Find $sin x$ and $cos x$ knowing $tan x$
$begingroup$
I know $tan x=-2sqrt2$. How to find $sin x$ and $cos x$ if $xin[-frac{pi}{2},0]$? They probably would be $-frac{2sqrt2}{3}$ and $frac{1}{3}$ respectively but I don't know how to prove it.
trigonometry
edited Dec 3 '18 at 9:21
Boshu
703315
asked Dec 3 '18 at 9:14
MarkMark
278
$endgroup$
add a comment |
$begingroup$
I know $tan x=-2sqrt2$. How to find $sin x$ and $cos x$ if $xin[-frac{pi}{2},0]$? They probably would be $-frac{2sqrt2}{3}$ and $frac{1}{3}$ respectively but I don't know how to prove it.
trigonometry
edited Dec 3 '18 at 9:21
Boshu
703315
asked Dec 3 '18 at 9:14
MarkMark
278
$endgroup$
add a comment |
$begingroup$
I know $tan x=-2sqrt2$. How to find $sin x$ and $cos x$ if $xin[-frac{pi}{2},0]$? They probably would be $-frac{2sqrt2}{3}$ and $frac{1}{3}$ respectively but I don't know how to prove it.
trigonometry
edited Dec 3 '18 at 9:21
Boshu
703315
asked Dec 3 '18 at 9:14
MarkMark
278
$endgroup$
I know $tan x=-2sqrt2$. How to find $sin x$ and $cos x$ if $xin[-frac{pi}{2},0]$? They probably would be $-frac{2sqrt2}{3}$ and $frac{1}{3}$ respectively but I don't know how to prove it.
trigonometry
trigonometry
edited Dec 3 '18 at 9:21
Boshu
703315
asked Dec 3 '18 at 9:14
MarkMark
278
edited Dec 3 '18 at 9:21
Boshu
703315
asked Dec 3 '18 at 9:14
MarkMark
278
edited Dec 3 '18 at 9:21
Boshu
703315
edited Dec 3 '18 at 9:21
Boshu
703315
edited Dec 3 '18 at 9:21
Boshu
703315
703315
asked Dec 3 '18 at 9:14
MarkMark
278
asked Dec 3 '18 at 9:14
MarkMark
278
asked Dec 3 '18 at 9:14
MarkMark
278
278
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We know the signs.
We have $$sin x = -2sqrt2 cos x$$ and
$$sin^2 x+ cos^2 x = 1$$
Substitute the first equation into the second, and you can solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should be able to recover $sin x$ using the first equation.
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
Take $sin x=y$ which means that $cos x= pmsqrt{1-y^2}$ where the sign depends on the quadrant of interest. Now equate their ratio with $tan x$ and solve. Again, you choose the appropriate value of $y$ depending on your coordinate of interest.
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
$endgroup$
add a comment |
$begingroup$
The given suggestions are good and simpler, as an alternative for a direct calculation recall that for $thetain(-pi/2,pi/2)$
$$y=tan theta iff theta=arctan y$$
and therefore since in that case $xin(-pi/2,0)$ by $y=-2sqrt2$ we can use that by composition formulas
- $sin x= sin (arctan y)=frac{y}{sqrt{1+y^2}}$
- $cos x= cos (arctan y)=frac{1}{sqrt{1+y^2}}$
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
Use the basic relations
$$cos^2x=frac1{1+tan^2x},enspacetext{whence }quad sin^2x=tan^2xcos^2x=frac{tan^2x}{1+tan^2x}.$$
So $sin x$ is known up to its sign. On $bigl[-fracpi2,0bigr]$, it is negative or $0$.
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
$endgroup$
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%2f3023818%2ffind-sin-x-and-cos-x-knowing-tan-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know the signs.
We have $$sin x = -2sqrt2 cos x$$ and
$$sin^2 x+ cos^2 x = 1$$
Substitute the first equation into the second, and you can solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should be able to recover $sin x$ using the first equation.
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
We know the signs.
We have $$sin x = -2sqrt2 cos x$$ and
$$sin^2 x+ cos^2 x = 1$$
Substitute the first equation into the second, and you can solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should be able to recover $sin x$ using the first equation.
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
add a comment |
$begingroup$
We know the signs.
We have $$sin x = -2sqrt2 cos x$$ and
$$sin^2 x+ cos^2 x = 1$$
Substitute the first equation into the second, and you can solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should be able to recover $sin x$ using the first equation.
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
$endgroup$
We know the signs.
We have $$sin x = -2sqrt2 cos x$$ and
$$sin^2 x+ cos^2 x = 1$$
Substitute the first equation into the second, and you can solve for $cos x$. Remember $cos x>0$ in this quadrant. After that you should be able to recover $sin x$ using the first equation.
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
answered Dec 3 '18 at 9:20
Siong Thye GohSiong Thye Goh
101k1466118
101k1466118
add a comment |
add a comment |
$begingroup$
Take $sin x=y$ which means that $cos x= pmsqrt{1-y^2}$ where the sign depends on the quadrant of interest. Now equate their ratio with $tan x$ and solve. Again, you choose the appropriate value of $y$ depending on your coordinate of interest.
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
$endgroup$
add a comment |
$begingroup$
Take $sin x=y$ which means that $cos x= pmsqrt{1-y^2}$ where the sign depends on the quadrant of interest. Now equate their ratio with $tan x$ and solve. Again, you choose the appropriate value of $y$ depending on your coordinate of interest.
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
$endgroup$
add a comment |
$begingroup$
Take $sin x=y$ which means that $cos x= pmsqrt{1-y^2}$ where the sign depends on the quadrant of interest. Now equate their ratio with $tan x$ and solve. Again, you choose the appropriate value of $y$ depending on your coordinate of interest.
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
$endgroup$
Take $sin x=y$ which means that $cos x= pmsqrt{1-y^2}$ where the sign depends on the quadrant of interest. Now equate their ratio with $tan x$ and solve. Again, you choose the appropriate value of $y$ depending on your coordinate of interest.
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
answered Dec 3 '18 at 9:18
BoshuBoshu
703315
703315
add a comment |
add a comment |
$begingroup$
The given suggestions are good and simpler, as an alternative for a direct calculation recall that for $thetain(-pi/2,pi/2)$
$$y=tan theta iff theta=arctan y$$
and therefore since in that case $xin(-pi/2,0)$ by $y=-2sqrt2$ we can use that by composition formulas
- $sin x= sin (arctan y)=frac{y}{sqrt{1+y^2}}$
- $cos x= cos (arctan y)=frac{1}{sqrt{1+y^2}}$
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
The given suggestions are good and simpler, as an alternative for a direct calculation recall that for $thetain(-pi/2,pi/2)$
$$y=tan theta iff theta=arctan y$$
and therefore since in that case $xin(-pi/2,0)$ by $y=-2sqrt2$ we can use that by composition formulas
- $sin x= sin (arctan y)=frac{y}{sqrt{1+y^2}}$
- $cos x= cos (arctan y)=frac{1}{sqrt{1+y^2}}$
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
The given suggestions are good and simpler, as an alternative for a direct calculation recall that for $thetain(-pi/2,pi/2)$
$$y=tan theta iff theta=arctan y$$
and therefore since in that case $xin(-pi/2,0)$ by $y=-2sqrt2$ we can use that by composition formulas
- $sin x= sin (arctan y)=frac{y}{sqrt{1+y^2}}$
- $cos x= cos (arctan y)=frac{1}{sqrt{1+y^2}}$
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
$endgroup$
The given suggestions are good and simpler, as an alternative for a direct calculation recall that for $thetain(-pi/2,pi/2)$
$$y=tan theta iff theta=arctan y$$
and therefore since in that case $xin(-pi/2,0)$ by $y=-2sqrt2$ we can use that by composition formulas
- $sin x= sin (arctan y)=frac{y}{sqrt{1+y^2}}$
- $cos x= cos (arctan y)=frac{1}{sqrt{1+y^2}}$
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
edited Dec 3 '18 at 9:35
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
answered Dec 3 '18 at 9:27
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
Use the basic relations
$$cos^2x=frac1{1+tan^2x},enspacetext{whence }quad sin^2x=tan^2xcos^2x=frac{tan^2x}{1+tan^2x}.$$
So $sin x$ is known up to its sign. On $bigl[-fracpi2,0bigr]$, it is negative or $0$.
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Use the basic relations
$$cos^2x=frac1{1+tan^2x},enspacetext{whence }quad sin^2x=tan^2xcos^2x=frac{tan^2x}{1+tan^2x}.$$
So $sin x$ is known up to its sign. On $bigl[-fracpi2,0bigr]$, it is negative or $0$.
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Use the basic relations
$$cos^2x=frac1{1+tan^2x},enspacetext{whence }quad sin^2x=tan^2xcos^2x=frac{tan^2x}{1+tan^2x}.$$
So $sin x$ is known up to its sign. On $bigl[-fracpi2,0bigr]$, it is negative or $0$.
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
$endgroup$
Use the basic relations
$$cos^2x=frac1{1+tan^2x},enspacetext{whence }quad sin^2x=tan^2xcos^2x=frac{tan^2x}{1+tan^2x}.$$
So $sin x$ is known up to its sign. On $bigl[-fracpi2,0bigr]$, it is negative or $0$.
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,5422928
5,5422928
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
answered Dec 3 '18 at 9:34
BernardBernard
121k740116
121k740116
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%2f3023818%2ffind-sin-x-and-cos-x-knowing-tan-x%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
