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Given that
cos A=(sqrt12)/(5) and
cos B=(sqrt28)/(6), 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 $\cos A=\frac{\sqrt{12}}{5}$ and $\cos B=\frac{\sqrt{28}}{6}$ , 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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Transformations of linear functions

Full solution

Q. Given that

\cos A=\frac{\sqrt{12}}{5}

and

\cos B=\frac{\sqrt{28}}{6}

, 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:

Use Cosine Difference Identity: To find the cosine of the difference between two angles, we use the cosine difference identity: $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ .
Find Sine Values: We already have the values for $\cos(A)$ and $\cos(B)$ . Now we need to find $\sin(A)$ and $\sin(B)$ . Since $A$ and $B$ are in Quadrant $I$ , their sine values will be positive.
Calculate $\sin(A)$ : To find $\sin(A)$ , we use the Pythagorean identity: $\sin^2(A) + \cos^2(A) = 1$ . We know $\cos(A) = (\sqrt{12})/5$ , so we can solve for $\sin(A)$ . $\newline$ $\sin^2(A) = 1 - \cos^2(A) = 1 - (\sqrt{12}/5)^2 = 1 - (12/25) = (25/25) - (12/25) = 13/25$ . $\newline$ Therefore, $\sin(A) = \sqrt{13/25} = (\sqrt{13})/5$ .
Calculate $\sin(B)$ : Similarly, to find $\sin(B)$ , we use the Pythagorean identity: $\sin^2(B) + \cos^2(B) = 1$ . We know $\cos(B) = (\sqrt{28})/6$ , so we can solve for $\sin(B)$ . $\newline$ $\sin^2(B) = 1 - \cos^2(B) = 1 - (\sqrt{28}/6)^2 = 1 - (28/36) = (36/36) - (28/36) = 8/36 = 2/9$ . $\newline$ Therefore, $\sin(B) = \sqrt{2/9} = (\sqrt{2})/3$ .
Calculate $\cos(A - B)$ : Now we can use the values we have found to calculate $\cos(A - B)$ using the cosine difference identity: $\newline$ $\cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B) = (\sqrt{12}/5)(\sqrt{28}/6) + (\sqrt{13}/5)(\sqrt{2}/3)$ .
Simplify Expression: Simplify the expression by multiplying the radicals and fractions: $\cos(A - B) = \frac{\sqrt{12 \times 28}}{5 \times 6} + \frac{\sqrt{13 \times 2}}{5 \times 3} = \frac{\sqrt{336}}{30} + \frac{\sqrt{26}}{15}$ .
Simplify Square Root: Simplify the square root of $336$ by factoring out perfect squares: $\sqrt{336} = \sqrt{16 \times 21} = 4 \times \sqrt{21}$ . So, $\cos(A - B) = \frac{4 \times \sqrt{21}}{30} + \frac{\sqrt{26}}{15}$ .
Combine Terms: Now we simplify the fractions: $(4 \cdot \sqrt{21})/30 = (2 \cdot \sqrt{21})/15$ . So, $\cos(A - B) = (2 \cdot \sqrt{21})/15 + (\sqrt{26})/15$ .
Combine Terms: Now we simplify the fractions: $(4 \cdot \sqrt{21})/30 = (2 \cdot \sqrt{21})/15$ . So, $\cos(A - B) = (2 \cdot \sqrt{21})/15 + (\sqrt{26})/15$ . Since both terms have a common denominator, we can combine them: $\cos(A - B) = ((2 \cdot \sqrt{21}) + \sqrt{26})/15$ .

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