- #1

chwala

Gold Member

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- Homework Statement
- See attached.

- Relevant Equations
- Trigonometry

##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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