Question 6.6.5.10: Writing a Trigonometric expression as an algebraic expressio......
Writing a Trigonometric expression as an algebraic expression
Write \sin\left(\sin^{-1}u+\cos^{-1}v\right) as an algebric expression containing u and v (that is, without any trigonometric functions). Give the restrictions on u and v.
First, for \sin^{-1} u, the restriction on u is -1\leq u\leq1, and for \cos^{-1}v, the restriction on v is -1\,\leq v\leq1. Now let \alpha=\sin^{-1}u and \beta=\cos^{-1}v. Then
\begin{array}{r l r l}{\sin\alpha=u}&{{}}&{-{\frac{\pi}{2}}\leq\alpha\leq{\frac{\pi}{2}}}&{-1\leq u\leq1}\end{array}\cos\beta=v\qquad{\begin{array}{l}{{0}\leq\beta\leq{\pi}\quad-1\leq v\leq1}\end{array}}
Because -{\frac{\pi}{2}}\leq\alpha\leq{\frac{\pi}{2}}, this means that \cos\alpha\geq0. As a result,
\cos\alpha={\sqrt{1-\sin^{2}\alpha}}={\sqrt{1-u^{2}}}Similarly, because 0\leq\beta\leq\pi, this means that \sin\beta\geq0. Then
\sin\beta={\sqrt{1-\cos^{2}\beta}}={\sqrt{1-\ v^{2}}}As a result,
{\sin\left(\sin^{-1}u+\cos^{-1}v\right)\,=\,\sin\left(\alpha+\beta\right)\,=\,\sin\alpha\cos\beta+\cos\alpha\sin\beta}\\ {=\,u v\,+\,{\sqrt{1-u^{2}}\,{\sqrt{1-v^{2}}}}}
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