanfracY2$

I tried expanding $frac sin X+sin Y-sin(X+Y)sin X+sin Y+sin(X+Y)$ and expanding $tanfracX2tanfracY2$ and then cross multiplying , but I can't seem to find the answer.

algebra-precalculus trigonometry

algebra-precalculus trigonometry

share|cite|improve this question

edited 1 hour ago

user21820

36.2k440140

asked 10 hours ago

mampuuu

654

share|cite|improve this question

edited 1 hour ago

user21820

36.2k440140

asked 10 hours ago

mampuuu

654

share|cite|improve this question

share|cite|improve this question

edited 1 hour ago

user21820

36.2k440140

edited 1 hour ago

user21820

36.2k440140

edited 1 hour ago

user21820

36.2k440140

36.2k440140

asked 10 hours ago

mampuuu

654

asked 10 hours ago

mampuuu

654

asked 10 hours ago

mampuuu

654

654

put on hold as off-topic by user21820, Xander Henderson, amWhy, alans, rschwieb 2 mins ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, amWhy, alans, rschwieb

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by user21820, Xander Henderson, amWhy, alans, rschwieb 2 mins ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Xander Henderson, amWhy, alans, rschwieb

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4 Answers
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up vote
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accepted

Write $t=tan(x/2)$ and $u=tan(y/2)$.
Then
$$sin x=frac2t1+t^2,$$
$$sin y=frac2u1+u^2$$
and
$$sin(x+y)=sin xcos y+cos xsin y=frac2t(1-u^2)+2u(1-t^2)(1+t^2)(1+u^2).$$
Therefore
$$sin x+sin ypmsin(x+y)
=frac2t(1+u^2)+2u(1+t^2)pmleft(2t(1-u^2)+2u(1-t^2)right)(1+t^2)(1+u^2).$$
Now expand out both possibilities.

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answered 10 hours ago

Lord Shark the Unknown

89.1k955115

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up vote
6
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Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to
$$
fracsin(fracX+Y2)cos(fracX-Y2)-sin(fracX+Y2)cos(fracX+Y2)sin(fracX+Y2)cos(fracX-Y2)+sin(fracX+Y2)cos(fracX+Y2)
$$
cancelling sin((X+Y)/2) and use sum-to-product again,
$$
=
fracsinfracX2sinfracY2cosfracX2cosfracY2
$$
which is the RHS.

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answered 10 hours ago

user10354138

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  – lab bhattacharjee
  10 hours ago
  

  
  
  
  

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up vote
4
down vote

Denote: $x=a+b, y=a-b$, then:
$$frac sin (a+b)+sin (a-b)-sin(2a)sin (a+b)+sin (a-b)+sin(2a)=\
fracsin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b-sin (2a)sin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b+sin (2a)=\
frac2sin acos b-2sin acos a2sin acos b+2sin acos a=\
fraccos b-cos acos b+cos a=\
frac-sin fracb-a2sin fracb+a2cos fracb-a2cos fracb+a2=\
frac-sin (-frac y2)sin (frac x2)cos (-frac y2)cos (frac x2)=\
tanfracX2tanfracY2.$$

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answered 8 hours ago

farruhota

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beginalign*fracsin x+sin y-sin(x+y)sin x+sin y+sin (x+y)&=dfrac2sindfracx+y2cosdfracx-y2-2sindfracx+y2cos dfracx+y22sindfracx+y2cosdfracx-y2+2sindfracx+y2cos dfracx+y2\&=dfraccosdfracx-y2-cos dfracx+y2cosdfracx-y2+cos dfracx+y2\&=dfrac-2sindfracx2sindfrac-y22cosdfracx2cos dfrac-y2\&=tandfracx2tandfracy2.endalign*

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answered 9 hours ago

mengdie1982

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Write $t=tan(x/2)$ and $u=tan(y/2)$.
Then
$$sin x=frac2t1+t^2,$$
$$sin y=frac2u1+u^2$$
and
$$sin(x+y)=sin xcos y+cos xsin y=frac2t(1-u^2)+2u(1-t^2)(1+t^2)(1+u^2).$$
Therefore
$$sin x+sin ypmsin(x+y)
=frac2t(1+u^2)+2u(1+t^2)pmleft(2t(1-u^2)+2u(1-t^2)right)(1+t^2)(1+u^2).$$
Now expand out both possibilities.

share|cite|improve this answer

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

add a comment | 

up vote
4
down vote



accepted

Write $t=tan(x/2)$ and $u=tan(y/2)$.
Then
$$sin x=frac2t1+t^2,$$
$$sin y=frac2u1+u^2$$
and
$$sin(x+y)=sin xcos y+cos xsin y=frac2t(1-u^2)+2u(1-t^2)(1+t^2)(1+u^2).$$
Therefore
$$sin x+sin ypmsin(x+y)
=frac2t(1+u^2)+2u(1+t^2)pmleft(2t(1-u^2)+2u(1-t^2)right)(1+t^2)(1+u^2).$$
Now expand out both possibilities.

share|cite|improve this answer

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Write $t=tan(x/2)$ and $u=tan(y/2)$.
Then
$$sin x=frac2t1+t^2,$$
$$sin y=frac2u1+u^2$$
and
$$sin(x+y)=sin xcos y+cos xsin y=frac2t(1-u^2)+2u(1-t^2)(1+t^2)(1+u^2).$$
Therefore
$$sin x+sin ypmsin(x+y)
=frac2t(1+u^2)+2u(1+t^2)pmleft(2t(1-u^2)+2u(1-t^2)right)(1+t^2)(1+u^2).$$
Now expand out both possibilities.

share|cite|improve this answer

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

Write $t=tan(x/2)$ and $u=tan(y/2)$.
Then
$$sin x=frac2t1+t^2,$$
$$sin y=frac2u1+u^2$$
and
$$sin(x+y)=sin xcos y+cos xsin y=frac2t(1-u^2)+2u(1-t^2)(1+t^2)(1+u^2).$$
Therefore
$$sin x+sin ypmsin(x+y)
=frac2t(1+u^2)+2u(1+t^2)pmleft(2t(1-u^2)+2u(1-t^2)right)(1+t^2)(1+u^2).$$
Now expand out both possibilities.

share|cite|improve this answer

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

share|cite|improve this answer

share|cite|improve this answer

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

answered 10 hours ago

Lord Shark the Unknown

89.1k955115

89.1k955115

add a comment | 

add a comment | 

up vote
6
down vote

Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to
$$
fracsin(fracX+Y2)cos(fracX-Y2)-sin(fracX+Y2)cos(fracX+Y2)sin(fracX+Y2)cos(fracX-Y2)+sin(fracX+Y2)cos(fracX+Y2)
$$
cancelling sin((X+Y)/2) and use sum-to-product again,
$$
=
fracsinfracX2sinfracY2cosfracX2cosfracY2
$$
which is the RHS.

share|cite|improve this answer

answered 10 hours ago

user10354138

4243

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  – lab bhattacharjee
  10 hours ago
  

  
  
  
  

add a comment | 

up vote
6
down vote

Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to
$$
fracsin(fracX+Y2)cos(fracX-Y2)-sin(fracX+Y2)cos(fracX+Y2)sin(fracX+Y2)cos(fracX-Y2)+sin(fracX+Y2)cos(fracX+Y2)
$$
cancelling sin((X+Y)/2) and use sum-to-product again,
$$
=
fracsinfracX2sinfracY2cosfracX2cosfracY2
$$
which is the RHS.

share|cite|improve this answer

answered 10 hours ago

user10354138

4243

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  – lab bhattacharjee
  10 hours ago
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to
$$
fracsin(fracX+Y2)cos(fracX-Y2)-sin(fracX+Y2)cos(fracX+Y2)sin(fracX+Y2)cos(fracX-Y2)+sin(fracX+Y2)cos(fracX+Y2)
$$
cancelling sin((X+Y)/2) and use sum-to-product again,
$$
=
fracsinfracX2sinfracY2cosfracX2cosfracY2
$$
which is the RHS.

share|cite|improve this answer

answered 10 hours ago

user10354138

4243

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Start with the LHS, use sum-to-product on sin(X)+sin(Y), and the double angle formula on sin(X+Y), the expression is equal to
$$
fracsin(fracX+Y2)cos(fracX-Y2)-sin(fracX+Y2)cos(fracX+Y2)sin(fracX+Y2)cos(fracX-Y2)+sin(fracX+Y2)cos(fracX+Y2)
$$
cancelling sin((X+Y)/2) and use sum-to-product again,
$$
=
fracsinfracX2sinfracY2cosfracX2cosfracY2
$$
which is the RHS.

share|cite|improve this answer

answered 10 hours ago

user10354138

4243

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this answer

share|cite|improve this answer

answered 10 hours ago

user10354138

4243

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 10 hours ago

user10354138

4243

answered 10 hours ago

user10354138

4243

4243

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

user10354138 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  – lab bhattacharjee
  10 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  – lab bhattacharjee
  10 hours ago
  

  
  
  
  

mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
10 hours ago

mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
10 hours ago

add a comment | 

up vote
4
down vote

Denote: $x=a+b, y=a-b$, then:
$$frac sin (a+b)+sin (a-b)-sin(2a)sin (a+b)+sin (a-b)+sin(2a)=\
fracsin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b-sin (2a)sin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b+sin (2a)=\
frac2sin acos b-2sin acos a2sin acos b+2sin acos a=\
fraccos b-cos acos b+cos a=\
frac-sin fracb-a2sin fracb+a2cos fracb-a2cos fracb+a2=\
frac-sin (-frac y2)sin (frac x2)cos (-frac y2)cos (frac x2)=\
tanfracX2tanfracY2.$$

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answered 8 hours ago

farruhota

15.2k2734

add a comment | 

up vote
4
down vote

Denote: $x=a+b, y=a-b$, then:
$$frac sin (a+b)+sin (a-b)-sin(2a)sin (a+b)+sin (a-b)+sin(2a)=\
fracsin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b-sin (2a)sin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b+sin (2a)=\
frac2sin acos b-2sin acos a2sin acos b+2sin acos a=\
fraccos b-cos acos b+cos a=\
frac-sin fracb-a2sin fracb+a2cos fracb-a2cos fracb+a2=\
frac-sin (-frac y2)sin (frac x2)cos (-frac y2)cos (frac x2)=\
tanfracX2tanfracY2.$$

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answered 8 hours ago

farruhota

15.2k2734

add a comment | 

up vote
4
down vote

up vote
4
down vote

Denote: $x=a+b, y=a-b$, then:
$$frac sin (a+b)+sin (a-b)-sin(2a)sin (a+b)+sin (a-b)+sin(2a)=\
fracsin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b-sin (2a)sin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b+sin (2a)=\
frac2sin acos b-2sin acos a2sin acos b+2sin acos a=\
fraccos b-cos acos b+cos a=\
frac-sin fracb-a2sin fracb+a2cos fracb-a2cos fracb+a2=\
frac-sin (-frac y2)sin (frac x2)cos (-frac y2)cos (frac x2)=\
tanfracX2tanfracY2.$$

share|cite|improve this answer

answered 8 hours ago

farruhota

15.2k2734

Denote: $x=a+b, y=a-b$, then:
$$frac sin (a+b)+sin (a-b)-sin(2a)sin (a+b)+sin (a-b)+sin(2a)=\
fracsin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b-sin (2a)sin acos b+requirecancelcancelcos asin b+sin acos b-requirecancelcancelcos asin b+sin (2a)=\
frac2sin acos b-2sin acos a2sin acos b+2sin acos a=\
fraccos b-cos acos b+cos a=\
frac-sin fracb-a2sin fracb+a2cos fracb-a2cos fracb+a2=\
frac-sin (-frac y2)sin (frac x2)cos (-frac y2)cos (frac x2)=\
tanfracX2tanfracY2.$$

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answered 8 hours ago

farruhota

15.2k2734

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share|cite|improve this answer

answered 8 hours ago

farruhota

15.2k2734

answered 8 hours ago

farruhota

15.2k2734

answered 8 hours ago

farruhota

15.2k2734

15.2k2734

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up vote
1
down vote

beginalign*fracsin x+sin y-sin(x+y)sin x+sin y+sin (x+y)&=dfrac2sindfracx+y2cosdfracx-y2-2sindfracx+y2cos dfracx+y22sindfracx+y2cosdfracx-y2+2sindfracx+y2cos dfracx+y2\&=dfraccosdfracx-y2-cos dfracx+y2cosdfracx-y2+cos dfracx+y2\&=dfrac-2sindfracx2sindfrac-y22cosdfracx2cos dfrac-y2\&=tandfracx2tandfracy2.endalign*

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answered 9 hours ago

mengdie1982

3,683216

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up vote
1
down vote

beginalign*fracsin x+sin y-sin(x+y)sin x+sin y+sin (x+y)&=dfrac2sindfracx+y2cosdfracx-y2-2sindfracx+y2cos dfracx+y22sindfracx+y2cosdfracx-y2+2sindfracx+y2cos dfracx+y2\&=dfraccosdfracx-y2-cos dfracx+y2cosdfracx-y2+cos dfracx+y2\&=dfrac-2sindfracx2sindfrac-y22cosdfracx2cos dfrac-y2\&=tandfracx2tandfracy2.endalign*

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answered 9 hours ago

mengdie1982

3,683216

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up vote
1
down vote

up vote
1
down vote

beginalign*fracsin x+sin y-sin(x+y)sin x+sin y+sin (x+y)&=dfrac2sindfracx+y2cosdfracx-y2-2sindfracx+y2cos dfracx+y22sindfracx+y2cosdfracx-y2+2sindfracx+y2cos dfracx+y2\&=dfraccosdfracx-y2-cos dfracx+y2cosdfracx-y2+cos dfracx+y2\&=dfrac-2sindfracx2sindfrac-y22cosdfracx2cos dfrac-y2\&=tandfracx2tandfracy2.endalign*

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answered 9 hours ago

mengdie1982

3,683216

beginalign*fracsin x+sin y-sin(x+y)sin x+sin y+sin (x+y)&=dfrac2sindfracx+y2cosdfracx-y2-2sindfracx+y2cos dfracx+y22sindfracx+y2cosdfracx-y2+2sindfracx+y2cos dfracx+y2\&=dfraccosdfracx-y2-cos dfracx+y2cosdfracx-y2+cos dfracx+y2\&=dfrac-2sindfracx2sindfrac-y22cosdfracx2cos dfrac-y2\&=tandfracx2tandfracy2.endalign*

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answered 9 hours ago

mengdie1982

3,683216

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share|cite|improve this answer

answered 9 hours ago

mengdie1982

3,683216

answered 9 hours ago

mengdie1982

3,683216

answered 9 hours ago

mengdie1982

3,683216

3,683216

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