Prove that the given inverse trigonometry equation is correct

AI Thread Summary
The discussion centers on proving the equation 2 tan^(-1)(1/5) = sin^(-1)(3/5) - cos^(-1)(63/65). Participants derive values for m and n using trigonometric identities, ultimately showing that m - n equals tan^(-1)(5/12). They confirm that both sides of the equation yield the same tangent value, thus validating the original equation. Suggestions for improvement in notation and approach are also provided, emphasizing the use of right triangles and trigonometric functions.
chwala
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Homework Statement
See attached.
Relevant Equations
Trigonometry
1694082442229.png
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.
 
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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.
 
Last edited:
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 said:
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...
 
fresh_42 said:
How do you get ##\tan(\sin^{-1}(3/5))=3/4##?
Using a 3-4-5 right triangle and some basic right triangle trig.
 
@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 = .
 
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