find sin if cos(2è)=(4)/(5) and theta terminates in quadrant 1
cos 2Ø = 1 - 2sin^2 Ø
2sin^2 Ø = 1 - cos 2Ø = 1 - 4/5
sin^2 Ø = 1/10
sin Ø = 1/√10
To find the value of sin(theta) given that cos(2theta) = 4/5 and theta terminates in quadrant 1, we can use the identity cos(2theta) = 1 - 2sin^2(theta). Rearranging this equation, we have 2sin^2(theta) = 1 - cos(2theta).
First, let's find cos(theta) using the given information. Since theta terminates in quadrant 1, cosine is positive in this quadrant. We know that cos(2theta) = 4/5. Since 2theta is in quadrant 1, we can write:
cos(2theta) = cos^2(theta) - sin^2(theta).
Substituting cos(2theta) = 4/5 in the equation, we get:
(4/5) = cos^2(theta) - sin^2(theta).
Next, we can use the Pythagorean identity sin^2(theta) + cos^2(theta) = 1 to simplify the equation further. Rearranging the equation, we have:
sin^2(theta) = 1 - cos^2(theta).
Substituting this value into the previous equation, we get:
(4/5) = cos^2(theta) - (1 - cos^2(theta)).
Simplifying further, we have:
(4/5) = 2cos^2(theta) - 1.
Let's solve this equation for cos(theta):
2cos^2(theta) - 1 = 4/5
2cos^2(theta) = 4/5 + 1
2cos^2(theta) = 9/5
cos^2(theta) = (9/5) / 2
cos^2(theta) = 9/10
Taking the square root of both sides, we get:
cos(theta) = ±√(9/10)
Since theta terminates in quadrant 1 and cosine is positive in this quadrant, we can take the positive square root:
cos(theta) = √(9/10).
Finally, using the Pythagorean identity, we have:
sin^2(theta) + cos^2(theta) = 1
sin^2(theta) + (9/10) = 1
sin^2(theta) = 1 - (9/10)
sin^2(theta) = 1/10
Taking the square root of both sides, we get:
sin(theta) = ±√(1/10).
Since theta terminates in quadrant 1 and sine is positive in this quadrant, we can take the positive square root:
sin(theta) = √(1/10).
Therefore, the value of sin(theta) is √(1/10).