# guys, I need some more PURE maths help :DWatch

#1
This is the question

Express 8 cos theta - 5 sin theta in the form r cos (theta + alpha) , where R > 0 and 0 < α < π
Write R in surd form and give the value of α correct to 4 decimal places. (4 marks)

This part is straight forward : I got root 89 as R and 1.0122 radians. you are free to check if I am correct or not.

b)
The temperature of a kiln, T degrees Celsius, used to make pottery can be modelled by the equation T = 1100 + 5 cos (x/3)- 8 sin (x/3) where x is the time in hours since the pottery was placed in the kiln. range:0 > (or equal to) x >72

Calculate the maximum value of T predicted by this model and the value of x, to 2 decimal places, when this maximum first occurs. (4 marks)

I am aware that x/3 replaces theta. I also changed the range, from 0 to 24, but I don't know what to do from here.

c Calculate the times during the first 24 hours when the temperature is predicted, by this model, to be exactly 1097 °C. (4 marks)

HELP WILL BE KINDLY APPRECIATED!!!
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6 days ago
#2
(Original post by KatheO11)
Calculate the maximum value of T predicted by this model and the value of x, to 2 decimal places, when this maximum first occurs. (4 marks)

I am aware that x/3 replaces theta. I also changed the range, from 0 to 24, but I don't know what to do from here.
Rewrite the equation using part (a). What do you get?
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#3
(Original post by RDKGames)
Rewrite the equation using part (a). What do you get?
root 89 cos (x/3 + 1.0122)

But What do I do from here?
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6 days ago
#4
(Original post by KatheO11)
root 89 cos (x/3 + 1.0122)
Not quite... have you completely ignored the fact that there is a 1100? Also the fact that the rest of that expression is not *quite* ?

P.S. In your original post, did you mean to write T = 1100 + 5 sin (x/3)- 8 cos (x/3) ?? Because this is the form I'm assuming. [My point is, it currently doesn't connect to part (a) as you have it, but it's close enough to be a typo at some point of your post!]
Last edited by RDKGames; 6 days ago
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#5
(Original post by RDKGames)
Not quite... have you completely ignored the fact that there is a 1100? Also the fact that the rest of that expression is not *quite* ?

P.S. In your original post, did you mean to write T = 1100 + 5 sin (x/3)- 8 cos (x/3) ?? Because this is the form I'm assuming. [My point is, it currently doesn't connect to part (a) as you have it, but it's close enough to be a typo at some point of your post!]
No, I didn't ignore it, when I wrote it down,I included the 1100.Also, the form is the same T = 1100 + 5 cos (x/3)- 8 sin (x/3)

PS: The way you presented it digitally, is how I wrote it with the 1100. Can you explain why I don't include the radian angle?
Last edited by KatheO11; 6 days ago
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6 days ago
#6
Surely it’s just 1100+root89?
The maximum value of cos is 1 so that’s what the max value of T would be
Then to work out the value of X, you just take away 1100 from the value of the above to get what root 89cos(x/3+1.022) and that will be 1, then you just use the standard trig method to complete this and find x.
You do the same thing for thing for the fourth part except you will get the trig ratio to be a negative
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6 days ago
#7
(Original post by KatheO11)
No, I didn't ignore it, when I wrote it down,I included the 1100.Also, the form is the same T = 1100 + 5 cos (x/3)- 8 sin (x/3)

PS: The way you presented it digitally, is how I wrote it with the 1100. Can you explain why I don't include the radian angle?
Hang on...

You miswrote the beginning of the post then.

You said 8 cos theta - 5 sin theta but your solution corresponds to -8sin(theta)+5cos(theta) which, by the way, makes more sense when it comes to part (b).
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#8
(Original post by RDKGames)
Hang on...

You miswrote the beginning of the post then.

You said 8 cos theta - 5 sin theta but your solution corresponds to -8sin(theta)+5cos(theta) which, by the way, makes more sense when it comes to part (b).
Ahhh, I see what you're saying: sorry about that. What do I do from here, though? I understand another user said I can use the maximum value of T as 1 as the maximum cos value is 1, but does that effect (x/3 + 1.0122)? I kind of need more help, please if you don't mind, thank you.
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6 days ago
#9
(Original post by KatheO11)
Ahhh, I see what you're saying: sorry about that. What do I do from here, though? I understand another user said I can use the maximum value of T as 1 as the maximum cos value is 1, but does that effect (x/3 + 1.0122)? I kind of need more help, please if you don't mind, thank you.
Right then...

You first of all need to realise that you can rewrite T as .

Cosine oscillates taking a max value of 1, and a min value of -1. So the extreme values of are precisely when cosine takes on +/- 1.

This means the max value is obviously (*), whenever (**).

(*) tells you the max value, and (**) can yield the relevant x value(s) when it occurs.
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#10
(Original post by RDKGames)
Right then...

You first of all need to realise that you can rewrite T as .

Cosine oscillates taking a max value of 1, and a min value of -1. So the extreme values of are precisely when cosine takes on +/- 1.

This means the max value is obviously (*), whenever (**).

(*) tells you the max value, and (**) can yield the relevant x value(s) when it occurs.
Thank you so much! I really appreciate the help! Have a good night.
0
6 days ago
#11
(Original post by RDKGames)
Right then...

You first of all need to realise that you can rewrite T as .

Cosine oscillates taking a max value of 1, and a min value of -1. So the extreme values of are precisely when cosine takes on +/- 1.

This means the max value is obviously (*), whenever (**).

(*) tells you the max value, and (**) can yield the relevant x value(s) when it occurs.
That’s expressed a lot better than mine😂
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#12
(Original post by Khushi.S)
That’s expressed a lot better than mine😂
Khushi, Thank you for helping me too. I am sorry that I didn't quote you. I appreciate the help! have a good night.
0
6 days ago
#13
(Original post by KatheO11)
Khushi, Thank you for helping me too. I am sorry that I didn't quote you. I appreciate the help! have a good night.
No problem glad I could help! It was good revision for me too as this was one of the questions I messed up on in my mocks
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