Given that $\tan x=6$ and $\cos y=\frac{3}{\sqrt{13}}$ , 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=6

and

\cos y=\frac{3}{\sqrt{13}}

, 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:

Given Information: We are given $\tan x = 6$ and $\cos y = \frac{3}{\sqrt{13}}$ . Since $x$ and $y$ are in Quadrant I, all trigonometric functions are positive. We need to find $\cos(x+y)$ . To do this, we will use the cosine addition formula: $\cos(x+y) = \cos x \cdot \cos y - \sin x \cdot \sin y$ . First, we need to find $\cos x$ and $\sin y$ .
Finding cos x: To find $\cos x$ , we use the identity $1 + \tan^2 x = \sec^2 x$ . Since $\tan x = 6$ , we have $1 + 6^2 = \sec^2 x$ , which simplifies to $1 + 36 = \sec^2 x$ , so $\sec^2 x = 37$ . Therefore, $\sec x = \sqrt{37}$ , and since $\sec x$ is the reciprocal of $\cos x$ , we have $\cos x = \frac{1}{\sqrt{37}}$ .
Finding $\sin y$ : To find $\sin y$ , we use the Pythagorean identity $\sin^2 y + \cos^2 y = 1$ . We know $\cos y = \frac{3}{\sqrt{13}}$ , so $\cos^2 y = \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13}$ . Substituting into the identity, we get $\sin^2 y + \frac{9}{13} = 1$ . Solving for $\sin^2 y$ gives us $\sin^2 y = 1 - \frac{9}{13} = \frac{4}{13}$ . Taking the square root, we find $\sin y = \sqrt{\frac{4}{13}} = \frac{2}{\sqrt{13}}$ .
Substitute into Formula: Now that we have $\cos x = \frac{1}{\sqrt{37}}$ and $\sin y = \frac{2}{\sqrt{13}}$ , we can substitute these into the cosine addition formula. $\cos(x+y) = \left(\frac{1}{\sqrt{37}}\right) \cdot \left(\frac{3}{\sqrt{13}}\right) - (6) \cdot \left(\frac{2}{\sqrt{13}}\right)$ . Simplifying this expression, we get $\cos(x+y) = \frac{3}{(\sqrt{37} \cdot \sqrt{13})} - \frac{12}{\sqrt{13}}$ .
Combine Terms: To simplify the expression further, we find a common denominator. The common denominator for $\sqrt{37} \times \sqrt{13}$ and $\sqrt{13}$ is $\sqrt{37} \times \sqrt{13}$ . So, we rewrite the expression as $\cos(x+y) = \frac{3\sqrt{13}}{37 \times 13} - \frac{12\sqrt{37}}{37 \times 13}.$
Simplify Denominator: Now we combine the terms over the common denominator: $\cos(x+y) = \frac{3\sqrt{13} - 12\sqrt{37}}{37 \times 13}$ . This is the simplified expression for $\cos(x+y)$ in radical form.
Final Result: Finally, we simplify the denominator: $37 \times 13 = 481$ . So, $\cos(x+y) = (3\sqrt{13} - 12\sqrt{37}) / 481$ . This is the exact value of $\cos(x+y)$ in simplest radical form.