How do i find
cos(2arctan(7/24))
Thanks.
A right triangle with sides of 7 and 24 has a hypotenuse of 25. The cosine of arctan 7/24 is 24/25 and the sine is 7/25.
Now use the formula
cos2A = cos^2A - sin^A
to arrive at
cos(2arctan(7/24))
= (24/25)^2 - (7/25)^2 = 527/625
To find cos(2arctan(7/24)), we can use the double-angle identity for cosine. But first, let's simplify the expression inside the arctan function.
We know that arctan(7/24) = θ, where tan(θ) = 7/24.
To find θ, we can use the inverse tangent function (arctan) on a calculator, by inputting 7/24 or its decimal equivalent.
Using a calculator, we find that θ ≈ 0.2736 radians.
Now, the double-angle identity for cosine states that cos(2θ) = 1 - 2sin²(θ).
Since θ ≈ 0.2736, we can substitute this value into the double-angle identity to find cos(2arctan(7/24)).
cos(2(0.2736)) = 1 - 2sin²(0.2736).
To find sin²(0.2736), we can use the identity sin²(θ) = 1 - cos²(θ).
sin²(θ) = 1 - cos²(θ).
sin²(0.2736) = 1 - cos²(0.2736).
Now, find cos²(0.2736) using a calculator or by evaluating cos²(0.2736).
Using a calculator, we find that cos²(0.2736) ≈ 0.9646.
Substituting this back into the equation, we have:
sin²(0.2736) = 1 - cos²(0.2736).
sin²(0.2736) = 1 - 0.9646.
sin²(0.2736) ≈ 0.0354.
Now, we can substitute sin²(0.2736) into the double-angle identity for cosine:
cos(2arctan(7/24)) = 1 - 2(0.0354).
cos(2arctan(7/24)) ≈ 1 - 2(0.0354).
cos(2arctan(7/24)) ≈ 1 - 0.0708.
cos(2arctan(7/24)) ≈ 0.9292.
Therefore, cos(2arctan(7/24)) is approximately 0.9292.