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 $cos(2x)$
Small nick on power cord from an electric alarm clock, and copper wiring exposed but intact
Is it a bad idea to plug the other end of ESD strap to wall ground?
Is the 21st century's idea of "freedom of speech" based on precedent?
Free fall ellipse or parabola?
Calculate the Mean mean of two numbers
MT "will strike" & LXX "will watch carefully" (Gen 3:15)?
Can Sri Krishna be called 'a person'?
How to find if SQL server backup is encrypted with TDE without restoring the backup
How badly should I try to prevent a user from XSSing themselves?
Is it possible to create a QR code using text?
Car headlights in a world without electricity
"Eavesdropping" vs "Listen in on"
Prodigo = pro + ago?
Strange use of "whether ... than ..." in official text
Could a dragon use hot air to help it take off?
What steps are necessary to read a Modern SSD in Medieval Europe?
What is the difference between 'contrib' and 'non-free' packages repositories?
Could you use a laser beam as a modulated carrier wave for radio signal?
What is a typical Mizrachi Seder like?
How can a day be of 24 hours?
Creating a script with console commands
What happens if you break a law in another country outside of that country?
How to show a landlord what we have in savings?
Why did the Drakh emissary look so blurred in S04:E11 "Lines of Communication"?
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