Question

chapter ii.8 score: 4/6 answered: 5/6 question 6 take a and b to be angles in the first quadrant with sin(a) = 9/23 and cos(b) = 27/29. calculate sin(a + b).

Explanation:

Step1: Find $\cos(A)$

Using $\sin^{2}\theta+\cos^{2}\theta = 1$, for $\theta = A$, we have $\cos(A)=\sqrt{1 - \sin^{2}(A)}$. Since $\sin(A)=\frac{9}{23}$, then $\cos(A)=\sqrt{1-(\frac{9}{23})^{2}}=\sqrt{\frac{529 - 81}{529}}=\sqrt{\frac{448}{529}}=\frac{8\sqrt{7}}{23}$.

Step2: Find $\sin(B)$

Using $\sin^{2}\theta+\cos^{2}\theta = 1$, for $\theta = B$, we have $\sin(B)=\sqrt{1 - \cos^{2}(B)}$. Since $\cos(B)=\frac{27}{29}$, then $\sin(B)=\sqrt{1 - (\frac{27}{29})^{2}}=\sqrt{\frac{841-729}{841}}=\sqrt{\frac{112}{841}}=\frac{4\sqrt{7}}{29}$.

Step3: Use the sum - formula for sine

The formula for $\sin(A + B)=\sin(A)\cos(B)+\cos(A)\sin(B)$.
Substitute $\sin(A)=\frac{9}{23}$, $\cos(A)=\frac{8\sqrt{7}}{23}$, $\cos(B)=\frac{27}{29}$, $\sin(B)=\frac{4\sqrt{7}}{29}$ into the formula:
\[

$$\begin{align*} \sin(A + B)&=\frac{9}{23}\times\frac{27}{29}+\frac{8\sqrt{7}}{23}\times\frac{4\sqrt{7}}{29}\\ &=\frac{243}{667}+\frac{224}{667}\\ &=\frac{243 + 224}{667}\\ &=\frac{467}{667} \end{align*}$$

\]

Answer:

$\frac{467}{667}$