find the cosine for an angle that has a sine of 3√40/20 and is in quadrant II. use the Pythagorean identity sin^2 (0) + cos^2 (0) = 1 and the quadrant to solve.
In order to find the cosine of an angle given the sine, we can use the Pythagorean identity you provided:
sin^2(θ) + cos^2(θ) = 1
First, let's calculate sin^2(θ). Given that sin(θ) is 3√40/20, we square it to find sin^2(θ):
sin^2(θ) = (3√40/20)^2
= (3/20 * √40)^2
= (3/20 * √(4*10))^2
= (3/20 * 2√10)^2
= (6√10/20)^2
= (3√10/10)^2
= (3√10)^2 / 100
= (9 * 10) / 100
= 90 / 100
= 9 / 10
Now we have sin^2(θ) = 9/10. Let's use this to solve for cos^2(θ):
cos^2(θ) = 1 - sin^2(θ)
cos^2(θ) = 1 - 9/10
cos^2(θ) = 10/10 - 9/10
cos^2(θ) = 1/10
Since we're looking for the cosine of an angle in the second quadrant, where cosine is negative (because cosine corresponds to the x-coordinate on the unit circle, which is negative in the second quadrant), we take the negative square root of cos^2(θ):
cos(θ) = -√(1/10)
cos(θ) = -√1/√10
cos(θ) = -1/√10
cos(θ) = -1/√(10/10)
cos(θ) = -1/√(10)*√(10/10)
cos(θ) = -√10/√(100)
cos(θ) = -√10/10
Therefore, the cosine of the angle with sine 3√40/20 in the second quadrant is -√10/10.