Given that $\sin x=\frac{\sqrt{24}}{6}$ and $\sin y=\frac{\sqrt{5}}{3}$ , and that angles $x$ and $y$ are both in Quadrant I, find the exact value of $\sin (x+y)$ , in simplest radical form. $\newline$ Answer:

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Q. Given that

\sin x=\frac{\sqrt{24}}{6}

and

\sin y=\frac{\sqrt{5}}{3}

, and that angles

x

and

y

are both in Quadrant I, find the exact value of

\sin (x+y)

, in simplest radical form.

\newline

Answer:

Use Sine Addition Formula: Use the sine addition formula. $\newline$ The sine addition formula is $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ . $\newline$ We need to find $\cos(x)$ and $\cos(y)$ to use this formula.
Find Cos Values: Find $\cos(x)$ and $\cos(y)$ using the Pythagorean identity. $\newline$ Since $\sin^2(x) + \cos^2(x) = 1$ and $\sin^2(y) + \cos^2(y) = 1$ , we can find $\cos(x)$ and $\cos(y)$ . $\newline$ For $x$ : $\sin(x) = (\sqrt{24})/6 = 2(\sqrt{6})/6 = (\sqrt{6})/3$ , so $\sin^2(x) = 6/9 = 2/3$ . $\newline$ Therefore, $\cos^2(x) = 1 - \sin^2(x) = 1 - 2/3 = 1/3$ . $\newline$ Since $x$ is in Quadrant I, $\cos(y)$ $1$ .
Substitute Values: Find $\cos(y)$ using the same method. $\newline$ For $y$ : $\sin(y) = \frac{\sqrt{5}}{3}$ , so $\sin^2(y) = \frac{5}{9}$ . $\newline$ Therefore, $\cos^2(y) = 1 - \sin^2(y) = 1 - \frac{5}{9} = \frac{4}{9}$ . $\newline$ Since $y$ is in Quadrant I, $\cos(y) = \sqrt{\frac{4}{9}} = \frac{2}{3}$ .
Simplify Expression: Substitute the values into the sine addition formula. $\newline$ $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ $\newline$ $\sin(x+y) = \left(\frac{\sqrt{6}}{3}\right)\left(\frac{2}{3}\right) + \left(\frac{\sqrt{3}}{3}\right)\left(\frac{\sqrt{5}}{3}\right)$
Simplify Expression: Substitute the values into the sine addition formula. $\newline$ $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ $\newline$ $\sin(x+y) = (\frac{\sqrt{6}}{3})(\frac{2}{3}) + (\frac{\sqrt{3}}{3})(\frac{\sqrt{5}}{3})$ Simplify the expression. $\newline$ $\sin(x+y) = (\frac{2\sqrt{6}}{9}) + (\frac{\sqrt{15}}{9})$ $\newline$ $\sin(x+y) = (\frac{2\sqrt{6} + \sqrt{15}}{9})$ $\newline$ This is the simplest radical form.