Given that $\tan A=2$ and $\sin B=\frac{5}{\sqrt{41}}$ , 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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Q. Given that

\tan A=2

and

\sin B=\frac{5}{\sqrt{41}}

, 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 Angle Subtraction Formula: We will use the angle subtraction formula for sine, which is $\sin(A-B) = \sin(A)\cos(B) - \cos(A)\sin(B)$ . To use this formula, we need to find $\sin(A)$ , $\cos(A)$ , and $\cos(B)$ .
Find $\sin(A)$ , $\cos(A)$ , $\cos(B)$ : Since $\tan A = 2$ and we are in Quadrant I, we can create a right triangle where the opposite side to angle A is $2$ and the adjacent side is $1$ . Using the Pythagorean theorem, the hypotenuse is $\sqrt{1^2 + 2^2} = \sqrt{5}$ . Therefore, $\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}$ . To rationalize, we multiply by $\frac{\sqrt{5}}{\sqrt{5}}$ to get $\sin(A) = \frac{2\sqrt{5}}{5}$ .
Use Pythagorean Identity for $\sin(B)$ : Using the same triangle, $\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{5}}$ . Rationalizing, we get $\cos(A) = \frac{\sqrt{5}}{5}$ .
Calculate $\cos(B)$ : We are given $\sin B = \frac{5}{\sqrt{41}}$ . To find $\cos(B)$ , we use the Pythagorean identity $\sin^2(B) + \cos^2(B) = 1$ . Substituting $\sin B$ , we get $\left(\frac{5}{\sqrt{41}}\right)^2 + \cos^2(B) = 1$ . Simplifying, we get $\frac{25}{41} + \cos^2(B) = 1$ .
Substitute into Angle Subtraction Formula: Subtract $\frac{25}{41}$ from both sides to solve for $\cos^2(B)$ : $\cos^2(B) = 1 - \frac{25}{41}$ . This simplifies to $\cos^2(B) = \left(\frac{41}{41}\right) - \left(\frac{25}{41}\right) = \frac{16}{41}$ . Taking the square root, $\cos(B) = \sqrt{\frac{16}{41}} = \frac{4}{\sqrt{41}}$ . Since B is in Quadrant I, $\cos(B)$ is positive, so $\cos(B) = \frac{4}{\sqrt{41}}$ . Rationalizing, we get $\cos(B) = \frac{4\sqrt{41}}{41}$ .
Combine Terms: Now we have $\sin(A) = \frac{2\sqrt{5}}{5}$ , $\cos(A) = \frac{\sqrt{5}}{5}$ , and $\cos(B) = \frac{4\sqrt{41}}{41}$ . We can substitute these into the angle subtraction formula: $\sin(A-B) = \left(\frac{2\sqrt{5}}{5}\right)\left(\frac{4\sqrt{41}}{41}\right) - \left(\frac{\sqrt{5}}{5}\right)\left(\frac{5}{\sqrt{41}}\right)$ .
Rewrite Fraction: Multiplying the terms, we get $\sin(A-B) = \frac{8\sqrt{205}}{205} - \frac{5\sqrt{5}}{41}$ . To combine these terms, we need a common denominator, which is $205$ .
Subtract Fractions: We rewrite $(5\sqrt{5}/41)$ as $(25\sqrt{5}/205)$ to have the same denominator. Now, $\sin(A-B) = (8\sqrt{205}/205) - (25\sqrt{5}/205)$ .
Final Result: Subtracting the fractions, we get $\sin(A-B) = \frac{8\sqrt{205} - 25\sqrt{5}}{205}$ . This is the exact value of $\sin(A-B)$ in simplest radical form.