Given that sin A=(6)/(sqrt61) and cos B=(1)/(sqrt2), and that angles A and B are both in Quadrant I, find the exact value of cos(A-B), in simplest radical form. Answer:

Question

Given that  sin A=(6)/(sqrt61) and  cos B=(1)/(sqrt2), and that angles  A and  B are both in Quadrant I, find the exact value of  cos(A-B), in simplest radical form. Answer:

Bytelearn · Accepted Answer

Find cos(A-B): To find cos(A-B), we can use the cosine difference identity: cos(A-B) = cos(A)cos(B) + sin(A)sin(B). First, we need to find cos(A) and sin(B). Since sin A = 6/sqrt(61), we can find cos(A) using the Pythagorean identity sin^2(A) + cos^2(A) = 1.Calculate cos(A): Calculate cos(A) using the Pythagorean identity: cos^2(A) = 1 - sin^2(A) cos^2(A) = 1 - (6/sqrt(61))^2 cos^2(A) = 1 - 36/61 cos^2(A) = (61/61) - (36/61) cos^2(A) = 25/61 cos(A) = sqrt(25/61) cos(A) = 5/sqrt(61) Since A is in Quadrant I, cos(A) is positive.Find sin(B): Next, we need to find sin(B). We know that cos B = 1/sqrt(2), and using the Pythagorean identity sin^2(B) + cos^2(B) = 1, we can find sin(B). sin^2(B) = 1 - cos^2(B) sin^2(B) = 1 - (1/sqrt(2))^2 sin^2(B) = 1 - 1/2 sin^2(B) = 1/2 sin(B) = sqrt(1/2) sin(B) = 1/sqrt(2) Since B is in Quadrant I, sin(B) is also positive.Use cosine difference identity: Now that we have cos(A) and sin(B), we can use the cosine difference identity to find cos(A-B): cos(A-B) = cos(A)cos(B) + sin(A)sin(B) cos(A-B) = (5/sqrt(61))(1/sqrt(2)) + (6/sqrt(61))(1/sqrt(2)) cos(A-B) = (5/sqrt(122)) + (6/sqrt(122)) cos(A-B) = (5 + 6)/sqrt(122) cos(A-B) = 11/sqrt(122)Rationalize the denominator: To express the answer in simplest radical form, we rationalize the denominator: cos(A-B) = 11/sqrt(122) * sqrt(122)/sqrt(122) cos(A-B) = 11sqrt(122)/122

Given that $\sin A=\frac{6}{\sqrt{61}}$ and $\cos B=\frac{1}{\sqrt{2}}$ , 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:

Full solution

More problems from Find trigonometric ratios using multiple identities

Related topics

Given that sin⁡A=661 \sin A=\frac{6}{\sqrt{61}} sinA=61​6​ and cos⁡B=12 \cos B=\frac{1}{\sqrt{2}} cosB=2​1​, and that angles A A A and B B B are both in Quadrant I, find the exact value of cos⁡(A−B) \cos (A-B) cos(A−B), in simplest radical form.\newlineAnswer:

Given that $\sin A=\frac{6}{\sqrt{61}}$ and $\cos B=\frac{1}{\sqrt{2}}$ , 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: