If sin(θ) = 8/17 and θ terminates in QI, find cos(θ)
sinØ = 8/17
y = 8, r = 17
x^2 + 8^2 = 17^2
x^2 = 225
x = ± 15 , but we are told Ø is in quad I
so x = 15
cosØ = x/r = 15/17
To find cos(θ), we can use the trigonometric identity:
cos^2(θ) + sin^2(θ) = 1
Given that sin(θ) = 8/17, we can substitute this into the equation:
cos^2(θ) + (8/17)^2 = 1
Simplifying, we have:
cos^2(θ) + 64/289 = 1
Subtracting 64/289 from both sides, we get:
cos^2(θ) = 289/289 - 64/289
cos^2(θ) = 225/289
Taking the square root of both sides, we have:
cos(θ) = √(225/289)
cos(θ) = 15/17
Therefore, the value of cos(θ) is 15/17.
To find the value of cos(θ), we can use the Pythagorean identity:
sin^2(θ) + cos^2(θ) = 1
We already know that sin(θ) = 8/17. We can substitute this value into the equation to solve for cos(θ):
(8/17)^2 + cos^2(θ) = 1
Simplifying the equation:
64/289 + cos^2(θ) = 1
Now, subtracting 64/289 from both sides:
cos^2(θ) = 1 - 64/289
cos^2(θ) = (289 - 64)/289
cos^2(θ) = 225/289
Taking the square root of both sides:
cos(θ) = ±√(225/289)
Since θ terminates in the first quadrant (QI), the cosine of θ is positive. Therefore:
cos(θ) = √(225/289)
cos(θ) ≈ ±15/17
However, since θ terminates in QI, the cosine of θ is positive. So the final answer is:
cos(θ) = 15/17