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If $fracsin^2 xa = fraccos^2 xb$, then prove that they are also equal to $fracsin^2 x + cos^2 xa+b$

0

$begingroup$

If $ dfracsin^2 xa = dfraccos^2 xb$, then prove that they are also equal to $dfracsin^2 x + cos^2 xa+b$.

I don't know how to solve this.

I was thinking about componeto dividento, but it doesn't help either.

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 25 at 8:37

Blue

49.7k870158

asked Mar 25 at 8:29

swarnimswarnim

1409

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $fracsin^2 xa = fraccos^2 xb=k$, then find sine and cosine in terms of $k$ and substitute into the required fraction to show it is also equal to $k$.
  $endgroup$
  – farruhota
  Mar 25 at 8:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See VII: Addendo Property of math-only-math.com/properties-of-ratio-and-proportion.html
  $endgroup$
  – lab bhattacharjee
  Mar 25 at 8:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I restored your statement about what you considered (along with making minor formatting tweaks), as the Math.SE community likes to know that questioners have given thought to their questions. (Of course, you're free to remove it again if it doesn't accurately reflect your thoughts, but if so: please tell what you know about the problem so that answerers can tailor their responses appropriately.)
  $endgroup$
  – Blue
  Mar 25 at 8:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check out my question here - math.meta.stackexchange.com/questions/29995/…
  $endgroup$
  – user619699
  Mar 25 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a general property of proportions, no trigonometry involved.
  $endgroup$
  – Yves Daoust
  Mar 25 at 8:56
  
  

  
  
  
  

add a comment | 

0

$begingroup$

If $ dfracsin^2 xa = dfraccos^2 xb$, then prove that they are also equal to $dfracsin^2 x + cos^2 xa+b$.

I don't know how to solve this.

I was thinking about componeto dividento, but it doesn't help either.

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 25 at 8:37

Blue

49.7k870158

asked Mar 25 at 8:29

swarnimswarnim

1409

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $fracsin^2 xa = fraccos^2 xb=k$, then find sine and cosine in terms of $k$ and substitute into the required fraction to show it is also equal to $k$.
  $endgroup$
  – farruhota
  Mar 25 at 8:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See VII: Addendo Property of math-only-math.com/properties-of-ratio-and-proportion.html
  $endgroup$
  – lab bhattacharjee
  Mar 25 at 8:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I restored your statement about what you considered (along with making minor formatting tweaks), as the Math.SE community likes to know that questioners have given thought to their questions. (Of course, you're free to remove it again if it doesn't accurately reflect your thoughts, but if so: please tell what you know about the problem so that answerers can tailor their responses appropriately.)
  $endgroup$
  – Blue
  Mar 25 at 8:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check out my question here - math.meta.stackexchange.com/questions/29995/…
  $endgroup$
  – user619699
  Mar 25 at 8:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a general property of proportions, no trigonometry involved.
  $endgroup$
  – Yves Daoust
  Mar 25 at 8:56
  
  

  
  
  
  

add a comment | 

$begingroup$

If $ dfracsin^2 xa = dfraccos^2 xb$, then prove that they are also equal to $dfracsin^2 x + cos^2 xa+b$.

I don't know how to solve this.

I was thinking about componeto dividento, but it doesn't help either.

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 25 at 8:37

Blue

49.7k870158

asked Mar 25 at 8:29

swarnimswarnim

1409

$endgroup$

share|cite|improve this question

edited Mar 25 at 8:37

Blue

49.7k870158

asked Mar 25 at 8:29

swarnimswarnim

1409

share|cite|improve this question

edited Mar 25 at 8:37

Blue

49.7k870158

asked Mar 25 at 8:29

swarnimswarnim

1409

edited Mar 25 at 8:37

Blue

49.7k870158

edited Mar 25 at 8:37

Blue

49.7k870158

asked Mar 25 at 8:29

swarnimswarnim

1409

asked Mar 25 at 8:29

swarnimswarnim

1409

3 Answers
3

active

oldest

votes

1

$begingroup$

It is generally true that $frac a b =frac c d$ implies $frac a b =frac c d=frac a+c b+d$. For a proof let $r=frac a b =frac c d$, write $a=rb,c=rd$ and calculate $frac a+c b+d$.

share|cite|improve this answer

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

add a comment | 

1

$begingroup$

We will prove a proposition in general.

Proposition 1. If $a/b=c/d$ then they are also equal to $fracapm cbpm d$.

Proof. We know that $ad=bc$. So
$$bc=ad$$
$$abpm bc= abpm ad$$
$$(apm c)b=a(bpm d)$$
$$(apm c)/(bpm d)=a/b$$
Actually I wrote these equations from bottom to top, and reversed them (since it is easier that way as you can see).

share|cite|improve this answer

answered Mar 25 at 8:40

TreborTrebor

1,01315

$endgroup$

add a comment | 

1

$begingroup$

$$dfracsin^2 xa = dfraccos^2 xbimplies\dfracasin^2 x = dfracbcos^2 ximplies\dfracasin^2 x+dfracbsin^2 x = dfracbsin^2 x+dfracbcos^2 ximplies\a+bover sin^2 x=b(sin^2 x+cos^2 x)over sin^2 xcos^2 x=bover sin^2 xcos^2 ximplies\a+b=bover cos^2 ximplies \sin^2 xover a=cos^2xover b=1over a+b=sin^2x+cos^2xover a+b$$

share|cite|improve this answer

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

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1

$begingroup$

It is generally true that $frac a b =frac c d$ implies $frac a b =frac c d=frac a+c b+d$. For a proof let $r=frac a b =frac c d$, write $a=rb,c=rd$ and calculate $frac a+c b+d$.

share|cite|improve this answer

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

add a comment | 

1

$begingroup$

It is generally true that $frac a b =frac c d$ implies $frac a b =frac c d=frac a+c b+d$. For a proof let $r=frac a b =frac c d$, write $a=rb,c=rd$ and calculate $frac a+c b+d$.

share|cite|improve this answer

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

add a comment | 

$begingroup$

It is generally true that $frac a b =frac c d$ implies $frac a b =frac c d=frac a+c b+d$. For a proof let $r=frac a b =frac c d$, write $a=rb,c=rd$ and calculate $frac a+c b+d$.

share|cite|improve this answer

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

share|cite|improve this answer

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

answered Mar 25 at 8:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

1

$begingroup$

We will prove a proposition in general.

Proposition 1. If $a/b=c/d$ then they are also equal to $fracapm cbpm d$.

Proof. We know that $ad=bc$. So
$$bc=ad$$
$$abpm bc= abpm ad$$
$$(apm c)b=a(bpm d)$$
$$(apm c)/(bpm d)=a/b$$
Actually I wrote these equations from bottom to top, and reversed them (since it is easier that way as you can see).

share|cite|improve this answer

answered Mar 25 at 8:40

TreborTrebor

1,01315

$endgroup$

add a comment | 

1

$begingroup$

We will prove a proposition in general.

Proposition 1. If $a/b=c/d$ then they are also equal to $fracapm cbpm d$.

Proof. We know that $ad=bc$. So
$$bc=ad$$
$$abpm bc= abpm ad$$
$$(apm c)b=a(bpm d)$$
$$(apm c)/(bpm d)=a/b$$
Actually I wrote these equations from bottom to top, and reversed them (since it is easier that way as you can see).

share|cite|improve this answer

answered Mar 25 at 8:40

TreborTrebor

1,01315

$endgroup$

add a comment | 

$begingroup$

We will prove a proposition in general.

Proposition 1. If $a/b=c/d$ then they are also equal to $fracapm cbpm d$.

Proof. We know that $ad=bc$. So
$$bc=ad$$
$$abpm bc= abpm ad$$
$$(apm c)b=a(bpm d)$$
$$(apm c)/(bpm d)=a/b$$
Actually I wrote these equations from bottom to top, and reversed them (since it is easier that way as you can see).

share|cite|improve this answer

answered Mar 25 at 8:40

TreborTrebor

1,01315

$endgroup$

share|cite|improve this answer

answered Mar 25 at 8:40

TreborTrebor

1,01315

answered Mar 25 at 8:40

TreborTrebor

1,01315

answered Mar 25 at 8:40

TreborTrebor

1,01315

1

$begingroup$

$$dfracsin^2 xa = dfraccos^2 xbimplies\dfracasin^2 x = dfracbcos^2 ximplies\dfracasin^2 x+dfracbsin^2 x = dfracbsin^2 x+dfracbcos^2 ximplies\a+bover sin^2 x=b(sin^2 x+cos^2 x)over sin^2 xcos^2 x=bover sin^2 xcos^2 ximplies\a+b=bover cos^2 ximplies \sin^2 xover a=cos^2xover b=1over a+b=sin^2x+cos^2xover a+b$$

share|cite|improve this answer

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

1

$begingroup$

$$dfracsin^2 xa = dfraccos^2 xbimplies\dfracasin^2 x = dfracbcos^2 ximplies\dfracasin^2 x+dfracbsin^2 x = dfracbsin^2 x+dfracbcos^2 ximplies\a+bover sin^2 x=b(sin^2 x+cos^2 x)over sin^2 xcos^2 x=bover sin^2 xcos^2 ximplies\a+b=bover cos^2 ximplies \sin^2 xover a=cos^2xover b=1over a+b=sin^2x+cos^2xover a+b$$

share|cite|improve this answer

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

$begingroup$

$$dfracsin^2 xa = dfraccos^2 xbimplies\dfracasin^2 x = dfracbcos^2 ximplies\dfracasin^2 x+dfracbsin^2 x = dfracbsin^2 x+dfracbcos^2 ximplies\a+bover sin^2 x=b(sin^2 x+cos^2 x)over sin^2 xcos^2 x=bover sin^2 xcos^2 ximplies\a+b=bover cos^2 ximplies \sin^2 xover a=cos^2xover b=1over a+b=sin^2x+cos^2xover a+b$$

share|cite|improve this answer

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 25 at 8:45

Mostafa AyazMostafa Ayaz

18.1k31040