Evaluate and Simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$
up vote
1
down vote
favorite
I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.
I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$
My Process:
$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$
$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$
$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$
$frac{6}{10} + frac{5}{10}$
$frac{11}{10}$
algebra-precalculus trigonometry
add a comment |
up vote
1
down vote
favorite
I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.
I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$
My Process:
$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$
$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$
$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$
$frac{6}{10} + frac{5}{10}$
$frac{11}{10}$
algebra-precalculus trigonometry
1
Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 at 18:05
1
@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 at 19:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.
I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$
My Process:
$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$
$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$
$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$
$frac{6}{10} + frac{5}{10}$
$frac{11}{10}$
algebra-precalculus trigonometry
I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.
I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$
My Process:
$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$
$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$
$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$
$frac{6}{10} + frac{5}{10}$
$frac{11}{10}$
algebra-precalculus trigonometry
algebra-precalculus trigonometry
asked Nov 22 at 17:56
LuminousNutria
1709
1709
1
Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 at 18:05
1
@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 at 19:28
add a comment |
1
Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 at 18:05
1
@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 at 19:28
1
1
Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 at 18:05
Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 at 18:05
1
1
@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 at 19:28
@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 at 19:28
add a comment |
5 Answers
5
active
oldest
votes
up vote
3
down vote
accepted
You cannot separate out the $cos$ function as you have done in step two.
You can remember this identity.
$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$
Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
1
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
|
show 1 more comment
up vote
1
down vote
The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.
1
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
add a comment |
up vote
0
down vote
$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$
add a comment |
up vote
0
down vote
By compound-angle formula,
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
& = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
end{eqnarray*}
To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
$sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
It follows that
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
& = & frac{3}{10}-frac{2sqrt{3}}{5}
end{eqnarray*}
add a comment |
up vote
-1
down vote
Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You cannot separate out the $cos$ function as you have done in step two.
You can remember this identity.
$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$
Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
1
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
|
show 1 more comment
up vote
3
down vote
accepted
You cannot separate out the $cos$ function as you have done in step two.
You can remember this identity.
$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$
Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
1
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
|
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You cannot separate out the $cos$ function as you have done in step two.
You can remember this identity.
$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$
Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
You cannot separate out the $cos$ function as you have done in step two.
You can remember this identity.
$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$
Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)
Using thus result you should get the desired answer.
edited Nov 22 at 18:07
Mark Viola
130k1273170
130k1273170
answered Nov 22 at 18:00
Akash Roy
1
1
1
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
|
show 1 more comment
1
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
1
1
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
(+1) ... for discussing the error in the OP
– Mark Viola
Nov 22 at 18:05
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
@MarkViola thanks for your upvote.
– Akash Roy
Nov 22 at 18:06
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thanks once again for your valuable edits @Mark, will bear these in mind.
– Akash Roy
Nov 22 at 18:08
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
– LuminousNutria
Nov 22 at 18:09
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
@Luminous you can accept my answer if you understood.
– Akash Roy
Nov 22 at 18:10
|
show 1 more comment
up vote
1
down vote
The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.
1
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
add a comment |
up vote
1
down vote
The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.
1
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
add a comment |
up vote
1
down vote
up vote
1
down vote
The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.
The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.
answered Nov 22 at 17:59
tarit goswami
1,7001421
1,7001421
1
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
add a comment |
1
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
1
1
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
(+1) for mentioning the error in the OP
– Mark Viola
Nov 22 at 18:10
add a comment |
up vote
0
down vote
$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$
add a comment |
up vote
0
down vote
$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$
add a comment |
up vote
0
down vote
up vote
0
down vote
$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$
$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$
answered Nov 22 at 18:03
Yiorgos S. Smyrlis
62.3k1383162
62.3k1383162
add a comment |
add a comment |
up vote
0
down vote
By compound-angle formula,
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
& = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
end{eqnarray*}
To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
$sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
It follows that
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
& = & frac{3}{10}-frac{2sqrt{3}}{5}
end{eqnarray*}
add a comment |
up vote
0
down vote
By compound-angle formula,
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
& = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
end{eqnarray*}
To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
$sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
It follows that
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
& = & frac{3}{10}-frac{2sqrt{3}}{5}
end{eqnarray*}
add a comment |
up vote
0
down vote
up vote
0
down vote
By compound-angle formula,
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
& = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
end{eqnarray*}
To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
$sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
It follows that
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
& = & frac{3}{10}-frac{2sqrt{3}}{5}
end{eqnarray*}
By compound-angle formula,
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
& = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
end{eqnarray*}
To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
$sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
It follows that
begin{eqnarray*}
cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
& = & frac{3}{10}-frac{2sqrt{3}}{5}
end{eqnarray*}
answered Nov 22 at 18:05
Danny Pak-Keung Chan
2,11038
2,11038
add a comment |
add a comment |
up vote
-1
down vote
Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
add a comment |
up vote
-1
down vote
Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.
Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.
answered Nov 24 at 9:59
Siddharth Dhiman
1
1
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
add a comment |
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
– Saucy O'Path
Nov 24 at 10:09
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
... and $cos 97^circ$ is not $-0.39$.
– Saucy O'Path
Nov 24 at 11:17
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1
Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 at 18:05
1
@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 at 19:28