find the exact value of the expressions \\( \\cos ( \\alpha + \\beta ) \\), \\( \\sin ( \\alpha + \\beta )…

Explanation:

Step1: Find $\cos\alpha$

Since $\sin\alpha=\frac{12}{13}$ and $\alpha$ is in quadrant I, using the identity $\sin^{2}\theta+\cos^{2}\theta = 1$, we have $\cos\alpha=\sqrt{1-\sin^{2}\alpha}=\sqrt{1 - (\frac{12}{13})^{2}}=\sqrt{\frac{169 - 144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$.

Step2: Find $\cos\beta$

Since $\sin\beta=\frac{4}{5}$ and $\beta$ is in quadrant II, using the identity $\sin^{2}\theta+\cos^{2}\theta = 1$, we have $\cos\beta=-\sqrt{1-\sin^{2}\beta}=-\sqrt{1 - (\frac{4}{5})^{2}}=-\sqrt{\frac{25 - 16}{25}}=-\sqrt{\frac{9}{25}}=-\frac{3}{5}$.

Step3: Use the sum formula for $\sin(\alpha+\beta)$

The sum formula for sine is $\sin(A + B)=\sin A\cos B+\cos A\sin B$. Substitute $A=\alpha$ and $B = \beta$: $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$. We know that $\sin\alpha=\frac{12}{13}$, $\cos\alpha=\frac{5}{13}$, $\sin\beta=\frac{4}{5}$, and $\cos\beta=-\frac{3}{5}$. $\sin(\alpha+\beta)=\frac{12}{13}\times(-\frac{3}{5})+\frac{5}{13}\times\frac{4}{5}$. First, calculate each product: $\frac{12}{13}\times(-\frac{3}{5})=-\frac{36}{65}$ and $\frac{5}{13}\times\frac{4}{5}=\frac{20}{65}$. Then, add the two results: $\sin(\alpha+\beta)=-\frac{36}{65}+\frac{20}{65}=\frac{-36 + 20}{65}=-\frac{16}{65}$.

Answer:

$-\frac{16}{65}$