Let x=asin theta in sqrt{a^{2}-x^{2}}. Then find cos theta tan theta

Question
Trigonometric Functions
asked 2021-01-24
Let \(x=a\sin \theta in \sqrt{a^{2}-x^{2}}\). Then find \(\cos \theta \ \tan \theta\)

Answers (1)

2021-01-25
\(\sqrt{a^{2}-s^{2}\sin^{2}(\theta)}=\)
Factor out common term \(a^{2}\)
\(=\sqrt{a^{2}(1-\sin^{2}(\theta))}\)
Apply radical rule \(\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}\), assuming \(a\geq0 / b\geq0\)
\(=\sqrt{a^{2}}\sqrt{-\sin^{2}(\theta)+1}\)
Apply radical rule \(\sqrt[n]{a^{n}}=a\), assuming \(a\geq0\) Use the following identity: \(\cos^{2}(x)+\sin^{2}(x)=1\)
Therefore \(1-\sin^{2}(x)=\cos^{2}(x)\)
\(=a\sqrt{\cos^{2}(\theta)}=a\cos(\theta)\)
Therefore \(\sqrt(a^{2}-a^{2}\sin^{2}(\theta))=a\cos(\theta)\)
\(\implies \cos \theta=\sqrt{a^{2}-a^{2}\frac{\sin^{2}(\theta)}}{a}\)
\(\implies\frac{a\sin(\theta)}{\sqrt{a^{2}-a^{2}\sin^{2}(\theta)}}=\tan(\theta)\)
\(\sqrt{a^{2}-a^{2}\sin^{2}(\theta)}=a\cos(\theta)\)
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