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Help prove this basic trig identity please!

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 $Trig equation help pleaseHelp With Double Angles And Trig Identity ProblemProve Trig IdentityHow to solve this trigonometric identity?Prove the following Trig Identity with reciprocalsHelp needed in verifying a trigonometric identityHow do I prove this seemingly simple trigonometric identityTrig Identity Proof $frac1 + sinthetacostheta + fraccostheta1 - sintheta = 2tanleft(fractheta2 + fracpi4right)$If $x costheta+ysintheta=a$ and $xsintheta-ycostheta=b$, then $tantheta=fracbx+ayax-by$. (Math Olympiad)Precalc Trig Identity, verify: $1 + cos(x) + cos(2x) = frac 12 + fracsin(5x/2)2sin(x/2)$

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.

trigonometry

share|cite|improve this question

edited 58 mins ago

wchargin

1,1161025

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

New contributor

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

$endgroup$

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  $begingroup$
  Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
  $endgroup$
  – J. W. Tanner
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
  $endgroup$
  – lab bhattacharjee
  3 hours ago
  

  
  
  
  

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.

trigonometry

share|cite|improve this question

edited 58 mins ago

wchargin

1,1161025

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
  $endgroup$
  – J. W. Tanner
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Set $theta=dfracpi2-psi$ and use math.stackexchange.com/questions/1833153/… or math.stackexchange.com/questions/2021356/…
  $endgroup$
  – lab bhattacharjee
  3 hours ago
  

  
  
  
  

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.

trigonometry

share|cite|improve this question

edited 58 mins ago

wchargin

1,1161025

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

New contributor

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

$endgroup$

share|cite|improve this question

edited 58 mins ago

wchargin

1,1161025

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

New contributor

Avinash Shastri 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 question

edited 58 mins ago

wchargin

1,1161025

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

New contributor

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

edited 58 mins ago

wchargin

1,1161025

edited 58 mins ago

wchargin

1,1161025

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

New contributor

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

asked 5 hours ago

Avinash ShastriAvinash Shastri

234

1 Answer
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5

$begingroup$

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

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$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 4 hours ago

StAKmodStAKmod

416110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

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

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$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 4 hours ago

StAKmodStAKmod

416110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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

$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$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 4 hours ago

StAKmodStAKmod

416110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 hours ago
  

  
  
  
  

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 4 hours ago

StAKmodStAKmod

416110

$endgroup$

share|cite|improve this answer

answered 4 hours ago

StAKmodStAKmod

416110

answered 4 hours ago

StAKmodStAKmod

416110

answered 4 hours ago

StAKmodStAKmod

416110