inverse trigonometric function

Aug 2011
146
1
The problem is

Show that Sin-1(12/13) + Cos-1(4/5) + Tan-1(63/16) = Pi

I converted Sin-1(12/13) as Tan-1(12/5)

Cos-1(4/5) as Tan-1(3/4)

Then I took the above problem as


Tan-1(12/5) + Tan-1(3/4) = Tan-1{[12/5 + 3/4]/[1-(12/5)(3/4)]} = Tan-1(-63/16) = - Tan-1(63/16)

The above problem becomes zero (0).

How I could conclude that the answer is Pi?

or What is wrong with my working.

Kindly guide me.

with warm regards,

Aranga
 
Aug 2011
146
1
I could infer from your reply they are not equal.

Thanks. Now I revisited the formula given in the book.

They have mentioned that

tan-1x + tan-1y = tan-1{[x + y]/[1-xy]} only when xy<1.

Why such condition they have given I did not understand.

in this case, xy is greater than 1. So it fails. That much I understand.

Now I do not know how to solve this problem or how to proceed further.

Kindly guide me.

with warm regards,

Aranga
 
Jun 2013
1,127
601
Lebanon
if \(\displaystyle x y>1\)

\(\displaystyle \tan ^{-1}x+\tan ^{-1}y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) + \pi \)

P.S. I am not sure if this is correct
 
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Aug 2011
146
1
Thanks. As you said If I put I am getting the answer.

Whether it is correct?

If it is so, how it is correct?

with warm regards,

Aranga
 
Dec 2011
2,314
916
St. Augustine, FL.
I think I would arrange the equation as:

\(\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)\)

Take the sine of both sides (apply the angle sum identity for sine on the left and the identity \(\displaystyle \sin(\pi-\theta)=\sin(\theta)\) on the right):

\(\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)\)

\(\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}\)

\(\displaystyle \frac{48+15}{65}=\frac{63}{65}\)

\(\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark\)
 
Aug 2011
146
1
please see the post of IDEA.

Whether what was said is correct?
 
Jun 2013
1,127
601
Lebanon
I think I would arrange the equation as:

\(\displaystyle \arcsin\left(\frac{12}{13}\right)+ \arccos\left(\frac{4}{5}\right)= \pi-\arctan\left(\frac{63}{16}\right)\)

Take the sine of both sides (apply the angle sum identity for sine on the left and the identity \(\displaystyle \sin(\pi-\theta)=\sin(\theta)\) on the right):

\(\displaystyle \sin\left(\arcsin\left(\frac{12}{13}\right)\right) \cos\left(\arccos\left(\frac{4}{5}\right)\right)+ \cos\left(\arcsin\left(\frac{12}{13}\right)\right) \sin\left(\arccos\left(\frac{4}{5}\right)\right)= \sin\left(\arctan\left(\frac{63}{16}\right)\right)\)

\(\displaystyle \frac{12}{13}\cdot\frac{4}{5}+ \frac{5}{13}\cdot\frac{3}{5}=\frac{63}{65}\)

\(\displaystyle \frac{48+15}{65}=\frac{63}{65}\)

\(\displaystyle \frac{63}{65}=\frac{63}{65}\quad\checkmark\)
This argument is not valid.

To see this, start with the following equation which is obviously not correct

\(\displaystyle \sin ^{-1}\frac{12}{13}+\cos ^{-1}\frac{4}{5}=\tan ^{-1}\frac{63}{16}\)

and apply $sin$ to both sides to get

\(\displaystyle \frac{63}{65}=\frac{63}{65}\)

so taking sin of both sides and getting equal answers does not prove the original assertion
 
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