Given that $\sin A=\frac{6}{\sqrt{61}}$ and $\cos B=\frac{1}{\sqrt{2}}$ , and that angles $A$ and $B$ are both in Quadrant I, find the exact value of $\cos (A-B)$ , in simplest radical form. $\newline$ Answer:

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Algebra 2

Find trigonometric ratios using multiple identities

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Q. Given that

\sin A=\frac{6}{\sqrt{61}}

and

\cos B=\frac{1}{\sqrt{2}}

, and that angles

A

and

B

are both in Quadrant I, find the exact value of

\cos (A-B)

, in simplest radical form.

\newline

Answer:

Find $\cos(A-B)$ : To find $\cos(A-B)$ , we can use the cosine difference identity: $\cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ . First, we need to find $\cos(A)$ and $\sin(B)$ . Since $\sin A = \frac{6}{\sqrt{61}}$ , we can find $\cos(A)$ using the Pythagorean identity $\sin^2(A) + \cos^2(A) = 1$ .
Calculate $\cos(A)$ : Calculate $\cos(A)$ using the Pythagorean identity: $\newline$ $\cos^2(A) = 1 - \sin^2(A)$ $\newline$ $\cos^2(A) = 1 - (\frac{6}{\sqrt{61}})^2$ $\newline$ $\cos^2(A) = 1 - \frac{36}{61}$ $\newline$ $\cos^2(A) = \frac{61}{61} - \frac{36}{61}$ $\newline$ $\cos^2(A) = \frac{25}{61}$ $\newline$ $\cos(A) = \sqrt{\frac{25}{61}}$ $\newline$ $\cos(A) = \frac{5}{\sqrt{61}}$ $\newline$ Since $A$ is in Quadrant I, $\cos(A)$ is positive.
Find $\sin(B)$ : Next, we need to find $\sin(B)$ . We know that $\cos B = \frac{1}{\sqrt{2}}$ , and using the Pythagorean identity $\sin^2(B) + \cos^2(B) = 1$ , we can find $\sin(B)$ . $\newline$ $\sin^2(B) = 1 - \cos^2(B)$ $\newline$ $\sin^2(B) = 1 - \left(\frac{1}{\sqrt{2}}\right)^2$ $\newline$ $\sin^2(B) = 1 - \frac{1}{2}$ $\newline$ $\sin^2(B) = \frac{1}{2}$ $\newline$ $\sin(B) = \sqrt{\frac{1}{2}}$ $\newline$ $\sin(B)$ $0$ $\newline$ Since B is in Quadrant I, $\sin(B)$ is also positive.
Use cosine difference identity: Now that we have $\cos(A)$ and $\sin(B)$ , we can use the cosine difference identity to find $\cos(A-B)$ : $\cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ $\cos(A-B) = \left(\frac{5}{\sqrt{61}}\right)\left(\frac{1}{\sqrt{2}}\right) + \left(\frac{6}{\sqrt{61}}\right)\left(\frac{1}{\sqrt{2}}\right)$ $\cos(A-B) = \left(\frac{5}{\sqrt{122}}\right) + \left(\frac{6}{\sqrt{122}}\right)$ $\cos(A-B) = \frac{5 + 6}{\sqrt{122}}$ $\cos(A-B) = \frac{11}{\sqrt{122}}$
Rationalize the denominator: To express the answer in simplest radical form, we rationalize the denominator: $\newline$ $\cos(A-B) = \frac{11}{\sqrt{122}} \times \frac{\sqrt{122}}{\sqrt{122}}$ $\newline$ $\cos(A-B) = \frac{11\sqrt{122}}{122}$