Help prove this basic trig identity please!












4












$begingroup$


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 $cos(theta-phi)=frac{2ab}{a^2+b^2}$.



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







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New contributor




Avinash Shastri is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    4 hours ago










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
















4












$begingroup$


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 $cos(theta-phi)=frac{2ab}{a^2+b^2}$.



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









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
    4 hours ago










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














4












4








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 $cos(theta-phi)=frac{2ab}{a^2+b^2}$.



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









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.







$endgroup$




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 $cos(theta-phi)=frac{2ab}{a^2+b^2}$.



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




trigonometry






share|cite|improve this question









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









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




share|cite|improve this question








edited 27 mins ago









wchargin

1,1161025




1,1161025






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




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.





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.






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












  • $begingroup$
    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
    4 hours ago










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


















  • $begingroup$
    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
    4 hours ago










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
















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




$begingroup$
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
4 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
4 hours ago












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




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










1 Answer
1






active

oldest

votes


















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









$endgroup$









  • 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











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









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









$endgroup$









  • 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
















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









$endgroup$









  • 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














5












5








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









$endgroup$



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



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












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









StAKmodStAKmod

416110




416110








  • 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














  • 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








1




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




$begingroup$
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










Avinash Shastri is a new contributor. Be nice, and check out our Code of Conduct.










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