Given that $\cos A=\frac{\sqrt{2}}{3}$ and $\cos B=\frac{\sqrt{30}}{6}$ , and that angles $A$ and $B$ are both in Quadrant I, find the exact value of $\sin (A-B)$ , in simplest radical form. $\newline$ Answer:

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Q. Given that

\cos A=\frac{\sqrt{2}}{3}

and

\cos B=\frac{\sqrt{30}}{6}

, and that angles

A

and

B

are both in Quadrant I, find the exact value of

\sin (A-B)

, in simplest radical form.

\newline

Answer:

Use Cosine Subtraction Formula: Use the cosine subtraction formula to find $\sin(A-B)$ . The cosine subtraction formula is $\cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ . Since we are looking for $\sin(A-B)$ , we need to use the sine and cosine values to find $\sin(A)$ and $\sin(B)$ first.
Find Sine Values: Find $\sin(A)$ and $\sin(B)$ using the Pythagorean identity. $\newline$ The Pythagorean identity is $\sin^2(\theta) + \cos^2(\theta) = 1$ . $\newline$ For angle A: $\newline$ $\sin^2(A) = 1 - \cos^2(A)$ $\newline$ $\sin^2(A) = 1 - (\frac{\sqrt{2}}{3})^2$ $\newline$ $\sin^2(A) = 1 - \frac{2}{9}$ $\newline$ $\sin^2(A) = \frac{9}{9} - \frac{2}{9}$ $\newline$ $\sin^2(A) = \frac{7}{9}$ $\newline$ $\sin(A) = \sqrt{\frac{7}{9}}$ $\newline$ $\sin(A) = \frac{\sqrt{7}}{3}$ $\newline$ For angle B: $\newline$ $\sin(B)$ $0$ $\newline$ $\sin(B)$ $1$ $\newline$ $\sin(B)$ $2$ $\newline$ $\sin(B)$ $3$ $\newline$ $\sin(B)$ $4$ $\newline$ $\sin(B)$ $5$ $\newline$ $\sin(B)$ $6$ $\newline$ $\sin(B)$ $7$
Apply Sine Subtraction Formula: Apply the sine subtraction formula. The sine subtraction formula is $\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B)$ . Substitute the values we found for $\sin(A)$ , $\cos(A)$ , $\sin(B)$ , and $\cos(B)$ : $\sin(A-B) = (\sqrt{7}/3)\cdot(\sqrt{30})/6 - (\sqrt{2})/3\cdot(\sqrt{6})/6$
Simplify Expression: Simplify the expression. $\newline$ $\sin(A-B) = \frac{\sqrt{7}\cdot\sqrt{30}}{3\cdot6} - \frac{\sqrt{2}\cdot\sqrt{6}}{3\cdot6}$ $\newline$ $\sin(A-B) = \frac{\sqrt{210}}{18} - \frac{\sqrt{12}}{18}$ $\newline$ $\sin(A-B) = \frac{\sqrt{210} - \sqrt{12}}{18}$ $\newline$ $\sin(A-B) = \frac{\sqrt{210} - 2\cdot\sqrt{3}}{18}$ $\newline$ $\sin(A-B) = \frac{\sqrt{210} - 2\cdot\sqrt{3}}{18}$
Simplify Radicals: Simplify the radicals if possible. $\sqrt{210}$ can be simplified to $\sqrt{2\times3\times5\times7} = \sqrt{2}\times\sqrt{3}\times\sqrt{5}\times\sqrt{7} = \sqrt{6}\times\sqrt{35}$ So, $\sin(A-B) = (\sqrt{6}\times\sqrt{35} - 2\times\sqrt{3})/18$ $\sin(A-B) = (\sqrt{6}\times\sqrt{35})/18 - (2\times\sqrt{3})/18$ $\sin(A-B) = (\sqrt{210})/18 - (2\times\sqrt{3})/18$ We already simplified $\sqrt{210}$ earlier, so this step does not change the expression.