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Given that
sin A=(sqrt7)/(6) and
tan B=sqrt3, and that angles
A and
B are both in Quadrant I, find the exact value of
cos(A-B), in simplest radical form.
Answer:

Given that $\sin A=\frac{\sqrt{7}}{6}$ and $\tan B=\sqrt{3}$ , 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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Find trigonometric ratios using multiple identities

Full solution

Q. Given that

\sin A=\frac{\sqrt{7}}{6}

and

\tan B=\sqrt{3}

, 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 $\cos A$ : Use the Pythagorean identity to find $\cos A$ . $\newline$ Since $\sin A = \frac{\sqrt{7}}{6}$ , we can use the identity $\sin^2 A + \cos^2 A = 1$ to find $\cos A$ . $\newline$ $\cos^2 A = 1 - \sin^2 A$ $\newline$ $\cos^2 A = 1 - \left(\frac{\sqrt{7}}{6}\right)^2$ $\newline$ $\cos^2 A = 1 - \frac{7}{36}$ $\newline$ $\cos^2 A = \frac{36}{36} - \frac{7}{36}$ $\newline$ $\cos^2 A = \frac{29}{36}$ $\newline$ $\cos A$ $0$ $\newline$ $\cos A$ $1$
Find $\cos B$ : Use the identity $1 + \tan^2 B = \sec^2 B$ to find $\cos B$ . Since $\tan B = \sqrt{3}$ , we can find $\sec B$ and then use the reciprocal identity to find $\cos B$ . $\sec^2 B = 1 + \tan^2 B$ $\sec^2 B = 1 + (\sqrt{3})^2$ $\sec^2 B = 1 + 3$ $\sec^2 B = 4$ $1 + \tan^2 B = \sec^2 B$ $0$ $1 + \tan^2 B = \sec^2 B$ $1$ $1 + \tan^2 B = \sec^2 B$ $2$ $1 + \tan^2 B = \sec^2 B$ $3$
Find $\cos(A-B)$ : Use the cosine difference identity to find $\cos(A-B)$ . The identity is $\cos(A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B$ . $\cos(A-B) = \frac{\sqrt{29}}{6} \cdot \frac{1}{2} + \frac{\sqrt{7}}{6} \cdot \frac{\sqrt{3}}{2}$ $\cos(A-B) = \frac{\sqrt{29}}{12} + \frac{\sqrt{7} \cdot \sqrt{3}}{12}$ $\cos(A-B) = \frac{\sqrt{29}}{12} + \frac{\sqrt{21}}{12}$ $\cos(A-B) = \frac{\sqrt{29} + \sqrt{21}}{12}$

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