c3 trig :( Watch

Rukia12
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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?
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NotNotBatman
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(Original post by Rukia12)
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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.
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Rukia12
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(Original post by NotNotBatman)
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?
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NotNotBatman
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(Original post by Rukia12)
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) .
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Rukia12
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(Original post by NotNotBatman)
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 ?
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NotNotBatman
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(Original post by Rukia12)
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)
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Rukia12
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(Original post by NotNotBatman)
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)
Oh i see!
Thank you very much !
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