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Maths question

I got 4sin(x+48.59) for part 11a but have no clue how to do 11b) could someone please help
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Reply 1
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Zacken
22
Original post by Jism1495
I got 4sin(x+48.59) for part 11a but have no clue how to do 11b) could someone please help


Use the fact that −4≤4sin⁡(x+48.59∘)≤4-4 \leq 4 \sin (x + 48.59^{\circ}) \leq 4 since you know that sin⁡x\sin x oscillates between −1-1 and 11. The minimum value of ff will be when the denominator is the largest. And you know what the largest the denominator can get... look at my inequality.
Reply 2
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Zacken
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Original post by Jism1495
I got 4sin(x+48.59) for part 11a but have no clue how to do 11b) could someone please help


I've posted a detailed explanation of what's going on (for a different question but same thing really) here. It might be beneficial for you to have a read over it, just in case. :smile:
Reply 3
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Jism1495
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Original post by Zacken
I've posted a detailed explanation of what's going on (for a different question but same thing really) here. It might be beneficial for you to have a read over it, just in case. :smile:


Thank you, do you have a link to it?
Reply 4
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Zacken
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Original post by Jism1495
Thank you, do you have a link to it?


****, I'm dumb - http://www.thestudentroom.co.uk/showpost.php?p=62893967&postcount=2361 here you go.

Basically, explot the fact that sin⁡\sin is boundede between −1-1 and 11.
Reply 5
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Jism1495
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Original post by Zacken
****, I'm dumb - http://www.thestudentroom.co.uk/showpost.php?p=62893967&postcount=2361 here you go.

Basically, explot the fact that sin⁡\sin is boundede between −1-1 and 11.

I'm still confused, could you tell me the answer you got and I shall try to work towards it, thank yiu
Reply 6
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Zacken
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Original post by Jism1495
I'm still confused, could you tell me the answer you got and I shall try to work towards it, thank yiu


The denominator is maximum with a max value of 4. Max (f) = 20/4.
Reply 7
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Jism1495
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Original post by Zacken
The denominator is maximum with a max value of 4. Max (f) = 20/4.


Thanks, I knew it wouldn't be that hard it's just too late at night nothing's making sense and it's so frustrating
Reply 8
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Jism1495
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Original post by Jism1495
Thanks, I knew it wouldn't be that hard it's just too late at night nothing's making sense and it's so frustrating

Do you happen to know how to do this question? It's the last one I have to do and it looks nothing like trig equations I've done before where I can substitute something in
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Reply 9
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Zacken
22
Original post by Jism1495
Do you happen to know how to do this question? It's the last one I have to do and it looks nothing like trig equations I've done before where I can substitute something in


Do the Rsin (x+ alpha) thing on the two trig terms and then solve it = -3.
Reply 10
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Jism1495
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Original post by Zacken
Do the Rsin (x+ alpha) thing on the two trig terms and then solve it = -3.


Okay thanks, for future questions how would I know whether to use rcos or rsin?
You can use either. The value of alpha will just be different in each case.

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