Given that $\tan x=\frac{1}{\sqrt{3}}$ and $\sin y=\frac{\sqrt{28}}{6}$ , 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{1}{\sqrt{3}}

and

\sin y=\frac{\sqrt{28}}{6}

, 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 $\tan x$ : We know that $\tan x = \frac{1}{\sqrt{3}}$ . Since $x$ is in Quadrant I, both sine and cosine are positive. We can find $\cos x$ using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ , and the fact that $\tan x = \frac{\sin x}{\cos x}$ .
Express $\sin x$ : First, let's express $\sin x$ in terms of $\tan x$ . We have $\tan x = \frac{\sin x}{\cos x}$ , so $\sin x = \tan x \cdot \cos x$ . We know $\tan x = \frac{1}{\sqrt{3}}$ , so $\sin x = \left(\frac{1}{\sqrt{3}}\right) \cdot \cos x$ .
Use Pythagorean identity: Now, we use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ . Substituting $\sin x$ with $(1/\sqrt{3}) \cdot \cos x$ , we get $((1/\sqrt{3}) \cdot \cos x)^2 + \cos^2 x = 1$ .
Solve for $\cos x$ : Simplifying the equation, we have $(1/3) \cdot \cos^2 x + \cos^2 x = 1$ . Combining like terms, we get $(1/3 + 1) \cdot \cos^2 x = 1$ , which simplifies to $(4/3) \cdot \cos^2 x = 1$ .
Find $\cos y$ : To find $\cos x$ , we solve for $\cos^2 x$ by multiplying both sides by $\frac{3}{4}$ , which gives us $\cos^2 x = \left(\frac{3}{4}\right)$ . Taking the square root of both sides, we get $\cos x = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ , since $x$ is in Quadrant I and cosine is positive.
Substitute $\sin y$ : Next, we need to find $\cos y$ . We are given $\sin y = \sqrt{28}/6$ . Using the Pythagorean identity $\sin^2 y + \cos^2 y = 1$ , we can find $\cos y$ .
Simplify the equation: Substitute $\sin y$ with $\sqrt{28}/6$ into the identity: $(\sqrt{28}/6)^2 + \cos^2 y = 1$ . Simplifying, we get $(28/36) + \cos^2 y = 1$ .
Find $\cos y$ : Subtracting $\frac{28}{36}$ from both sides, we get $\cos^2 y = 1 - \frac{28}{36}$ . Simplifying further, we get $\cos^2 y = \frac{36}{36} - \frac{28}{36} = \frac{8}{36}$ .
Use cosine sum formula: Taking the square root of both sides to find $\cos y$ , we get $\cos y = \sqrt{\frac{8}{36}} = \frac{\sqrt{8}}{6}$ . Since $y$ is in Quadrant I, $\cos y$ is positive, so $\cos y = \frac{\sqrt{8}}{6} = \frac{\sqrt{2*4}}{6} = \frac{2*\sqrt{2}}{6} = \frac{\sqrt{2}}{3}$ .
Find $\sin x$ : Now we have both $\cos x = \sqrt{3}/2$ and $\cos y = \sqrt{2}/3$ . We can use the cosine sum formula to find $\cos(x+y)$ : $\cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y$ .
Substitute all values: We already have $\cos x$ and $\cos y$ . We need to find $\sin x$ , which we can get from $\tan x = \frac{1}{\sqrt{3}}$ and $\cos x = \frac{\sqrt{3}}{2}$ . Since $\tan x = \frac{\sin x}{\cos x}$ , we have $\sin x = \tan x \cdot \cos x = \left(\frac{1}{\sqrt{3}}\right) \cdot \left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2}$ .
Simplify the expression: We already have $\sin y = \frac{\sqrt{28}}{6}$ . Now we can substitute all values into the cosine sum formula: $\cos(x+y) = \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{\sqrt{2}}{3}\right) - \left(\frac{1}{2}\right) \cdot \left(\frac{\sqrt{28}}{6}\right)$ .
Combine the terms: Simplifying the expression, we get $\cos(x+y) = (\sqrt{3}\cdot\sqrt{2})/(2\cdot3) - (1\cdot\sqrt{28})/(2\cdot6) = (\sqrt{6}/6) - (\sqrt{28}/12)$ .
Simplify the numerator: To combine the terms, we need a common denominator. Multiplying the first term by $\frac{2}{2}$ , we get $\frac{2\sqrt{6}}{12}$ - $\frac{\sqrt{28}}{12}$ = $\frac{2\sqrt{6} - \sqrt{28}}{12}$ .
Final simplification: Simplifying the numerator, we get $(2\sqrt{6} - \sqrt{4\times7})/12 = (2\sqrt{6} - 2\sqrt{7})/12$ .
Final simplification: Simplifying the numerator, we get $(2\sqrt{6} - \sqrt{4\times7})/12 = (2\sqrt{6} - 2\sqrt{7})/12$ .Finally, we can simplify the expression by dividing both numerator and denominator by $2$ , which gives us $(\sqrt{6} - \sqrt{7})/6$ as the exact value of $\cos(x+y)$ in simplest radical form.