Given that $\cos x=\frac{3}{\sqrt{10}}$ and $\tan y=\frac{3}{2}$ , and that angles $x$ and $y$ are both in Quadrant I, find the exact value of $\sin (x-y)$ , in simplest radical form. $\newline$ Answer:

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Q. Given that

\cos x=\frac{3}{\sqrt{10}}

and

\tan y=\frac{3}{2}

, and that angles

x

and

y

are both in Quadrant I, find the exact value of

\sin (x-y)

, in simplest radical form.

\newline

Answer:

Find $\sin x$ : Use the Pythagorean identity to find $\sin x$ . Since $\cos x = \frac{3}{\sqrt{10}}$ , we can find $\sin x$ using the identity $\sin^2(x) + \cos^2(x) = 1$ . $\sin^2(x) = 1 - \cos^2(x)$ $\sin^2(x) = 1 - \left(\frac{3}{\sqrt{10}}\right)^2$ $\sin^2(x) = 1 - \frac{9}{10}$ $\sin^2(x) = \frac{10}{10} - \frac{9}{10}$ $\sin^2(x) = \frac{1}{10}$ $\sin x$ $0$ $\sin x$ $1$ $\sin x$ $2$
Find $\sin y$ : Use the definition of tangent to find $\sin y$ . Since $\tan y = \frac{3}{2}$ , and $\tan y = \frac{\sin y}{\cos y}$ , we can find $\sin y$ using the Pythagorean identity $\sin^2(y) + \cos^2(y) = 1$ . Let's find $\cos y$ first: $\tan^2(y) + 1 = \sec^2(y)$ $\left(\frac{3}{2}\right)^2 + 1 = \frac{1}{\cos^2(y)}$ $\frac{9}{4} + 1 = \frac{1}{\cos^2(y)}$ $\sin y$ $0$ $\sin y$ $1$ $\sin y$ $2$ $\sin y$ $3$ Now, we can find $\sin y$ : $\sin y$ $5$ $\sin y$ $6$ $\sin y$ $7$
Find $\sin(x-y)$ : Use the angle difference identity for sine to find $\sin(x-y)$ . The identity is $\sin(x-y) = \sin x \cdot \cos y - \cos x \cdot \sin y$ . $\sin(x-y) = \frac{1}{\sqrt{10}} \cdot \frac{2}{\sqrt{13}} - \frac{3}{\sqrt{10}} \cdot \frac{3}{\sqrt{13}}$ $\sin(x-y) = \frac{2}{\sqrt{130}} - \frac{9}{\sqrt{130}}$ $\sin(x-y) = \frac{2 - 9}{\sqrt{130}}$ $\sin(x-y) = -\frac{7}{\sqrt{130}}$
Rationalize $\sin(x-y)$ : Rationalize the denominator of $\sin(x-y)$ . To rationalize the denominator, multiply the numerator and the denominator by $\sqrt{130}$ . $\sin(x-y) = \frac{-7}{\sqrt{130}} \times \frac{\sqrt{130}}{\sqrt{130}}$ $\sin(x-y) = \frac{-7\times\sqrt{130}}{130}$