Prove that the given inverse trigonometry equation is correct

[chwala]

Homework Statement

See attached.

Relevant Equations

Trigonometry

Ok in my approach i have,

$2 \tan^{-1} \left(\dfrac{1}{5}\right)= \sin^{-1} \left(\dfrac{3}{5}\right) - \cos^{-1} \left(\dfrac{63}{65}\right)$Consider the rhs,

Let

$\sin^{-1} \left(\dfrac{3}{5}\right)= m$ then $\tan m =\dfrac{3}{4}$

also

let

$\cos^{-1} \left(\dfrac{63}{65}\right)= n$ then $\tan n=\dfrac{16}{63}$

Then,

$m-n=\tan^{-1} \left[\dfrac{\frac{3}{4}-\frac{16}{63}}{1+\dfrac{3}{4}⋅\dfrac{16}{63}}\right]$

$m-n=\tan^{-1}\left[\dfrac{\frac{125}{252}}{\frac{300}{252}}\right]$

$m-n=\tan^{-1}\left[\dfrac{125}{252}×\dfrac{252}{300}\right]$

$m-n= \tan^{-1}\left(\dfrac{5}{12}\right) = 22.6^0$ to one decimal place.

on the lhs, we let

$2 \tan^{-1} \left(\dfrac{1}{5}\right) = p$

$\tan^{-1} \left(\dfrac{1}{5}\right)=\dfrac{p}{2}$

$\tan \dfrac{p}{2}=\dfrac{1}{5}$

let $\dfrac{p}{2} = θ$

$\tan θ = \dfrac{1}{5}$

$θ = \tan^{-1} \left(\dfrac{1}{5}\right) = 11.3^0$

$p= 2 ×11.3=22.6^0$

I hope this has been done correctly, ... otherwise, your correction is welcome...there may be a better approach.

 

[anuttarasammyak]

From rectangle triangles
$$\cos^{-1}\frac{63}{65}=\tan^{-1}\frac{16}{63}$$
$$\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4}$$
Then evaluate tan of
$$\tan^{-1}\frac{16}{63}+\tan^{-1}\frac{1}{5}$$
and tan of
$$\tan^{-1}\frac{3}{4}-\tan^{-1}\frac{1}{5}$$
They coincide to be 11/23.

 

[fresh_42]

How do you get $\tan(\sin^{-1}(3/5))=3/4$?

I would use the Weierstraß substitutions and see whether I could solve the resulting polynomial equation.

 

[chwala]

How do you get $\tan(\sin^{-1}(3/5))=3/4$?

I would use the Weierstraß substitutions and see whether I could solve the resulting polynomial equation.

@fresh_42 I used trigonometry and Pythagoras theorem for that part...

 

[Mark44]

How do you get $\tan(\sin^{-1}(3/5))=3/4$?

Using a 3-4-5 right triangle and some basic right triangle trig.

 

[Mark44]

@chwala, your work is OK but could be improved.

The original equation is equivalent to this one:
$2 \tan^{-1} \left(\dfrac{1}{5}\right)= \sin^{-1} \left(\dfrac{3}{5}\right) - \cos^{-1} \left(\dfrac{63}{65}\right)$

Let $\theta = \tan^{-1}(1/5), \alpha = \sin^{-1}(3/5), \beta = \cos^{-1}(63/65)$
Then the equation we're trying to verify can be written as $2\theta = \alpha - \beta$

Take the tangent of both sides:
$\tan(2\theta) = \tan(\alpha - \beta)$
If we can verify that this is a true statement, we will have verified that the original equation is also a true statement.

With a bit of right-triangle trig and the use of the double-angle and difference of angles formulas for the tangent, the LHS of the equation just above equals 5/12, and the RHS equals the same value.

Comments
1. There is no need to find p ($2\tan^{-1}(1/5)$).
2. You have started several lines with "m - n = ..." You don't have to keep writing the left side of an equation -- instead, just continue the right side with = .