Speed of a pendulum at lowest point

[skate_nerd]

Homework Statement

I am told to find the speed of a pendulum bob at the lowest point of the swing where the initial angle of displacement is θ=5.7° and the period of the pendulum is 2.5 seconds.

Homework Equations

K.E.=.5mv², T=2∏√(L/g), P.E.=mgh, h=L-Lcosθ=L(1-cosθ)

The Attempt at a Solution

So I found the length of the pendulum first by plugging in 9.81 m/s² for g and 2.5 s for T into the equation T=2∏√(L/g) to get 1.552 m.

Next, we know that the height displacement in a pendulum bob is equal to L-Lcosθ, so I plug this into P.E.=mgh, and also from conservation of energy we know P.E. can be set as .5mv², so the final equation I used is:
v=√(2gL(1-cosθ))=√(2(9.81)(1.553(1-cos(5.7)))) which got me the answer v=2.244 m/s.

However the answer is apparently v=0.388 m/s. My teacher is kind of notorious for being wrong a lot of the time, so if somebody could let me know which answer is right and why that would be awesome, thanks.

 

[mfb]

Did you set your calculator to degrees (DEG) to calculate sin(5.7°)?

 

[skate_nerd]

Wow I feel stupid. Got the answer now, thanks.