Given that cos A=(sqrt20)/(5) and cos B=(sqrt6)/(3), 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=(sqrt20)/(5) and  cos B=(sqrt6)/(3), 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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Apply Cosine Subtraction Formula: Use the cosine subtraction formula. The cosine subtraction formula is cos(A - B) = cos(A)cos(B) + sin(A)sin(B). We will use this formula to find sin(A - B) by expressing sin(A) and sin(B) in terms of cos(A) and cos(B).Find sin(A) and sin(B): Find sin(A) and sin(B) using the Pythagorean identity. Since A and B are in Quadrant I, sin(A) and sin(B) will be positive. 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 - (sqrt20/5)^2 sin^2(A) = 1 - (20/25) sin^2(A) = 1 - 4/5 sin^2(A) = 1/5 sin(A) = sqrt(1/5) sin(A) = sqrt(1)/sqrt(5) sin(A) = 1/sqrt(5) sin(A) = sqrt(5)/5 (Rationalizing the denominator)Calculate sin(B): Calculate sin(B) using the same method. For angle B: sin^2(B) = 1 - cos^2(B) sin^2(B) = 1 - (sqrt6/3)^2 sin^2(B) = 1 - (6/9) sin^2(B) = 1 - 2/3 sin^2(B) = 1/3 sin(B) = sqrt(1/3) sin(B) = sqrt(1)/sqrt(3) sin(B) = 1/sqrt(3) sin(B) = sqrt(3)/3 (Rationalizing the denominator)Use sin(A) and sin(B): Use the values of sin(A) and sin(B) to find sin(A - B). We know that sin(A - B) = sin(A)cos(B) - cos(A)sin(B). Substitute the values we found: sin(A - B) = (sqrt(5)/5)(sqrt(6)/3) - (sqrt(20)/5)(sqrt(3)/3) sin(A - B) = (sqrt(30)/15) - (sqrt(60)/15) sin(A - B) = (sqrt(30) - sqrt(60))/15 sin(A - B) = (sqrt(30) - 2*sqrt(15))/15 sin(A - B) = (sqrt(30) - 2*sqrt(3*5))/15 sin(A - B) = (sqrt(30) - 2*sqrt(3)*sqrt(5))/15 sin(A - B) = (sqrt(30) - 2*sqrt(3)*sqrt(5))/15 sin(A - B) = (sqrt(30) - 2*sqrt(15))/15

Given that $\cos A=\frac{\sqrt{20}}{5}$ and $\cos B=\frac{\sqrt{6}}{3}$ , 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=205 \cos A=\frac{\sqrt{20}}{5} cosA=520​​ and cos⁡B=63 \cos B=\frac{\sqrt{6}}{3} cosB=36​​, 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 $\cos A=\frac{\sqrt{20}}{5}$ and $\cos B=\frac{\sqrt{6}}{3}$ , 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: