Given that $\cos x=\frac{\sqrt{5}}{4}$ and $\sin y=\frac{1}{4}$ , 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

\cos x=\frac{\sqrt{5}}{4}

and

\sin y=\frac{1}{4}

, 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 Trigonometric Values: We are given that $\cos x = \frac{\sqrt{5}}{4}$ and $\sin y = \frac{1}{4}$ , and we need to find $\cos(x-y)$ . We can use the cosine difference identity, which states that $\cos(x-y) = \cos x \cdot \cos y + \sin x \cdot \sin y$ . First, we need to find $\cos y$ and $\sin x$ .
Find $\cos y$ : Since $\sin y = \frac{1}{4}$ and $y$ is in Quadrant I, where all trigonometric functions are positive, we can find $\cos y$ using the Pythagorean identity $\sin^2y + \cos^2y = 1$ . We calculate $\cos y$ as follows: $\newline$ $\cos^2y = 1 - \sin^2y$ $\newline$ $\cos^2y = 1 - \left(\frac{1}{4}\right)^2$ $\newline$ $\cos^2y = 1 - \frac{1}{16}$ $\newline$ $\cos^2y = \frac{15}{16}$ $\newline$ $\sin y = \frac{1}{4}$ $0$ $\newline$ $\sin y = \frac{1}{4}$ $1$
Find $\sin x$ : Similarly, we can find $\sin x$ using the Pythagorean identity $\sin^2x + \cos^2x = 1$ . We calculate $\sin x$ as follows: $\newline$ $\sin^2x = 1 - \cos^2x$ $\newline$ $\sin^2x = 1 - (\sqrt{5}/4)^2$ $\newline$ $\sin^2x = 1 - 5/16$ $\newline$ $\sin^2x = 11/16$ $\newline$ $\sin x = \sqrt{11/16}$ $\newline$ $\sin x = \sqrt{11}/4$
Calculate $\cos(x-y)$ : Now that we have $\cos y = \frac{\sqrt{15}}{4}$ and $\sin x = \frac{\sqrt{11}}{4}$ , we can substitute these values into the cosine difference identity: $\newline$ $\cos(x-y) = \cos x \cdot \cos y + \sin x \cdot \sin y$ $\newline$ $\cos(x-y) = \left(\frac{\sqrt{5}}{4}\right) \cdot \left(\frac{\sqrt{15}}{4}\right) + \left(\frac{\sqrt{11}}{4}\right) \cdot \left(\frac{1}{4}\right)$ $\newline$ $\cos(x-y) = \frac{\sqrt{5} \cdot \sqrt{15}}{4 \cdot 4} + \frac{\sqrt{11}}{4 \cdot 4}$ $\newline$ $\cos(x-y) = \frac{\sqrt{75} + \sqrt{11}}{16}$
Simplify $\sqrt{75}$ : We can simplify $\sqrt{75}$ by factoring out the perfect square: $\newline$ $\sqrt{75} = \sqrt{25 \times 3}$ $\newline$ $\sqrt{75} = \sqrt{25} \times \sqrt{3}$ $\newline$ $\sqrt{75} = 5 \times \sqrt{3}$
Substitute and Simplify: Substitute the simplified form of $\sqrt{75}$ back into the expression for $\cos(x-y)$ : $\cos(x-y) = \frac{5 \cdot \sqrt{3} + \sqrt{11}}{16}$ This is the exact value of $\cos(x-y)$ in simplest radical form.