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$cos(theta-phi)=frac2aba^2+b^2$ where $a=sin(theta)+cos(phi)$ and $b=cos(theta)+sin(phi)$

If $sintheta + sinphi = a$ and $costheta + cosphi = b$, then $sin(theta+phi) = ???$If $ sin alpha + sin beta = a $ and $ cos alpha + cos beta = b $ , then show that $sin(alpha + beta) = frac 2ab a^2 + b^2 $If $sintheta+sinphi=a$ and $costheta+ cosphi=b$, then find $tan fractheta-phi2$.How to solve $3 - 2 cos theta - 4 sin theta - cos 2theta + sin 2theta = 0$Solution to $bsin(theta)cos(phi)+acos(theta)sin(phi)=0$ for $phi$use De Moivre's theorem to show that: $cos(3theta) = 4cos(3theta)-3cos(theta)$ and $sin(3theta) = 3sin(theta) - 4sin(3theta)$Prove $fracsintheta-costheta+1sintheta+costheta-1=frac1+sinthetacostheta$Prove that $fracsin(x+theta)sin(x+phi) = cos(theta - phi) +cot(x+phi)sin(theta-phi)$.If $asin^2theta+bcos^2theta=m, acos^2phi+bsin^2phi=n, atantheta=btanphi$, prove that $frac1n+frac1m=frac1a+frac1b$Trig Identity Proof $frac1 + sinthetacostheta + fraccostheta1 - sintheta = 2tanleft(fractheta2 + fracpi4right)$Find $cos2theta+cos2phi$, given $sintheta + sinphi = a$ and $costheta+cosphi = b$If $x costheta+ysintheta=a$ and $xsintheta-ycostheta=b$, then $tantheta=fracbx+ayax-by$. (Math Olympiad)

4

2

$begingroup$

I'm really stuck trying to answer this question and have spent endless hours doing so.

If $a=sin(theta)+cos(phi)$ and $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.

I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.

I've also tried going $ab=ldots$ and then trying to get it from there, that didn't come to fruition either.

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 18 at 8:17

user21820

39.8k544158

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
  $endgroup$
  – J. W. Tanner
  Mar 17 at 23:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
  $endgroup$
  – Avinash Shastri
  Mar 17 at 23:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
  $endgroup$
  – lab bhattacharjee
  Mar 18 at 1:17
  

  
  
  
  

add a comment | 

4

2

$begingroup$

I'm really stuck trying to answer this question and have spent endless hours doing so.

If $a=sin(theta)+cos(phi)$ and $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.

I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.

I've also tried going $ab=ldots$ and then trying to get it from there, that didn't come to fruition either.

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 18 at 8:17

user21820

39.8k544158

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
  $endgroup$
  – J. W. Tanner
  Mar 17 at 23:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, $cos(theta-phi)=cos(theta)cos(phi)+sin(theta)sin(phi)$, and I went much further but was not able to poduce anything useful in the sense of the proof.
  $endgroup$
  – Avinash Shastri
  Mar 17 at 23:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
  $endgroup$
  – lab bhattacharjee
  Mar 18 at 1:17
  

  
  
  
  

add a comment | 

$begingroup$

I'm really stuck trying to answer this question and have spent endless hours doing so.

If $a=sin(theta)+cos(phi)$ and $b=cos(theta)+sin(phi)$, prove that $cos(theta-phi)=frac2aba^2+b^2$.

I've tried working LHS to RHS and couldn't get, I've also tried RHS to LHS and still couldn't get it, and advice or help would be much appreciated please.

I've also tried going $ab=ldots$ and then trying to get it from there, that didn't come to fruition either.

algebra-precalculus trigonometry

share|cite|improve this question

edited Mar 18 at 8:17

user21820

39.8k544158

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

$endgroup$

share|cite|improve this question

edited Mar 18 at 8:17

user21820

39.8k544158

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

share|cite|improve this question

edited Mar 18 at 8:17

user21820

39.8k544158

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

edited Mar 18 at 8:17

user21820

39.8k544158

edited Mar 18 at 8:17

user21820

39.8k544158

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

asked Mar 17 at 23:04

Avinash ShastriAvinash Shastri

264

2 Answers
2

active

oldest

votes

6

$begingroup$

$$(i).a=sin(theta)+cos(phi)$$

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$

share|cite|improve this answer

answered Mar 17 at 23:53

StAKmodStAKmod

458111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
  $endgroup$
  – Avinash Shastri
  Mar 17 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One possible intuition: we need $colorblueab$ and $colorbluea^2 + b^2$ in the final thing we are calculating ($frac2aba^2 + b^2$), so it makes sense that we should examine what these are and see if dividing them gets us to what we want.
  $endgroup$
  – Minus One-Twelfth
  Mar 18 at 8:24
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Here's a different approach, possibly more geometric one.

Let $P$ and $Q$ be on the unit circle with $angle POX=theta$ and $angle QOX=fracpi2-phi$, respectively. Plot $A=(a,b)$ on the plane so that $vecOA=vecOP+vecOQ$, and let $B=(c,d)$ be the intersection of the unit circle and the half line $oversetlongrightarrowOA$. Then it holds $$
c=fracasqrta^2+b^2,quad d=fracbsqrta^2+b^2.
$$ Now, since the quadrilateral $OPAQ$ is a Rhombus, $overlineOA$ is a angle bisector of $angle POQ$, which implies that $angle BOX=fracpi4+fractheta-phi2$. This gives $$(c,d)=left(cosleft(fracpi4+fractheta-phi2right),sinleft(fracpi4+fractheta-phi2right)right),$$ hence it follows by double angle formula
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$

share|cite|improve this answer

answered Mar 18 at 8:27

SongSong

18.5k21651

$endgroup$

add a comment | 

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6

$begingroup$

$$(i).a=sin(theta)+cos(phi)$$

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$

share|cite|improve this answer

answered Mar 17 at 23:53

StAKmodStAKmod

458111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
  $endgroup$
  – Avinash Shastri
  Mar 17 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One possible intuition: we need $colorblueab$ and $colorbluea^2 + b^2$ in the final thing we are calculating ($frac2aba^2 + b^2$), so it makes sense that we should examine what these are and see if dividing them gets us to what we want.
  $endgroup$
  – Minus One-Twelfth
  Mar 18 at 8:24
  
  

  
  
  
  

add a comment | 

6

$begingroup$

$$(i).a=sin(theta)+cos(phi)$$

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$

share|cite|improve this answer

answered Mar 17 at 23:53

StAKmodStAKmod

458111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, thank you, you've made it seem so simple. Was the path intuitive or how did you decide to choose this path?
  $endgroup$
  – Avinash Shastri
  Mar 17 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One possible intuition: we need $colorblueab$ and $colorbluea^2 + b^2$ in the final thing we are calculating ($frac2aba^2 + b^2$), so it makes sense that we should examine what these are and see if dividing them gets us to what we want.
  $endgroup$
  – Minus One-Twelfth
  Mar 18 at 8:24
  
  

  
  
  
  

add a comment | 

$begingroup$

$$(i).a=sin(theta)+cos(phi)$$

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$so
$$sin(theta+phi) =(a^2+b^2)over 2-1$$.
$$(i)*(ii)=sin(2theta)+sin(2phi) over 2+cos(theta-phi)=sin(theta+phi)cos(theta-phi)+cos(theta-phi)$$
so$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$

share|cite|improve this answer

answered Mar 17 at 23:53

StAKmodStAKmod

458111

$endgroup$

share|cite|improve this answer

answered Mar 17 at 23:53

StAKmodStAKmod

458111

answered Mar 17 at 23:53

StAKmodStAKmod

458111

answered Mar 17 at 23:53

StAKmodStAKmod

458111

1

$begingroup$

Here's a different approach, possibly more geometric one.

Let $P$ and $Q$ be on the unit circle with $angle POX=theta$ and $angle QOX=fracpi2-phi$, respectively. Plot $A=(a,b)$ on the plane so that $vecOA=vecOP+vecOQ$, and let $B=(c,d)$ be the intersection of the unit circle and the half line $oversetlongrightarrowOA$. Then it holds $$
c=fracasqrta^2+b^2,quad d=fracbsqrta^2+b^2.
$$ Now, since the quadrilateral $OPAQ$ is a Rhombus, $overlineOA$ is a angle bisector of $angle POQ$, which implies that $angle BOX=fracpi4+fractheta-phi2$. This gives $$(c,d)=left(cosleft(fracpi4+fractheta-phi2right),sinleft(fracpi4+fractheta-phi2right)right),$$ hence it follows by double angle formula
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$

share|cite|improve this answer

answered Mar 18 at 8:27

SongSong

18.5k21651

$endgroup$

add a comment | 

1

$begingroup$

Here's a different approach, possibly more geometric one.

Let $P$ and $Q$ be on the unit circle with $angle POX=theta$ and $angle QOX=fracpi2-phi$, respectively. Plot $A=(a,b)$ on the plane so that $vecOA=vecOP+vecOQ$, and let $B=(c,d)$ be the intersection of the unit circle and the half line $oversetlongrightarrowOA$. Then it holds $$
c=fracasqrta^2+b^2,quad d=fracbsqrta^2+b^2.
$$ Now, since the quadrilateral $OPAQ$ is a Rhombus, $overlineOA$ is a angle bisector of $angle POQ$, which implies that $angle BOX=fracpi4+fractheta-phi2$. This gives $$(c,d)=left(cosleft(fracpi4+fractheta-phi2right),sinleft(fracpi4+fractheta-phi2right)right),$$ hence it follows by double angle formula
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$

share|cite|improve this answer

answered Mar 18 at 8:27

SongSong

18.5k21651

$endgroup$

add a comment | 

$begingroup$

Here's a different approach, possibly more geometric one.

Let $P$ and $Q$ be on the unit circle with $angle POX=theta$ and $angle QOX=fracpi2-phi$, respectively. Plot $A=(a,b)$ on the plane so that $vecOA=vecOP+vecOQ$, and let $B=(c,d)$ be the intersection of the unit circle and the half line $oversetlongrightarrowOA$. Then it holds $$
c=fracasqrta^2+b^2,quad d=fracbsqrta^2+b^2.
$$ Now, since the quadrilateral $OPAQ$ is a Rhombus, $overlineOA$ is a angle bisector of $angle POQ$, which implies that $angle BOX=fracpi4+fractheta-phi2$. This gives $$(c,d)=left(cosleft(fracpi4+fractheta-phi2right),sinleft(fracpi4+fractheta-phi2right)right),$$ hence it follows by double angle formula
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$

share|cite|improve this answer

answered Mar 18 at 8:27

SongSong

18.5k21651

$endgroup$

share|cite|improve this answer

answered Mar 18 at 8:27

SongSong

18.5k21651

answered Mar 18 at 8:27

SongSong

18.5k21651

answered Mar 18 at 8:27

SongSong

18.5k21651