Question

find the exact value of the trigonometric expression without the use of a calculator.
\\( \sin \left( \sin ^ { - 1 } \left( \frac { 1 } { 2 } \
ight) + \cos ^ { - 1 } \left( - \frac { 3 } { 4 } \
ight) \
ight) \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. do n
\\( \sin \left( \sin ^ { - 1 } \left( \frac { 1 } { 2 } \
ight) + \cos ^ { - 1 } \left( - \frac { 3 } { 4 } \
ight) \
ight) = \square \\)

Explanation:

Step1: Let \(\alpha=\sin^{-1}(\frac{2}{3})\) and \(\beta = \cos^{-1}(-\frac{3}{4})\)

By the definition of inverse - sine function, if \(\alpha=\sin^{-1}(\frac{2}{3})\), then \(\sin\alpha=\frac{2}{3}\) and \(\cos\alpha=\sqrt{1 - (\frac{2}{3})^2}=\sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}\) (since \(\cos\alpha=\sqrt{1-\sin^{2}\alpha}\) and \(\alpha\in[-\frac{\pi}{2},\frac{\pi}{2}]\), so \(\cos\alpha\geq0\)).
By the definition of inverse - cosine function, if \(\beta=\cos^{-1}(-\frac{3}{4})\), then \(\cos\beta=-\frac{3}{4}\) and \(\sin\beta=\sqrt{1 - (-\frac{3}{4})^2}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}\) (since \(\sin\beta=\sqrt{1 - \cos^{2}\beta}\) and \(\beta\in[0,\pi]\), so \(\sin\beta\geq0\)).

Step2: Use the sum formula for sine \(\sin(A + B)=\sin A\cos B+\cos A\sin B\)

Here \(A=\alpha\) and \(B = \beta\), so \(\sin(\sin^{-1}(\frac{2}{3})+\cos^{-1}(-\frac{3}{4}))=\sin(\alpha+\beta)\).
Substitute \(\sin\alpha=\frac{2}{3},\cos\alpha=\frac{\sqrt{5}}{3},\cos\beta=-\frac{3}{4},\sin\beta=\frac{\sqrt{7}}{4}\) into the sum formula:

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Answer:

\(\frac{-6+\sqrt{35}}{12}\)