if sin where defined=1/5 and where defined is in quadrant I find the exact value of sin2 where defined
make a diagram and use Pythagoras to find the third side x
x^2 + 1^2 = 5^2
x = sqrt(24)
so cos(theta)= sqrt(24)/5
sin(2theta) = 2sin(theta)cos(theta)
= 2(1/5)(sqrt(24)/5)
= 2sqrt(24)/25
or
= 4sqrt(6)/25
To find the exact value of sin^2(x) in quadrant I, where sin(x) = 1/5, we can use the Pythagorean identity: sin^2(x) + cos^2(x) = 1.
First, let's find the value of cos(x) in quadrant I. In quadrant I, cosine is positive, so we can find cos(x) using the Pythagorean identity: sin^2(x) + cos^2(x) = 1.
Given that sin(x) = 1/5, we can solve for cos(x):
(1/5)^2 + cos^2(x) = 1
1/25 + cos^2(x) = 1
cos^2(x) = 1 - 1/25
cos^2(x) = 24/25
Taking the square root of both sides, we get:
cos(x) = ± √(24/25)
Since we are in quadrant I and cosine is positive, we take the positive value:
cos(x) = √(24/25) = √24/5
Now, we can solve for sin^2(x):
sin^2(x) = 1 - cos^2(x)
sin^2(x) = 1 - (√24/5)^2
sin^2(x) = 1 - 24/25
sin^2(x) = (25 - 24)/25
sin^2(x) = 1/25
So, the exact value of sin^2(x) where sin(x) = 1/5 in quadrant I is 1/25.