Question

How do you find the exact value of $\cos \left( \arctan \left( \dfrac{3}{4} \right) \right)$ ?

Answer 1

In this question we have been asked to find the exact value of $\cos \left( \arctan \left( \dfrac{3}{4} \right) \right)$ we will do that by using the formulae $\arctan \left( \dfrac{x}{y} \right)=\arccos \left( \dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right)$ and $\cos \left( \arccos x \right)=x$ . After that we will perform some simple arithmetic calculations to get it in the reduced form.

Complete step by step answer:
To answer this question we need to find the exact value of $\cos \left( \arctan \left( \dfrac{3}{4} \right) \right)$.
First we will use our basic inverse trigonometric concepts and express $\arctan \left( \dfrac{3}{4} \right)$ in terms of $\arccos$ by using the formulae $\arctan \left( \dfrac{x}{y} \right)=\arccos \left( \dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right)$ .
After applying this formulae we will have $\arctan \left( \dfrac{3}{4} \right)=\arccos \left( \dfrac{4}{\sqrt{{{3}^{2}}+{{4}^{2}}}} \right)$ .
By further arithmetic simplifying this we will have $\arctan \left( \dfrac{3}{4} \right)=\arccos \left( \dfrac{4}{5} \right)$ .
By substituting this value we will have $\cos \left( \arccos \left( \dfrac{4}{5} \right) \right)$ .
As we know that $\cos \left( \arccos x \right)=x$ by using this we will have $\cos \left( \arccos \left( \dfrac{4}{5} \right) \right)=\dfrac{4}{5}$ .
The $\arctan$ can be also expressed as ${{\tan }^{-1}}$ conveniently and similarly all the other inverse trigonometric ratios can be expressed as ${{\tan }^{-1}}x=\theta \Rightarrow \tan \theta =x$ similarly ${{\cos }^{-1}}x=\theta \Rightarrow \cos \theta =x$ and all other formulae can be expressed similarly.

Hence we can conclude that the exact value of $\cos \left( \arctan \left( \dfrac{3}{4} \right) \right)$ is $\dfrac{4}{5}$ that is $0.8$

Note: We should be sure with the calculations and inverse trigonometric concepts for answering questions of this type. The $\arctan$ can be also expressed as ${{\tan }^{-1}}$ conveniently and similarly all the other inverse trigonometric ratios can be expressed. We will prove the formulae that we have used in the given question using ${{\tan }^{-1}}x=\theta \Rightarrow \tan \theta =x$ similarly ${{\cos }^{-1}}x=\theta \Rightarrow \cos \theta =x$. By applying $\cos$ on both sides we will have $\cos \left( {{\cos }^{-1}}x \right)\Rightarrow \cos \theta =x$ . Similarly we can prove the formula $\arctan \left( \dfrac{x}{y} \right)=\arccos \left( \dfrac{y}{\sqrt{{{x}^{2}}+{{y}^{2}}}} \right)$ by using ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ .