Given that $\tan x=\frac{2}{\sqrt{5}}$ and $\sin y=\frac{4}{5}$ , and that angles $x$ and $y$ are both in Quadrant I, find the exact value of $\cos (x+y)$ , in simplest radical form. $\newline$ Answer:

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Q. Given that

\tan x=\frac{2}{\sqrt{5}}

and

\sin y=\frac{4}{5}

, and that angles

x

and

y

are both in Quadrant I, find the exact value of

\cos (x+y)

, in simplest radical form.

\newline

Answer:

Find $\cos x$ : We know that $\tan x = \frac{2}{\sqrt{5}}$ . To find $\cos x$ , we use the identity $\tan^2 x + 1 = \sec^2 x$ , which can be rewritten as $\frac{1}{\cos^2 x}$ . We can then solve for $\cos x$ . $\tan^2 x = \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5}$ $1 + \tan^2 x = 1 + \frac{4}{5} = \frac{9}{5}$ Therefore, $\sec^2 x = \frac{9}{5}$ , which means $\cos^2 x = \frac{5}{9}$ .Since $\tan x = \frac{2}{\sqrt{5}}$ $0$ is in Quadrant I, $\cos x$ is positive, so $\tan x = \frac{2}{\sqrt{5}}$ $2$ .
Find $\cos y$ : We are given $\sin y = \frac{4}{5}$ . To find $\cos y$ , we use the Pythagorean identity $\sin^2 y + \cos^2 y = 1$ . We can then solve for $\cos y$ . $\newline$ $\sin^2 y = (\frac{4}{5})^2 = \frac{16}{25}$ $\newline$ $1 - \sin^2 y = 1 - \frac{16}{25} = \frac{9}{25}$ $\newline$ Therefore, $\cos^2 y = \frac{9}{25}$ , which means $\cos y = \sqrt{\frac{9}{25}} = \frac{3}{5}$ . $\newline$ Since $y$ is in Quadrant I, $\cos y$ is positive, so $\sin y = \frac{4}{5}$ $1$ .
Calculate $\cos(x+y)$ : Now we need to find $\cos(x+y)$ . We use the angle sum identity for cosine, which is $\cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y$ . We already have $\cos x = \sqrt{5}/3$ and $\cos y = 3/5$ . To find $\sin x$ , we use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ . $\cos^2 x = (\sqrt{5}/3)^2 = 5/9$ $1 - \cos^2 x = 1 - 5/9 = 4/9$ Therefore, $\sin^2 x = 4/9$ , which means $\cos(x+y)$ $0$ . Since $\cos(x+y)$ $1$ is in Quadrant I, $\sin x$ is positive, so $\cos(x+y)$ $3$ .
Calculate $\cos(x+y)$ : Now we need to find $\cos(x+y)$ . We use the angle sum identity for cosine, which is $\cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y$ . We already have $\cos x = \sqrt{5}/3$ and $\cos y = 3/5$ . To find $\sin x$ , we use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ . $\cos^2 x = (\sqrt{5}/3)^2 = 5/9$ $1 - \cos^2 x = 1 - 5/9 = 4/9$ Therefore, $\sin^2 x = 4/9$ , which means $\cos(x+y)$ $0$ . Since x is in Quadrant I, $\sin x$ is positive, so $\cos(x+y)$ $2$ .Now we can calculate $\cos(x+y)$ using the values we have found: $\cos(x+y)$ $4$ $\cos(x+y)$ $5$ To combine these fractions, we need a common denominator, which is $\cos(x+y)$ $6$ . $\cos(x+y)$ $7$ $\cos(x+y)$ $8$