You can put this solution on YOUR website! if cos(t) = 5/13, then t = arccos(5/13) = 67.38013505 if t is in Q1.
t will be equal to 360 - 67.38013505 if t is in Q4.
that means that t = 292.6198649 degrees.
there are several ways to find cos(2t).
the first way is to double the angle and then find its cosine.
when t = 292.6198649 degrees, 2t = 585.2397299 degrees.
cos(585.2397299) = -.7041420118
the second way is to use the formula that tells you that cos(2t) = cos^2t - sin^2t.
this is derived from the formula for cos(a + b) = cos(a)cos(b) - sin(a)sin(b).
if a is equal to t and b is equal to t, then this formula becomes cos(t)cos(t) - sin(t)sin(t) which becomes cos^2(t) - sin^2(t).
since sin^2(t) is equal to 1 - cos^2(t), then this formula becomes cos^2(t) - (1 - cos^2(t) which becomes cos^2(t) - 1 + cos^2(t) which becomes 2cos^2(t) - 1.
the formula is therefore cos(2t) = 2cos^2(t) - 1.
applying this formula to t, we get:
cos(2t) = 2cos^2(t) - 1 becomes cos(2*292.6198649) = 2cos^2(292.6198649) - 1 which becomes 2*(cos(292.6198649))^2 - 1 which becomes 2*(.3846153846)^2 - 1 which becomes -.7041420118.
your answer should be that cos(2t) = -.7041420118.
there are other ways to derive it but these are the ones that i am familiar with so far.
i'm still experimenting with the first way to determine if it will always find the correct answer.
so far it's been doing good, but the verdict is not in yet, at least not for me.
if in doubt, go with the second way.
that is the method using trigonometric identities that is normally taught.
