Why does sin(x) - sin(y) equal this? The Next CEO of Stack OverflowProve that $sin(2A)+sin(2B)+sin(2C)=4sin(A)sin(B)sin(C)$ when $A,B,C$ are angles of a triangleWhy $sin(pi)$ sometimes equal to $0$?Understanding expanding trig identitiesWhy does this always equal $1$?When does this equation $cos(alpha + beta) = cos(alpha) + cos(beta)$ hold?Solve $ cos 2x - sin x +1=0$Writing equation in terms of sin and cosSolve Trigonometric Equality, Multiple Angle TrigonometryFinding relationships between angles, a, b and c when $sin a - sin b - sin c = 0$Does $sin^2x-cos^2x$ equal $cos(2x)$
Does the Idaho Potato Commission associate potato skins with healthy eating?
How dangerous is XSS
How can a day be of 24 hours?
Car headlights in a world without electricity
Is a distribution that is normal, but highly skewed, considered Gaussian?
Gödel's incompleteness theorems - what are the religious implications?
MT "will strike" & LXX "will watch carefully" (Gen 3:15)?
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Mathematica command that allows it to read my intentions
Does Germany produce more waste than the US?
What difference does it make matching a word with/without a trailing whitespace?
Is it reasonable to ask other researchers to send me their previous grant applications?
How to implement Comparable so it is consistent with identity-equality
How to compactly explain secondary and tertiary characters without resorting to stereotypes?
pgfplots: How to draw a tangent graph below two others?
Strange use of "whether ... than ..." in official text
Could you use a laser beam as a modulated carrier wave for radio signal?
Can I hook these wires up to find the connection to a dead outlet?
That's an odd coin - I wonder why
What did the word "leisure" mean in late 18th Century usage?
How to coordinate airplane tickets?
Was the Stack Exchange "Happy April Fools" page fitting with the 90s code?
Shortening a title without changing its meaning
Upgrading From a 9 Speed Sora Derailleur?
Why does sin(x) - sin(y) equal this?
The Next CEO of Stack OverflowProve that $sin(2A)+sin(2B)+sin(2C)=4sin(A)sin(B)sin(C)$ when $A,B,C$ are angles of a triangleWhy $sin(pi)$ sometimes equal to $0$?Understanding expanding trig identitiesWhy does this always equal $1$?When does this equation $cos(alpha + beta) = cos(alpha) + cos(beta)$ hold?Solve $ cos 2x - sin x +1=0$Writing equation in terms of sin and cosSolve Trigonometric Equality, Multiple Angle TrigonometryFinding relationships between angles, a, b and c when $sin a - sin b - sin c = 0$Does $sin^2x-cos^2x$ equal $cos(2x)$
$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
add a comment |
$begingroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Why does this equality hold?
$sin x - sin y = 2 cos(fracx+y2) sin(fracx-y2)$.
My professor was saying that since
(i) $sin(A+B)=sin A cos B+ sin B cos A$
and
(ii) $sin(A-B) = sin A cos B - sin B cos A$
we just let $A=fracx+y2$ and $B=fracx-y2$. But I tried to write this out and could not figure it out. Any help would be appreciated
real-analysis analysis trigonometry
real-analysis analysis trigonometry
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ryan DuranRyan Duran
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Ryan DuranRyan Duran
111
asked 2 hours ago
Ryan DuranRyan Duran
111
111
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ryan Duran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
add a comment |
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered 1 hour ago
AdmuthAdmuth
685
$endgroup$
add a comment |
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
);
);
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3171404%2fwhy-does-sinx-siny-equal-this%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
add a comment |
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
add a comment |
$begingroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered 1 hour ago
MarianDMarianD
2,0831617
$endgroup$
The main trick is here:
beginalign
colorred x = x+yover2 + x-yover2\[1em]
colorbluey = x+yover2 - x-yover2
endalign
(You may evaluate the right-hand sides of them to verify that these strange equations are correct.)
Substituting the right-hand sides for $colorredx$ and $colorbluey,,$ you will obtain
beginalign
sin colorred x - sin colorblue y = sin left(colorredx+yover2 + x-yover2 right) - sin left(colorblue x+yover2 - x-yover2 right) \[1em]
endalign
All the rest is then only a routine calculation:
beginalign
requireenclose
&= sin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-left[sin left(x+yover2right) cosleft( x-yover2 right) -
sin left(x-yover2right) cosleft( x+yover2 right)right]\[3em]
&= encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)\[1em]
&-encloseupdiagonalstrikesin left(x+yover2right) cosleft( x-yover2 right) +
sin left(x-yover2right) cosleft( x+yover2 right)
\[3em]
&=2sin left(x-yover2right) cosleft( x+yover2 right)\
endalign
answered 1 hour ago
MarianDMarianD
2,0831617
edited 1 hour ago
answered 1 hour ago
MarianDMarianD
2,0831617
answered 1 hour ago
MarianDMarianD
2,0831617
answered 1 hour ago
MarianDMarianD
2,0831617
2,0831617
add a comment |
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
add a comment |
$begingroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered 2 hours ago
John DoeJohn Doe
11.4k11239
$endgroup$
Following your professor's advice, let $A=fracx+y2$, $B=fracx-y2$. Then $$x=A+B\y=A-B$$So the LHS of your equation becomes $$sin(A+B)-sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2cos Asin B$ as required.
answered 2 hours ago
John DoeJohn Doe
11.4k11239
answered 2 hours ago
John DoeJohn Doe
11.4k11239
answered 2 hours ago
John DoeJohn Doe
11.4k11239
answered 2 hours ago
John DoeJohn Doe
11.4k11239
11.4k11239
add a comment |
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered 1 hour ago
AdmuthAdmuth
685
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered 1 hour ago
AdmuthAdmuth
685
$endgroup$
add a comment |
$begingroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered 1 hour ago
AdmuthAdmuth
685
$endgroup$
Following your notation, let $A=dfracx+y2$ and $B=dfracx-y2$.
Note that $A+B=x$ and $A-B=y$.
Now, $sin x=sin(A+B)=sin Acos B+cos Asin B$ and $sin y=sin(A-B)=sin Acos B - cos Asin B$ from your professor's advice.
To get the LHS, $sin x-sin y = 2cos Asin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
answered 1 hour ago
AdmuthAdmuth
685
answered 1 hour ago
AdmuthAdmuth
685
answered 1 hour ago
AdmuthAdmuth
685
answered 1 hour ago
AdmuthAdmuth
685
685
add a comment |
add a comment |
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Duran is a new contributor. Be nice, and check out our Code of Conduct.
Ryan Duran 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.
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%2f3171404%2fwhy-does-sinx-siny-equal-this%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
$begingroup$
Let A and B be as you defined. Then $sin(A+B)=sin(fracx+y2+fracx-y2)$. Evaluate this and use the given identities.
$endgroup$
– Newman
2 hours ago
$begingroup$
After substituting for A and B in the equations (i) and (ii) you have to calculate (i) - (ii)
$endgroup$
– R_D
2 hours ago