Writing a Trigonometric expression as an algebraic expression Write

Question

Writing a Trigonometric expression as an algebraic expression

Write [latex]\sin\left(\sin^{-1}u+\cos^{-1}v\right)[/latex] as an algebric expression containing u and v (that is, without any trigonometric functions). Give the restrictions on u and v.

Accepted Answer

First, for [latex]\sin^{-1} u[/latex], the restriction on u is [latex]-1\leq u\leq1,[/latex] and for [latex]\cos^{-1}v,[/latex] the restriction on v is [latex]-1\,\leq v\leq1.[/latex] Now let [latex]\alpha=\sin^{-1}u[/latex] and [latex]\beta=\cos^{-1}v.[/latex] Then

[latex]\begin{array}{r l r l}{\sin\alpha=u}&{{}}&{-{\frac{\pi}{2}}\leq\alpha\leq{\frac{\pi}{2}}}&{-1\leq u\leq1}\end{array}[/latex]

[latex]\cos\beta=v\qquad{\begin{array}{l}{{0}\leq\beta\leq{\pi}\quad-1\leq v\leq1}\end{array}}[/latex]

Because [latex]-{\frac{\pi}{2}}\leq\alpha\leq{\frac{\pi}{2}},[/latex] this means that [latex]\cos\alpha\geq0.[/latex] As a result,

[latex]\cos\alpha={\sqrt{1-\sin^{2}\alpha}}={\sqrt{1-u^{2}}}[/latex]

Similarly, because [latex]0\leq\beta\leq\pi,[/latex] this means that [latex]\sin\beta\geq0.[/latex] Then

[latex]\sin\beta={\sqrt{1-\cos^{2}\beta}}={\sqrt{1-\ v^{2}}}[/latex]

As a result,

[latex]{\sin\left(\sin^{-1}u+\cos^{-1}v ight)\,=\,\sin\left(\alpha+\beta ight)\,=\,\sin\alpha\cos\beta+\cos\alpha\sin\beta}\ {=\,u v\,+\,{\sqrt{1-u^{2}}\,{\sqrt{1-v^{2}}}}}[/latex]

Question 6.6.5.10: Writing a Trigonometric expression as an algebraic expressio......

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