c3 trig :(

I've worked out a and b, which were correct:
(a) 5 cos(θ – 53.13)
(b) max = 5 and smallest value possible = 53.13

For c I got far as substituting:
f(t) = 10+ 5cos(15t-α)
But I don't know what to do after.. do i rearrange?

Your alpha is the same as in part(b), the minimum of cosx= -1, sub that in and work out the minimum temperature.

Hello, okay so f(t) = 10+ 5cos(15t-53.13), but how did you get cosx = -1?

You need to have a minimum of f(t), so make it easier by saying f(t) = 10 + cos(x) , where x = 15t-53.13.

Now, consider the variable factor cos(x), what's the lowest value of cos (x) ? It is -1 because of the graph it peaks at 1 and the bottom is at -1, so substitute the lowest value that cosx can be into f(t) to get the minimum of f(t) .

Erm so I did this:

cos (15t - 53) = -1
15t - 53 = 180
so t = 15.5

but this is the answer for d, and for c the min temp is 5 ?

You don't need to solve for t. Minimum of cos (15t - 53) = -1, substitute that in, so the minimum of f(t) = 10 + 5(-1)