Given that $\sin x=\frac{3}{\sqrt{58}}$ and $\tan y=1$ , 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

\sin x=\frac{3}{\sqrt{58}}

and

\tan y=1

, 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 $\sin x = \frac{3}{\sqrt{58}}$ and $\tan y = 1$ . Since both angles are in Quadrant I, all trigonometric functions are positive. We need to find $\cos(x-y)$ . To do this, we will use the cosine difference identity: $\cos(x-y) = \cos x \cos y + \sin x \sin y$ . First, we need to find $\cos x$ and $\cos y$ .
Finding $\cos x$ : To find $\cos x$ , we use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ . We have $\sin x = \frac{3}{\sqrt{58}}$ , so $\sin^2 x = \left(\frac{3}{\sqrt{58}}\right)^2 = \frac{9}{58}$ . Therefore, $\cos^2 x = 1 - \sin^2 x = 1 - \frac{9}{58} = \frac{49}{58}$ . Since $x$ is in Quadrant I, $\cos x$ is positive, so $\cos x = \sqrt{\frac{49}{58}} = \frac{7}{\sqrt{58}}$ .
Finding $\cos y$ : For angle $y$ , we have $\tan y = 1$ . In Quadrant I, this means that $\sin y = \cos y$ because $\tan y = \frac{\sin y}{\cos y}$ . Since $\tan y = 1$ , $\sin y = \cos y$ . To find the value of $\sin y$ and $\cos y$ , we use the Pythagorean identity $\sin^2 y + \cos^2 y = 1$ . Since $\sin y = \cos y$ , we have $y$ $1$ . Therefore, $y$ $2$ , and $y$ $3$ .
Using Cosine Difference Identity: Now that we have $\cos x = \frac{7}{\sqrt{58}}$ and $\cos y = \frac{\sqrt{2}}{2}$ , and we know $\sin x = \frac{3}{\sqrt{58}}$ and $\sin y = \frac{\sqrt{2}}{2}$ , we can use the cosine difference identity to find $\cos(x-y)$ . So, $\cos(x-y) = \cos x \cos y + \sin x \sin y = \left(\frac{7}{\sqrt{58}}\right)\left(\frac{\sqrt{2}}{2}\right) + \left(\frac{3}{\sqrt{58}}\right)\left(\frac{\sqrt{2}}{2}\right)$ .
Simplifying Expression: We simplify the expression: $\cos(x-y) = \left(\frac{$7$}{\sqrt{$58$}}\right)\left(\frac{\sqrt{$2$}}{$2$}\right) + \left(\frac{$3$}{\sqrt{$58$}}\right)\left(\frac{\sqrt{$2$}}{$2$}\right) = \left(\frac{$7$\sqrt{$2$}}{$2$\sqrt{$58$}}\right) + \left(\frac{$3$\sqrt{$2$}}{$2$\sqrt{$58$}}\right) = \frac{$7$\sqrt{$2$} + $3$\sqrt{$2$}}{$2$\sqrt{$58$}} = \frac{$10$\sqrt{$2$}}{$2$\sqrt{$58$}} = \frac{$5$\sqrt{$2$}}{\sqrt{$58$}}.
Rationalizing Denominator: To rationalize the denominator, we multiply the numerator and the denominator by $\sqrt{58}$: $\cos(x-y) = \frac{5\sqrt{2}}{\sqrt{58}} \cdot \frac{\sqrt{58}}{\sqrt{58}} = \frac{5\sqrt{2}\sqrt{58}}{\sqrt{58}\sqrt{58}} = \frac{5\sqrt{116}}{58}$.
Simplifying Square Root: We simplify the square root in the numerator: $\sqrt{116} = \sqrt{(4\cdot29)} = 2\sqrt{29}$. So, $\cos(x-y) = \frac{5\cdot2\sqrt{29}}{58} = \frac{10\sqrt{29}}{58}$.
Final Simplification: Finally, we simplify the fraction by dividing both the numerator and the denominator by $2$: $\cos(x-y) = \frac{10\sqrt{29}}{58} / \left(\frac{2}{2}\right) = \frac{5\sqrt{29}}{29}$.