Given that $\cos A=\frac{\sqrt{20}}{5}$ and $\cos B=\frac{\sqrt{6}}{3}$ , 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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Find trigonometric ratios using a Pythagorean or reciprocal identity

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Q. Given that

\cos A=\frac{\sqrt{20}}{5}

and

\cos B=\frac{\sqrt{6}}{3}

, 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:

Apply Cosine Subtraction Formula: Use the cosine subtraction formula. $\newline$ The cosine subtraction formula is $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ . $\newline$ We will use this formula to find $\sin(A - B)$ by expressing $\sin(A)$ and $\sin(B)$ in terms of $\cos(A)$ and $\cos(B)$ .
Find $\sin(A)$ and $\sin(B)$ : Find $\sin(A)$ and $\sin(B)$ using the Pythagorean identity. $\newline$ Since $A$ and $B$ are in Quadrant I, $\sin(A)$ and $\sin(B)$ will be positive. $\newline$ The Pythagorean identity is $\sin^2(\theta) + \cos^2(\theta) = 1$ . $\newline$ For angle $A$ : $\newline$ $\sin(B)$ $0$ $\newline$ $\sin(B)$ $1$ $\newline$ $\sin(B)$ $2$ $\newline$ $\sin(B)$ $3$ $\newline$ $\sin(B)$ $4$ $\newline$ $\sin(B)$ $5$ $\newline$ $\sin(B)$ $6$ $\newline$ $\sin(B)$ $7$ $\newline$ $\sin(B)$ $8$ (Rationalizing the denominator)
Calculate sin(B): Calculate $\sin(B)$ using the same method. $\newline$ For angle B: $\newline$ $\sin^2(B) = 1 - \cos^2(B)$ $\newline$ $\sin^2(B) = 1 - (\sqrt{6}/3)^2$ $\newline$ $\sin^2(B) = 1 - (6/9)$ $\newline$ $\sin^2(B) = 1 - 2/3$ $\newline$ $\sin^2(B) = 1/3$ $\newline$ $\sin(B) = \sqrt{1/3}$ $\newline$ $\sin(B) = \sqrt{1}/\sqrt{3}$ $\newline$ $\sin(B) = 1/\sqrt{3}$ $\newline$ $\sin(B) = \sqrt{3}/3$ (Rationalizing the denominator)
Use $\sin(A)$ and $\sin(B)$ : Use the values of $\sin(A)$ and $\sin(B)$ to find $\sin(A - B)$ . We know that $\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B)$ . Substitute the values we found: $\sin(A - B) = (\sqrt{5}/5)(\sqrt{6}/3) - (\sqrt{20}/5)(\sqrt{3}/3)$ $\sin(A - B) = (\sqrt{30}/15) - (\sqrt{60}/15)$ $\sin(A - B) = (\sqrt{30} - \sqrt{60})/15$ $\sin(A - B) = (\sqrt{30} - 2\sqrt{15})/15$ $\sin(B)$ $0$ $\sin(B)$ $1$ $\sin(B)$ $1$ $\sin(A - B) = (\sqrt{30} - 2\sqrt{15})/15$