find the exact value of the trigonometric expression without the use of a calculator.\n\\( \\sin \\left(…

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: [ \begin{align*} \sin(\alpha+\beta)&=\sin\alpha\cos\beta+\cos\alpha\sin\beta\ &=\frac{2}{3}\times(-\frac{3}{4})+\frac{\sqrt{5}}{3}\times\frac{\sqrt{7}}{4}\ &=-\frac{6}{12}+\frac{\sqrt{35}}{12}\ &=\frac{-6 + \sqrt{35}}{12} \end{align*} ]

Answer:

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