Find the value of $3tan^{-1}frac{1}{2}+2tan^{-1}frac{1}{5}+sin^{-1}frac{142}{65sqrt{5}}$
$begingroup$
Find the value of $3tan^{-1}left(dfrac{1}{2}right)+2tan^{-1}left(dfrac{1}{5}right)+sin^{-1}left(dfrac{142}{65sqrt{5}}right)$
My reference gives the solution $0$ to this problem.
My Attempt
$$
|x_1|=frac{1}{2}leqfrac{1}{sqrt{3}}implies
3tan^{-1}frac{1}{2}=tan^{-1}frac{frac{3}{2}-frac{1}{8}}{1-frac{3}{4}}=tan^{-1}frac{11}{2}\
|x_2|=frac{1}{5}<1implies2tan^{-1}dfrac{1}{5}=tan^{-1}frac{frac{2}{5}}{1-frac{1}{25}}=tan^{-1}frac{10}{24}=tan^{-1}frac{5}{12}\
XY=frac{11}{2}frac{5}{12}=frac{55}{24}>1quad&quad X,Y>0\
3tan^{-1}dfrac{1}{2}+2tan^{-1}dfrac{1}{5}=tan^{-1}frac{11}{2}+tan^{-1}frac{5}{12}=pi+tan^{-1}frac{frac{11}{2}+frac{5}{12}}{1-frac{11}{2}frac{5}{12}}\
=pi-tan^{-1}frac{142}{31}
$$
So I seem to get $pi$ as the answer, what is going wrong here ?
trigonometry inverse-function
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
$endgroup$
add a comment |
$begingroup$
Find the value of $3tan^{-1}left(dfrac{1}{2}right)+2tan^{-1}left(dfrac{1}{5}right)+sin^{-1}left(dfrac{142}{65sqrt{5}}right)$
My reference gives the solution $0$ to this problem.
My Attempt
$$
|x_1|=frac{1}{2}leqfrac{1}{sqrt{3}}implies
3tan^{-1}frac{1}{2}=tan^{-1}frac{frac{3}{2}-frac{1}{8}}{1-frac{3}{4}}=tan^{-1}frac{11}{2}\
|x_2|=frac{1}{5}<1implies2tan^{-1}dfrac{1}{5}=tan^{-1}frac{frac{2}{5}}{1-frac{1}{25}}=tan^{-1}frac{10}{24}=tan^{-1}frac{5}{12}\
XY=frac{11}{2}frac{5}{12}=frac{55}{24}>1quad&quad X,Y>0\
3tan^{-1}dfrac{1}{2}+2tan^{-1}dfrac{1}{5}=tan^{-1}frac{11}{2}+tan^{-1}frac{5}{12}=pi+tan^{-1}frac{frac{11}{2}+frac{5}{12}}{1-frac{11}{2}frac{5}{12}}\
=pi-tan^{-1}frac{142}{31}
$$
So I seem to get $pi$ as the answer, what is going wrong here ?
trigonometry inverse-function
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
$endgroup$
2
$begingroup$
That definitely looks positive to me!
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 19:03
$begingroup$
Your reference is wrong.You should compute the value numerically. This cannot prove your answer is right but it can show your reference is wrong.
$endgroup$
– achille hui
Dec 11 '18 at 19:10
add a comment |
$begingroup$
Find the value of $3tan^{-1}left(dfrac{1}{2}right)+2tan^{-1}left(dfrac{1}{5}right)+sin^{-1}left(dfrac{142}{65sqrt{5}}right)$
My reference gives the solution $0$ to this problem.
My Attempt
$$
|x_1|=frac{1}{2}leqfrac{1}{sqrt{3}}implies
3tan^{-1}frac{1}{2}=tan^{-1}frac{frac{3}{2}-frac{1}{8}}{1-frac{3}{4}}=tan^{-1}frac{11}{2}\
|x_2|=frac{1}{5}<1implies2tan^{-1}dfrac{1}{5}=tan^{-1}frac{frac{2}{5}}{1-frac{1}{25}}=tan^{-1}frac{10}{24}=tan^{-1}frac{5}{12}\
XY=frac{11}{2}frac{5}{12}=frac{55}{24}>1quad&quad X,Y>0\
3tan^{-1}dfrac{1}{2}+2tan^{-1}dfrac{1}{5}=tan^{-1}frac{11}{2}+tan^{-1}frac{5}{12}=pi+tan^{-1}frac{frac{11}{2}+frac{5}{12}}{1-frac{11}{2}frac{5}{12}}\
=pi-tan^{-1}frac{142}{31}
$$
So I seem to get $pi$ as the answer, what is going wrong here ?
trigonometry inverse-function
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
$endgroup$
Find the value of $3tan^{-1}left(dfrac{1}{2}right)+2tan^{-1}left(dfrac{1}{5}right)+sin^{-1}left(dfrac{142}{65sqrt{5}}right)$
My reference gives the solution $0$ to this problem.
My Attempt
$$
|x_1|=frac{1}{2}leqfrac{1}{sqrt{3}}implies
3tan^{-1}frac{1}{2}=tan^{-1}frac{frac{3}{2}-frac{1}{8}}{1-frac{3}{4}}=tan^{-1}frac{11}{2}\
|x_2|=frac{1}{5}<1implies2tan^{-1}dfrac{1}{5}=tan^{-1}frac{frac{2}{5}}{1-frac{1}{25}}=tan^{-1}frac{10}{24}=tan^{-1}frac{5}{12}\
XY=frac{11}{2}frac{5}{12}=frac{55}{24}>1quad&quad X,Y>0\
3tan^{-1}dfrac{1}{2}+2tan^{-1}dfrac{1}{5}=tan^{-1}frac{11}{2}+tan^{-1}frac{5}{12}=pi+tan^{-1}frac{frac{11}{2}+frac{5}{12}}{1-frac{11}{2}frac{5}{12}}\
=pi-tan^{-1}frac{142}{31}
$$
So I seem to get $pi$ as the answer, what is going wrong here ?
trigonometry inverse-function
trigonometry inverse-function
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
edited Dec 11 '18 at 19:36
Lorenzo B.
1,8602520
1,8602520
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
asked Dec 11 '18 at 18:53
ss1729ss1729
1,9721923
1,9721923
2
$begingroup$
That definitely looks positive to me!
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 19:03
$begingroup$
Your reference is wrong.You should compute the value numerically. This cannot prove your answer is right but it can show your reference is wrong.
$endgroup$
– achille hui
Dec 11 '18 at 19:10
add a comment |
2
$begingroup$
That definitely looks positive to me!
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 19:03
$begingroup$
Your reference is wrong.You should compute the value numerically. This cannot prove your answer is right but it can show your reference is wrong.
$endgroup$
– achille hui
Dec 11 '18 at 19:10
2
2
$begingroup$
That definitely looks positive to me!
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 19:03
$begingroup$
That definitely looks positive to me!
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 19:03
$begingroup$
Your reference is wrong.You should compute the value numerically. This cannot prove your answer is right but it can show your reference is wrong.
$endgroup$
– achille hui
Dec 11 '18 at 19:10
$begingroup$
Your reference is wrong.You should compute the value numerically. This cannot prove your answer is right but it can show your reference is wrong.
$endgroup$
– achille hui
Dec 11 '18 at 19:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your reference is clearly wrong without calculating anything. Functions like $tan^{-1}$ and $sin^{-1}$ generate positive values for positive arguments, by convention. All your arguments are positive so the result must be positive too. Numeric check proves that your result is correct.
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
$endgroup$
add a comment |
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$begingroup$
Your reference is clearly wrong without calculating anything. Functions like $tan^{-1}$ and $sin^{-1}$ generate positive values for positive arguments, by convention. All your arguments are positive so the result must be positive too. Numeric check proves that your result is correct.
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
$endgroup$
add a comment |
$begingroup$
Your reference is clearly wrong without calculating anything. Functions like $tan^{-1}$ and $sin^{-1}$ generate positive values for positive arguments, by convention. All your arguments are positive so the result must be positive too. Numeric check proves that your result is correct.
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
$endgroup$
add a comment |
$begingroup$
Your reference is clearly wrong without calculating anything. Functions like $tan^{-1}$ and $sin^{-1}$ generate positive values for positive arguments, by convention. All your arguments are positive so the result must be positive too. Numeric check proves that your result is correct.
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
$endgroup$
Your reference is clearly wrong without calculating anything. Functions like $tan^{-1}$ and $sin^{-1}$ generate positive values for positive arguments, by convention. All your arguments are positive so the result must be positive too. Numeric check proves that your result is correct.
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
answered Dec 11 '18 at 19:17
OldboyOldboy
8,4521936
8,4521936
add a comment |
add a comment |
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2
$begingroup$
That definitely looks positive to me!
$endgroup$
– Lord Shark the Unknown
Dec 11 '18 at 19:03
$begingroup$
Your reference is wrong.You should compute the value numerically. This cannot prove your answer is right but it can show your reference is wrong.
$endgroup$
– achille hui
Dec 11 '18 at 19:10