Find $sin x$ and $cos x$ knowing $tan x$
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
705315
asked Dec 3 '18 at 9:14
Mark
275
add a comment |
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
705315
asked Dec 3 '18 at 9:14
Mark
275
add a comment |
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
705315
asked Dec 3 '18 at 9:14
Mark
275
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
705315
asked Dec 3 '18 at 9:14
Mark
275
edited Dec 3 '18 at 9:21
Boshu
705315
asked Dec 3 '18 at 9:14
Mark
275
edited Dec 3 '18 at 9:21
Boshu
705315
edited Dec 3 '18 at 9:21
Boshu
705315
edited Dec 3 '18 at 9:21
Boshu
705315
705315
asked Dec 3 '18 at 9:14
Mark
275
asked Dec 3 '18 at 9:14
Mark
275
asked Dec 3 '18 at 9:14
Mark
275
275
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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 Goh
99.5k1464117
add a comment |
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
Boshu
705315
add a comment |
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
gimusi
1
add a comment |
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
5,3051826
answered Dec 3 '18 at 9:34
Bernard
118k639112
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Goh
99.5k1464117
add a comment |
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 Goh
99.5k1464117
add a comment |
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 Goh
99.5k1464117
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 Goh
99.5k1464117
answered Dec 3 '18 at 9:20
Siong Thye Goh
99.5k1464117
answered Dec 3 '18 at 9:20
Siong Thye Goh
99.5k1464117
answered Dec 3 '18 at 9:20
Siong Thye Goh
99.5k1464117
99.5k1464117
add a comment |
add a comment |
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
Boshu
705315
add a comment |
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
Boshu
705315
add a comment |
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
Boshu
705315
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
Boshu
705315
answered Dec 3 '18 at 9:18
Boshu
705315
answered Dec 3 '18 at 9:18
Boshu
705315
answered Dec 3 '18 at 9:18
Boshu
705315
705315
add a comment |
add a comment |
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
gimusi
1
add a comment |
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
gimusi
1
add a comment |
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
gimusi
1
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
gimusi
1
edited Dec 3 '18 at 9:35
answered Dec 3 '18 at 9:27
gimusi
1
answered Dec 3 '18 at 9:27
gimusi
1
answered Dec 3 '18 at 9:27
gimusi
1
1
add a comment |
add a comment |
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
5,3051826
answered Dec 3 '18 at 9:34
Bernard
118k639112
add a comment |
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
5,3051826
answered Dec 3 '18 at 9:34
Bernard
118k639112
add a comment |
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
5,3051826
answered Dec 3 '18 at 9:34
Bernard
118k639112
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
5,3051826
answered Dec 3 '18 at 9:34
Bernard
118k639112
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,3051826
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,3051826
edited Dec 3 '18 at 18:40
Chinnapparaj R
5,3051826
5,3051826
answered Dec 3 '18 at 9:34
Bernard
118k639112
answered Dec 3 '18 at 9:34
Bernard
118k639112
answered Dec 3 '18 at 9:34
Bernard
118k639112
118k639112
add a comment |
add a comment |
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