Given that $\cos A=\frac{\sqrt{23}}{5}$ and $\sin B=\frac{\sqrt{21}}{6}$ , 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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Algebra 2

Find trigonometric ratios using multiple identities

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Q. Given that

\cos A=\frac{\sqrt{23}}{5}

and

\sin B=\frac{\sqrt{21}}{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 Sine Addition Formula: Use the sine addition formula: $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$ . $\newline$ We need to find $\sin(A)$ and $\cos(B)$ .
Find $\sin(A)$ and $\cos(B)$ : Since $A$ is in Quadrant I, $\sin(A)$ is positive. We can find $\sin(A)$ using the Pythagorean identity: $\sin^2(A) + \cos^2(A) = 1$ . We know $\cos(A) = \frac{\sqrt{23}}{5}$ , so $\cos^2(A) = \frac{23}{25}$ . $\sin^2(A) = 1 - \cos^2(A) = 1 - \frac{23}{25} = \frac{25}{25} - \frac{23}{25} = \frac{2}{25}$ . $\sin(A) = \sqrt{\sin^2(A)} = \sqrt{\frac{2}{25}} = \frac{\sqrt{2}}{5}$ .
Find $\sin(B)$ and $\cos(B)$ : Since $B$ is in Quadrant I, $\cos(B)$ is also positive. We can find $\cos(B)$ using the Pythagorean identity: $\sin^2(B) + \cos^2(B) = 1$ . We know $\sin(B) = \frac{\sqrt{21}}{6}$ , so $\sin^2(B) = \frac{21}{36}$ . $\cos^2(B) = 1 - \sin^2(B) = 1 - \frac{21}{36} = \frac{36}{36} - \frac{21}{36} = \frac{15}{36} = \frac{5}{12}$ after simplifying. $\cos(B) = \sqrt{\cos^2(B)} = \sqrt{\frac{5}{12}} = \frac{\sqrt{5}}{\sqrt{12}} = \frac{\sqrt{5}}{2\sqrt{3}} = \frac{\sqrt{5}\sqrt{3}}{2\cdot3} = \frac{\sqrt{15}}{6}$ after rationalizing the denominator.
Plug in Values: Now we have $\sin(A) = \frac{\sqrt{2}}{5}$ and $\cos(B) = \frac{\sqrt{15}}{6}$ . Plug these values into the sine addition formula: $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$ . $\sin(A+B) = \left(\frac{\sqrt{2}}{5}\right)\left(\frac{\sqrt{15}}{6}\right) + \left(\frac{\sqrt{23}}{5}\right)\left(\frac{\sqrt{21}}{6}\right).$
Simplify Expression: Simplify the expression: $\sin(A+B) = \frac{\sqrt{2}\cdot\sqrt{15}}{5\cdot6} + \frac{\sqrt{23}\cdot\sqrt{21}}{5\cdot6}$ . $\newline$ $\sin(A+B) = \frac{\sqrt{30}}{30} + \frac{\sqrt{483}}{30}.$
Combine Terms: Combine the terms over a common denominator: $\sin(A+B) = (\sqrt{30} + \sqrt{483})/30$ . Since $\sqrt{483}$ can be simplified to $\sqrt{3\times161} = \sqrt{3}\times\sqrt{161} = \sqrt{3}\times\sqrt{7\times23} = \sqrt{3}\times\sqrt{7}\times\sqrt{23}$ , we can rewrite the expression. $\sin(A+B) = (\sqrt{30} + \sqrt{3}\times\sqrt{7}\times\sqrt{23})/30$ .
Simplify Further: Simplify the expression further: $\sin(A+B) = (\sqrt{30} + \sqrt{3}\cdot\sqrt{7}\cdot\sqrt{23})/30$ . $\sin(A+B) = (\sqrt{30} + \sqrt{3}\cdot\sqrt{7}\cdot\sqrt{23})/30 = (\sqrt{30} + \sqrt{3\cdot7\cdot23})/30$ . $\sin(A+B) = (\sqrt{30} + \sqrt{3\cdot7\cdot23})/30 = (\sqrt{30} + \sqrt{483})/30$ .
Final Expression: Notice that $\sqrt{483}$ was already simplified in a previous step, so we do not need to simplify it again. $\newline$ The final expression for $\sin(A+B)$ is $\frac{\sqrt{30} + \sqrt{483}}{30}$ .