Given that $\cos x=\frac{\sqrt{5}}{5}$ and $\sin y=\frac{1}{3}$ , 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

\cos x=\frac{\sqrt{5}}{5}

and

\sin y=\frac{1}{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: To find $\sin(x+y)$ , we will use the sine addition formula: $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ . First, we need to find $\sin(x)$ and $\cos(y)$ .
Find $\sin(x)$ : Since $x$ is in Quadrant I, $\sin(x)$ is positive. We can find $\sin(x)$ using the Pythagorean identity: $\sin^2(x) + \cos^2(x) = 1$ . We know $\cos(x) = \frac{\sqrt{5}}{5}$ , so we can solve for $\sin(x)$ . $\sin^2(x) = 1 - \cos^2(x)$ $\sin^2(x) = 1 - \left(\frac{\sqrt{5}}{5}\right)^2$ $\sin^2(x) = 1 - \frac{5}{25}$ $x$ $0$ $x$ $1$ $x$ $2$ $x$ $3$ $x$ $4$
Find $\cos(y)$ : Now we need to find $\cos(y)$ . Since $y$ is in Quadrant I, $\cos(y)$ is also positive. Using the Pythagorean identity again: $\sin^2(y) + \cos^2(y) = 1$ . We know $\sin(y) = \frac{1}{3}$ , so we can solve for $\cos(y)$ . $\newline$ $\cos^2(y) = 1 - \sin^2(y)$ $\newline$ $\cos^2(y) = 1 - \left(\frac{1}{3}\right)^2$ $\newline$ $\cos^2(y) = 1 - \frac{1}{9}$ $\newline$ $\cos(y)$ $0$ $\newline$ $\cos(y)$ $1$ $\newline$ $\cos(y)$ $2$ $\newline$ $\cos(y)$ $3$
Calculate $\sin(x+y)$ : Now that we have $\sin(x)$ and $\cos(y)$ , we can use the sine addition formula to find $\sin(x+y)$ . $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ $\sin(x+y) = \left[\frac{2\sqrt{5}}{5}\right] * \left[\frac{2\sqrt{2}}{3}\right] + \left[\frac{\sqrt5}{5}\right] * \left[\frac{1}{3}\right]$ $\sin(x+y) = \left[\frac{4\sqrt{10}}{15}\right] + \left[\frac{\sqrt5}{15}\right]$ $\sin(x+y) = \left[\frac{4\sqrt{10} + \sqrt5}{15}\right]$