Suppose that cos 𝜃 = 12/13 and 3𝜋/2 < 𝜃 < 2𝜋. Compute cos 2𝜃

Answer 1

so we're in quadrant 4 where cosine is positive...

cosine = adjacent/hypotenuse = 12/13

adjacent =12 and hypotenuse = 13

then by pythagorean, opposite=5

so the terminal point is (12,-5)

the angle is then 2*pi - inverse_cosine (12/13)

twice this angle is 4*pi - 2 * inverse_cosine(12/13)

the cosine of that is cos[ 4*pi - 2 * inverse_cosine(12/13) ]

= cos[ 4*pi - 2 * arccos(12/13)]

Angle subtraction formula for cosine is

cos(M-n) = cosM cosn + sin M sin n where M=4*pi and n = 2*arccos(12/13)

cosM = cos(4*pi) = 1

sinM = sin(4*pi) = 0

cos N = cos(2*arccos(12/13)) = cos(45+) = 0.70414201183431952662721893491124...< sqrt(2)/2

sin N = sin(2*arccos(12/13)) = sin(45+) = 0.71005917159763313609467455621302.... < sqrt(2)/2

in other words, twice the missing angle is ABOUT 45 degrees APPROXIMATELY

However, since sinM=0 and cosM=1, the last term in the angle subtraction expansion vanishes

and the first term simplifies to just cos n

so the bolded expression above shall be the final result

Answer 2

Cos 2θ = 2 cos ^2 θ -1

So that is 2(12/13 squared ) -1

288/169. -1. = 119/169