Writing a Trigonometric Expression as an Algebraic Expression

Write \(\sin ( \tan ^{-1}u) \) as an algebraic expression containing \(u.\)

Solution Let \(\theta =\tan ^{-1}u\) so that \(\tan \theta =u\), where \(-\dfrac{\pi }{2}<\theta <\dfrac{\pi }{2}\). We note that in the interval \(\left( -\dfrac{\pi }{2},\dfrac{\pi }{2}\right) ,\) \(\sec \theta >0.\) Then \[ \begin{array}{l} \sin (\tan ^{ - 1} u) = \sin \theta = \sin \theta \cdot \frac{{\cos \theta }}{{\cos \theta }} = \tan \theta \cos \theta \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{\tan \theta }}{{\sec \theta }} = \frac{{\tan \theta }}{{\sqrt {1 + \tan ^2 \theta } }} = \frac{u}{{\sqrt {1 + u^2 } }} \\ {\color{#0066A7}{ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec ^2 \theta = 1 + \mathop {\tan ^2 }\limits^ \uparrow \theta ;\sec \theta > 0}} \\ \end{array} \]

An alternate method of obtaining the solution to Example 2 uses right triangles. Let \(\theta =\tan ^{-1}u\) so that \(\tan \theta =u\), \(- \dfrac{\pi }{2}<\theta <\dfrac{\pi }{2}\), and label the right triangles drawn in Figure 88. Using the Pythagorean Theorem, the hypotenuse of each triangle is \(\sqrt{1+u^{2}}.\) Then \(\sin (\tan ^{-1}u) =\sin \theta =\dfrac{u}{\sqrt{1+u^{2}}}.\)