Given that $\tan x=\sqrt{3}$ and $\tan y=\frac{3}{\sqrt{7}}$ , 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

\tan x=\sqrt{3}

and

\tan y=\frac{3}{\sqrt{7}}

, 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:

Use Angle Subtraction Formula: We will use the angle subtraction formula for sine, which is $\sin(x-y) = \sin(x)\cos(y) - \cos(x)\sin(y)$ . To find $\sin(x)$ and $\cos(x)$ , we use the fact that $\tan x = \frac{\sin x}{\cos x}$ . Since $\tan x = \sqrt{3}$ , we can create a right triangle where the opposite side to angle $x$ is $\sqrt{3}$ and the adjacent side is $1$ , because $\tan x = \frac{\text{opposite}}{\text{adjacent}}$ . Then, using the Pythagorean theorem, we find the hypotenuse.
Find Triangle Sides: For angle $x$ , we have $\tan x = \frac{\sqrt{3}}{1}$ , so the sides of the triangle are opposite = $\sqrt{3}$ , adjacent = $1$ . Using the Pythagorean theorem, $\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$ , we get $\text{hypotenuse}^2 = (\sqrt{3})^2 + 1^2 = 3 + 1 = 4$ . Therefore, the hypotenuse is $\sqrt{4} = 2$ .
Calculate $\sin x$ and $\cos x$ : Now we can find $\sin x$ and $\cos x$ . Since $\sin x = \frac{\text{opposite}}{\text{hypotenuse}}$ , we have $\sin x = \frac{\sqrt{3}}{2}$ . And since $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$ , we have $\cos x = \frac{1}{2}$ .
Find Triangle Sides: Next, we find $\sin y$ and $\cos y$ using the same method. We have $\tan y = \frac{3}{\sqrt{7}}$ , so we can create a right triangle where the opposite side to angle $y$ is $3$ and the adjacent side is $\sqrt{7}$ . Again, using the Pythagorean theorem, we find the hypotenuse.
Calculate $\sin y$ and $\cos y$ : For angle $y$ , we have $\tan y = \frac{3}{\sqrt{7}}$ , so the sides of the triangle are opposite = $3$ , adjacent = $\sqrt{7}$ . Using the Pythagorean theorem, $\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$ , we get $\text{hypotenuse}^2 = 3^2 + (\sqrt{7})^2 = 9 + 7 = 16$ . Therefore, the hypotenuse is $\sqrt{16} = 4$ .
Substitute Values: Now we can find $\sin y$ and $\cos y$ . Since $\sin y = \frac{\text{opposite}}{\text{hypotenuse}}$ , we have $\sin y = \frac{3}{4}$ . And since $\cos y = \frac{\text{adjacent}}{\text{hypotenuse}}$ , we have $\cos y = \frac{\sqrt{7}}{4}$ .
Simplify Expression: We can now substitute the values of $\sin x$ , $\cos x$ , $\sin y$ , and $\cos y$ into the angle subtraction formula for sine: $\sin(x-y) = \sin(x)\cos(y) - \cos(x)\sin(y) = (\sqrt{3}/2)(\sqrt{7}/4) - (1/2)(3/4)$ .
Combine Terms: Simplify the expression: $\sin(x-y) = \frac{\sqrt{3} \cdot \sqrt{7}}{2 \cdot 4} - \frac{1 \cdot 3}{2 \cdot 4} = \frac{\sqrt{21}}{8} - \frac{3}{8}.$
Combine Terms: Simplify the expression: $\sin(x-y) = (\sqrt{3} \times \sqrt{7})/(2 \times 4) - (1 \times 3)/(2 \times 4) = (\sqrt{21}/8) - (3/8)$ .Combine the terms to get the final answer: $\sin(x-y) = (\sqrt{21} - 3)/8$ .