# Explain how we can evaluate the expression cos(arctan(frac{v}{a})) and what the evaluation is in terms of v and a.

Question
Trigonometry
Explain how we can evaluate the expression $$\displaystyle{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}$$ and what the evaluation is in terms of v and a.

2021-02-09
Solution: $$\displaystyle{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}$$
Let $$\displaystyle{\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}={x}\Rightarrow{\cos{{\left({x}\right)}}}={\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}$$
$$\displaystyle\Rightarrow{\frac{{{v}}}{{{a}}}}={\tan{{x}}}$$
$$\displaystyle\Rightarrow{\frac{{{v}}}{{{a}}}}={\frac{{{\sin{{x}}}}}{{{\cos{{x}}}}}}$$
$$\displaystyle\Rightarrow\text{Squaring both side}$$
$$\displaystyle{\frac{{{v}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}$$
$$\displaystyle\Rightarrow{\frac{{{v}^{{2}}}}{{{a}^{{2}}}}}+{1}={\frac{{{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}+{1}$$
$$\displaystyle\Rightarrow{\frac{{{v}^{{2}}+{a}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}$$
$$\displaystyle\therefore{\frac{{{v}^{{2}}+{a}^{{2}}}}{{{a}^{{2}}}}}={\frac{{{1}}}{{{{\cos}^{{2}}{x}}}}}$$
$$\displaystyle\Rightarrow{\cos{{x}}}=\sqrt{{{\frac{{{a}^{{2}}}}{{{v}^{{2}}+{a}^{{2}}}}}}}$$
$$\displaystyle\therefore{\cos{{\left({\arctan{{\left({\frac{{{v}}}{{{a}}}}\right)}}}\right)}}}=\sqrt{{{\frac{{{1}}}{{{\left({\frac{{{v}}}{{{a}}}}\right)}^{{2}}+{1}}}}}}$$

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