$cos(theta-phi)=frac2aba^2+b^2$ where $a=sin(theta)+cos(phi)$ and $b=cos(theta)+sin(phi)$
Clash Royale CLAN TAG#URR8PPP
$begingroup$
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
edited Mar 18 at 8:17
user21820
40.3k544163
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
$endgroup$
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
edited Mar 18 at 8:17
user21820
40.3k544163
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
edited Mar 18 at 8:17
user21820
40.3k544163
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
$endgroup$
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 $cos(theta-phi)=frac2aba^2+b^2$.
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
algebra-precalculus trigonometry
edited Mar 18 at 8:17
user21820
40.3k544163
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
edited Mar 18 at 8:17
user21820
40.3k544163
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
edited Mar 18 at 8:17
user21820
40.3k544163
edited Mar 18 at 8:17
user21820
40.3k544163
edited Mar 18 at 8:17
user21820
40.3k544163
40.3k544163
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
asked Mar 17 at 23:04
Avinash ShastriAvinash Shastri
264
264
$begingroup$
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 |
$begingroup$
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
$begingroup$
Welcome to Math Stack Exchange. Do you know the formula for cosine of a difference?
$endgroup$
– J. W. Tanner
Mar 17 at 23:10
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
2 Answers
2
active
oldest
votes
$begingroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$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$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered Mar 17 at 23:53
StAKmodStAKmod
481111
$endgroup$
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 |
$begingroup$
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
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$
answered Mar 18 at 8:27
SongSong
18.6k21651
$endgroup$
add a comment |
Your Answer
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3152200%2fcos-theta-phi-frac2aba2b2-where-a-sin-theta-cos-phi-and-b%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$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$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered Mar 17 at 23:53
StAKmodStAKmod
481111
$endgroup$
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$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered Mar 17 at 23:53
StAKmodStAKmod
481111
$endgroup$
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$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered Mar 17 at 23:53
StAKmodStAKmod
481111
$endgroup$
$$(i).a=sin(theta)+cos(phi)$$
$$(ii).b=cos(theta)+sin(phi)$$
$$(i)^2+(ii)^2=2+2sin(theta +phi)$$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$$cos(theta-phi)=abover 1+sin(theta+phi)=2abover a^2+b^2$$
answered Mar 17 at 23:53
StAKmodStAKmod
481111
answered Mar 17 at 23:53
StAKmodStAKmod
481111
answered Mar 17 at 23:53
StAKmodStAKmod
481111
answered Mar 17 at 23:53
StAKmodStAKmod
481111
481111
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 |
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
2
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
$begingroup$
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
$begingroup$
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 |
$begingroup$
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
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$
answered Mar 18 at 8:27
SongSong
18.6k21651
$endgroup$
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
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$
answered Mar 18 at 8:27
SongSong
18.6k21651
$endgroup$
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
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$
answered Mar 18 at 8:27
SongSong
18.6k21651
$endgroup$
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
$$
frac2aba^2+b^2=2cd = sinleft(fracpi2+theta-phiright)=cos(theta-phi).
$$
answered Mar 18 at 8:27
SongSong
18.6k21651
answered Mar 18 at 8:27
SongSong
18.6k21651
answered Mar 18 at 8:27
SongSong
18.6k21651
answered Mar 18 at 8:27
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3152200%2fcos-theta-phi-frac2aba2b2-where-a-sin-theta-cos-phi-and-b%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
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