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

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
asked 2021-02-08
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.

Answers (1)

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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