Trigonometric Identities and equations practice problems
Given that \(\alpha\) is the acute angle such that \(\tan\alpha=\frac{2}{5}\), find the exact value of \(\cos\alpha\).
On a triangle, if \(\tan\alpha=\frac{2}{5}\) then the opposite length is 2, and the adjacent length is 5, using pythagoras, we can find that the hypotenuse is \(\sqrt{ 29 }\).
Now using SOHCAHTOA, we can find that \(\cos\alpha=\frac{5}{\sqrt{ 29 }}\)
\(\tan\theta=\frac{1}{2}\) and \(\theta\) is acute. Show that \(\cos^2\theta=\frac{4}{5}\).
- This means that the opposite is \(1\) and the adjacent is \(2\). \(\therefore\) the hypotenuse is \(\sqrt{ 5 }\).
- \(\cos\theta=\frac{2}{\sqrt{ 5 }}\)
- \(\frac{2}{\sqrt{ 5 }}^2=\frac{4}{5}\)
- \(\therefore \cos^2\theta=\frac{4}{5}\)