# Prove that the given inverse trigonometry equation is correct

• chwala
chwala
Gold Member
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.

Last edited by a moderator:
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.

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 = .

chwala, berkeman and jim mcnamara

## What are inverse trigonometric functions?

Inverse trigonometric functions are the inverse operations of the basic trigonometric functions (sine, cosine, tangent, etc.). They allow us to find the angle that corresponds to a given trigonometric value. The main inverse trigonometric functions are arcsin (inverse sine), arccos (inverse cosine), and arctan (inverse tangent).

## How do you prove an inverse trigonometric equation?

To prove an inverse trigonometric equation, you typically start by expressing the given equation in terms of a variable. Then, use known identities and properties of trigonometric and inverse trigonometric functions to simplify and manipulate the equation. The goal is to show that both sides of the equation are equivalent.

## What are some common identities used in proving inverse trigonometric equations?

Some common identities include:1. $$\sin(\arcsin(x)) = x$$2. $$\cos(\arccos(x)) = x$$3. $$\tan(\arctan(x)) = x$$4. $$\arcsin(\sin(x)) = x$$ for $$x$$ in $$[- \frac{\pi}{2}, \frac{\pi}{2}]$$5. $$\arccos(\cos(x)) = x$$ for $$x$$ in $$[0, \pi]$$6. $$\arctan(\tan(x)) = x$$ for $$x$$ in $$(- \frac{\pi}{2}, \frac{\pi}{2})$$

## Can you give an example of proving an inverse trigonometric equation?

Sure, let's prove that $$\arcsin(\sin(x)) = x$$ for $$x$$ in $$[- \frac{\pi}{2}, \frac{\pi}{2}]$$:1. Let $$y = \arcsin(\sin(x))$$.2. By definition, $$\sin(y) = \sin(x)$$.3. Since $$y$$ must be in the range of $$\arcsin$$, which is $$[- \frac{\pi}{2}, \frac{\pi}{2}]$$, and $$\sin$$ is one-to-one in this interval, we have $$y = x$$.4. Hence, $$\arcsin(\sin(x)) = x$$ for $$x$$ in $$[- \frac{\pi}{2}, \frac{\pi}{2}]$$.

## What should I do if I get stuck while proving an inverse trigonometric equation?

If you get stuck, try the following steps:1. Revisit the definitions and properties of the

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