Given that cos A=(sqrt2)/(3) and cos B=(sqrt30)/(6), and that angles A and B are both in Quadrant I, find the exact value of sin(A-B), in simplest radical form. Answer:

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Given that  cos A=(sqrt2)/(3) and  cos B=(sqrt30)/(6), and that angles  A and  B are both in Quadrant I, find the exact value of  sin(A-B), in simplest radical form. Answer:

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Use Cosine Subtraction Formula: Use the cosine subtraction formula to find sin(A-B). The cosine subtraction formula is cos(A-B) = cos(A)cos(B) + sin(A)sin(B). Since we are looking for sin(A-B), we need to use the sine and cosine values to find sin(A) and sin(B) first.Find Sine Values: Find sin(A) and sin(B) using the Pythagorean identity. The Pythagorean identity is sin^2(θ) + cos^2(θ) = 1. For angle A: sin^2(A) = 1 - cos^2(A) sin^2(A) = 1 - ((sqrt2)/3)^2 sin^2(A) = 1 - (2/9) sin^2(A) = (9/9) - (2/9) sin^2(A) = 7/9 sin(A) = sqrt(7/9) sin(A) = sqrt(7)/3 For angle B: sin^2(B) = 1 - cos^2(B) sin^2(B) = 1 - ((sqrt30)/6)^2 sin^2(B) = 1 - (30/36) sin^2(B) = (36/36) - (30/36) sin^2(B) = 6/36 sin^2(B) = 1/6 sin(B) = sqrt(1/6) sin(B) = sqrt(6)/6Apply Sine Subtraction Formula: Apply the sine subtraction formula. The sine subtraction formula is sin(A-B) = sin(A)cos(B) - cos(A)sin(B). Substitute the values we found for sin(A), cos(A), sin(B), and cos(B): sin(A-B) = (sqrt(7)/3)*(sqrt30)/6 - (sqrt2)/3*(sqrt(6)/6)Simplify Expression: Simplify the expression. sin(A-B) = (sqrt(7)*sqrt(30))/(3*6) - (sqrt2*sqrt(6))/(3*6) sin(A-B) = (sqrt(210))/18 - (sqrt(12))/18 sin(A-B) = (sqrt(210) - sqrt(12))/18 sin(A-B) = (sqrt(210) - 2*sqrt(3))/18 sin(A-B) = (sqrt(210) - 2*sqrt(3))/18Simplify Radicals: Simplify the radicals if possible. sqrt(210) can be simplified to sqrt(2*3*5*7) = sqrt(2)*sqrt(3)*sqrt(5)*sqrt(7) = sqrt(6)*sqrt(35) So, sin(A-B) = (sqrt(6)*sqrt(35) - 2*sqrt(3))/18 sin(A-B) = (sqrt(6)*sqrt(35))/18 - (2*sqrt(3))/18 sin(A-B) = (sqrt(210))/18 - (2*sqrt(3))/18 We already simplified sqrt(210) earlier, so this step does not change the expression.

Given that $\cos A=\frac{\sqrt{2}}{3}$ and $\cos B=\frac{\sqrt{30}}{6}$ , 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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Given that cos⁡A=23 \cos A=\frac{\sqrt{2}}{3} cosA=32​​ and cos⁡B=306 \cos B=\frac{\sqrt{30}}{6} cosB=630​​, and that angles A A A and B B B are both in Quadrant I, find the exact value of sin⁡(A−B) \sin (A-B) sin(A−B), in simplest radical form.\newlineAnswer:

Given that $\cos A=\frac{\sqrt{2}}{3}$ and $\cos B=\frac{\sqrt{30}}{6}$ , 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: