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







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
    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
















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
    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














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 59 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
    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


















  • $begingroup$
    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
















$begingroup$
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
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




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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






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










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3152200%2fhelp-prove-this-basic-trig-identity-please%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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.










draft saved

draft discarded


















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













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












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
















Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3152200%2fhelp-prove-this-basic-trig-identity-please%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Knooppunt Holsloot

Image
Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
Read more

Altaar (religie)

Image
Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
Read more

Gregoriusmis

Image
Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
Read more