Trigonometry

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I've been stuck on part b for some time. I don't know why maximum isn't 2/11 and I have no idea why they've used theta+alpha= pi for the next part... Why is it pi??? Shouldn't it be 1???

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Reply 1

TenOfThem

18

Original post by ps1265A

I've been stuck on part b for some time. I don't know why maximum isn't 2/11 and I have no idea why they've used theta+alpha= pi for the next part... Why is it pi??? Shouldn't it be 1???

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You want Cos(theta + alpha) = -1

Surely you know that 4/2 is bigger that 4/22

edit - Mis-subtracted

(edited 9 years ago)

Reply 2

ps1265A

OP

14

Original post by TenOfThem

You want Cos(theta + alpha) = -1

Surely you know that 4/6 is bigger that 4/22

I understand b(I) but not. B(Ii)

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Reply 3

Flame Alchemist

13

Part a had you expressing the same trigonometric addition as a single cos function - you should have gotten R as 10, right? Use that same substitution in part b; it'll allow you to deal with the maximum more easily.

Reply 4

TenOfThem

18

Original post by ps1265A

I understand b(I) but not. B(Ii)

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So - you have a maximum of 2 for b(i)

Reply 5

ps1265A

OP

14

Original post by TenOfThem

So - you have a maximum of 2 for b(i)

Yes , 4/12-10

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Reply 6

TenOfThem

18

Original post by ps1265A

Yes , 4/12-10

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So Cos (....) = -1

So (....) = pi

Reply 7

Flame Alchemist

13

Original post by ps1265A

Yes , 4/12-10

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You already assumed that cos(theta - alpha) = -1 in order to work out the maximum. You already found alpha in part a, so just solve that for theta by taking arccos of both sides and rearranging.