Mixing
How can $2cos(x-frac{pi}2) = -2sin(x-frac{pi}2)$
$begingroup$
How can $2cos(x-dfrac{pi}2) = -2sin(x-dfrac{pi}2)$
I know that $cos(-x) = cos(x)$ and that $cos(dfrac{pi}2-x) = sin(x)$
From these two formulas I can get
$1.$ $2cos(x-dfrac{pi}2) = 2cos(dfrac{pi}2-x)$
$2.$ $2cos(dfrac{pi}2-x) = 2 sin(x)$
$3.$ $ 2sin(x) = -2sin(-x)$
$4.$ How can I get $-2sin(-x) = -2sin(x-dfrac{pi}2)$
It was part of calculation of limit:
$lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$
and from there in the next step we assumed the identity I mentioned above
algebra-precalculus trigonometry
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
asked Jan 27 at 10:42
cris14cris14
1338
$endgroup$
add a comment |
$begingroup$
How can $2cos(x-dfrac{pi}2) = -2sin(x-dfrac{pi}2)$
I know that $cos(-x) = cos(x)$ and that $cos(dfrac{pi}2-x) = sin(x)$
From these two formulas I can get
$1.$ $2cos(x-dfrac{pi}2) = 2cos(dfrac{pi}2-x)$
$2.$ $2cos(dfrac{pi}2-x) = 2 sin(x)$
$3.$ $ 2sin(x) = -2sin(-x)$
$4.$ How can I get $-2sin(-x) = -2sin(x-dfrac{pi}2)$
It was part of calculation of limit:
$lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$
and from there in the next step we assumed the identity I mentioned above
algebra-precalculus trigonometry
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
asked Jan 27 at 10:42
cris14cris14
1338
$endgroup$
1
$begingroup$
This equality is not true in general: $$-2sin(x-pi/2)=2sin(pi/2-x)=2cos xne LHS=2sin x$$It is only true for $x=npi+pi/4,ninBbb Z$.
$endgroup$
– Shubham Johri
Jan 27 at 10:47
1
$begingroup$
This is not an identity working for all $x$, $x=frac{pi}{2}$ trivially does not work. But rather an equation you need to solve
$endgroup$
– Thomas Lesgourgues
Jan 27 at 10:50
1
$begingroup$
Perhaps you had to solve an equation? This is obviously not an identity: just input $;x=0;$ to get a contradiction...
$endgroup$
– DonAntonio
Jan 27 at 10:52
$begingroup$
It was part of calculation of limit: $lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$ and from there in the next step we assumed the identity I mentioned in my original post
$endgroup$
– cris14
Jan 27 at 11:04
add a comment |
$begingroup$
How can $2cos(x-dfrac{pi}2) = -2sin(x-dfrac{pi}2)$
I know that $cos(-x) = cos(x)$ and that $cos(dfrac{pi}2-x) = sin(x)$
From these two formulas I can get
$1.$ $2cos(x-dfrac{pi}2) = 2cos(dfrac{pi}2-x)$
$2.$ $2cos(dfrac{pi}2-x) = 2 sin(x)$
$3.$ $ 2sin(x) = -2sin(-x)$
$4.$ How can I get $-2sin(-x) = -2sin(x-dfrac{pi}2)$
It was part of calculation of limit:
$lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$
and from there in the next step we assumed the identity I mentioned above
algebra-precalculus trigonometry
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
asked Jan 27 at 10:42
cris14cris14
1338
$endgroup$
How can $2cos(x-dfrac{pi}2) = -2sin(x-dfrac{pi}2)$
I know that $cos(-x) = cos(x)$ and that $cos(dfrac{pi}2-x) = sin(x)$
From these two formulas I can get
$1.$ $2cos(x-dfrac{pi}2) = 2cos(dfrac{pi}2-x)$
$2.$ $2cos(dfrac{pi}2-x) = 2 sin(x)$
$3.$ $ 2sin(x) = -2sin(-x)$
$4.$ How can I get $-2sin(-x) = -2sin(x-dfrac{pi}2)$
It was part of calculation of limit:
$lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$
and from there in the next step we assumed the identity I mentioned above
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
asked Jan 27 at 10:42
cris14cris14
1338
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
asked Jan 27 at 10:42
cris14cris14
1338
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
edited Jan 29 at 20:08
Martin Sleziak
44.9k10122276
44.9k10122276
asked Jan 27 at 10:42
cris14cris14
1338
asked Jan 27 at 10:42
cris14cris14
1338
asked Jan 27 at 10:42
cris14cris14
1338
1338
1
$begingroup$
This equality is not true in general: $$-2sin(x-pi/2)=2sin(pi/2-x)=2cos xne LHS=2sin x$$It is only true for $x=npi+pi/4,ninBbb Z$.
$endgroup$
– Shubham Johri
Jan 27 at 10:47
1
$begingroup$
This is not an identity working for all $x$, $x=frac{pi}{2}$ trivially does not work. But rather an equation you need to solve
$endgroup$
– Thomas Lesgourgues
Jan 27 at 10:50
1
$begingroup$
Perhaps you had to solve an equation? This is obviously not an identity: just input $;x=0;$ to get a contradiction...
$endgroup$
– DonAntonio
Jan 27 at 10:52
$begingroup$
It was part of calculation of limit: $lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$ and from there in the next step we assumed the identity I mentioned in my original post
$endgroup$
– cris14
Jan 27 at 11:04
add a comment |
1
$begingroup$
This equality is not true in general: $$-2sin(x-pi/2)=2sin(pi/2-x)=2cos xne LHS=2sin x$$It is only true for $x=npi+pi/4,ninBbb Z$.
$endgroup$
– Shubham Johri
Jan 27 at 10:47
1
$begingroup$
This is not an identity working for all $x$, $x=frac{pi}{2}$ trivially does not work. But rather an equation you need to solve
$endgroup$
– Thomas Lesgourgues
Jan 27 at 10:50
1
$begingroup$
Perhaps you had to solve an equation? This is obviously not an identity: just input $;x=0;$ to get a contradiction...
$endgroup$
– DonAntonio
Jan 27 at 10:52
$begingroup$
It was part of calculation of limit: $lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$ and from there in the next step we assumed the identity I mentioned in my original post
$endgroup$
– cris14
Jan 27 at 11:04
1
1
$begingroup$
This equality is not true in general: $$-2sin(x-pi/2)=2sin(pi/2-x)=2cos xne LHS=2sin x$$It is only true for $x=npi+pi/4,ninBbb Z$.
$endgroup$
– Shubham Johri
Jan 27 at 10:47
$begingroup$
This equality is not true in general: $$-2sin(x-pi/2)=2sin(pi/2-x)=2cos xne LHS=2sin x$$It is only true for $x=npi+pi/4,ninBbb Z$.
$endgroup$
– Shubham Johri
Jan 27 at 10:47
1
1
$begingroup$
This is not an identity working for all $x$, $x=frac{pi}{2}$ trivially does not work. But rather an equation you need to solve
$endgroup$
– Thomas Lesgourgues
Jan 27 at 10:50
$begingroup$
This is not an identity working for all $x$, $x=frac{pi}{2}$ trivially does not work. But rather an equation you need to solve
$endgroup$
– Thomas Lesgourgues
Jan 27 at 10:50
1
1
$begingroup$
Perhaps you had to solve an equation? This is obviously not an identity: just input $;x=0;$ to get a contradiction...
$endgroup$
– DonAntonio
Jan 27 at 10:52
$begingroup$
Perhaps you had to solve an equation? This is obviously not an identity: just input $;x=0;$ to get a contradiction...
$endgroup$
– DonAntonio
Jan 27 at 10:52
$begingroup$
It was part of calculation of limit: $lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$ and from there in the next step we assumed the identity I mentioned in my original post
$endgroup$
– cris14
Jan 27 at 11:04
$begingroup$
It was part of calculation of limit: $lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$ and from there in the next step we assumed the identity I mentioned in my original post
$endgroup$
– cris14
Jan 27 at 11:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$2cosleft(x-dfracpi2right)=-2sinleft(x-dfracpi2right)$$
$$impliestanleft(x-dfracpi2right)=-1=tanleft(-dfracpi4right)$$
$$implies x-dfracpi2=npi-dfracpi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
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
});
}
});
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%2f3089395%2fhow-can-2-cosx-frac-pi2-2-sinx-frac-pi2%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
$begingroup$
$$2cosleft(x-dfracpi2right)=-2sinleft(x-dfracpi2right)$$
$$impliestanleft(x-dfracpi2right)=-1=tanleft(-dfracpi4right)$$
$$implies x-dfracpi2=npi-dfracpi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
$$2cosleft(x-dfracpi2right)=-2sinleft(x-dfracpi2right)$$
$$impliestanleft(x-dfracpi2right)=-1=tanleft(-dfracpi4right)$$
$$implies x-dfracpi2=npi-dfracpi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
add a comment |
$begingroup$
$$2cosleft(x-dfracpi2right)=-2sinleft(x-dfracpi2right)$$
$$impliestanleft(x-dfracpi2right)=-1=tanleft(-dfracpi4right)$$
$$implies x-dfracpi2=npi-dfracpi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
$endgroup$
$$2cosleft(x-dfracpi2right)=-2sinleft(x-dfracpi2right)$$
$$impliestanleft(x-dfracpi2right)=-1=tanleft(-dfracpi4right)$$
$$implies x-dfracpi2=npi-dfracpi4$$ where $n$ is any integer
Clearly, the given relationship is not an identity
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
answered Jan 27 at 10:45
lab bhattacharjeelab bhattacharjee
227k15158278
227k15158278
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%2f3089395%2fhow-can-2-cosx-frac-pi2-2-sinx-frac-pi2%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
1
$begingroup$
This equality is not true in general: $$-2sin(x-pi/2)=2sin(pi/2-x)=2cos xne LHS=2sin x$$It is only true for $x=npi+pi/4,ninBbb Z$.
$endgroup$
– Shubham Johri
Jan 27 at 10:47
1
$begingroup$
This is not an identity working for all $x$, $x=frac{pi}{2}$ trivially does not work. But rather an equation you need to solve
$endgroup$
– Thomas Lesgourgues
Jan 27 at 10:50
1
$begingroup$
Perhaps you had to solve an equation? This is obviously not an identity: just input $;x=0;$ to get a contradiction...
$endgroup$
– DonAntonio
Jan 27 at 10:52
$begingroup$
It was part of calculation of limit: $lim_{x->(dfrac{pi}{2})} dfrac{1-e^{2cosx}}{2cosx} dfrac{2cos(x-dfrac{pi}{2})}{sin(4(x-dfrac{pi}{x}))}$ and from there in the next step we assumed the identity I mentioned in my original post
$endgroup$
– cris14
Jan 27 at 11:04