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Urgent help C4 trig addition formulae & integration

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Reply 20
Avatar for jordanwu
jordanwu
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Original post by SeanFM
Great :borat:

So yes, you can take the half outside of the integral and that leaves you with (1/sin(...)). What's the integral of that? :h:


ln(sin(theta-(pi/3)))? Not really sure..
Reply 21
Avatar for Zacken
Zacken
22
Original post by jordanwu
ln(sin(theta-(pi/3)))? Not really sure..


No, what does cosec x integrate to? You have a formula booklet. Use it.
Original post by jordanwu
ln(sin(theta-(pi/3)))? Not really sure..


Not quite. You may be able to use something from your formula book or things that you know.

How can you rewrite 1/sin(x) where x is theta-(pi/3) as something that you know the integral of?
Reply 23
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jordanwu
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Original post by SeanFM
Not quite. You may be able to use something from your formula book or things that you know.

How can you rewrite 1/sin(x) where x is theta-(pi/3) as something that you know the integral of?


image.jpg This?
Original post by jordanwu
image.jpg This?


Almost there, but the integral of cosecx is ln|cosecx + cotx|.
Reply 25
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jordanwu
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Original post by SeanFM
Almost there, but the integral of cosecx is ln|cosecx + cotx|.


From this http://filestore.aqa.org.uk/subjects/FORMULAE.PDF it says that it's -ln|cosecx + cotx|?
Original post by jordanwu
From this http://filestore.aqa.org.uk/subjects/FORMULAE.PDF it says that it's -ln|cosecx + cotx|?


Oops, forgot the -, sorry! (it was next to an equals sign where I was looking it up :s-smilie:)
Reply 27
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jordanwu
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Original post by SeanFM
Oops, forgot the -, sorry! (it was next to an equals sign where I was looking it up :s-smilie:)


From the Aqa formula book it also says that the -ln|cosecx + cotx| = ln|tan(1/2x)|?
Original post by jordanwu
From the Aqa formula book it also says that the -ln|cosecx + cotx| = ln|tan(1/2x)|?


Oh, sorry, then you are right! I've never seen that equality before :tongue:
Reply 29
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jordanwu
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Original post by SeanFM
Oh, sorry, then you are right! I've never seen that equality before :tongue:


Lol really? XD But you're studying maths at uni right?
Reply 30
Avatar for Zacken
Zacken
22
Original post by SeanFM
Oh, sorry, then you are right! I've never seen that equality before :tongue:


csc⁡x+cot⁡x=1sin⁡x+cos⁡xsin⁡x=1+cos⁡xsin⁡x=1+2cos⁡2x/2−12cos⁡x/2sin⁡x/2=cos⁡x/2sin⁡x/2\displaystyle \csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1+ \cos x}{\sin x} = \frac{1 + 2 \cos^2 x/2 - 1}{2 \cos x/2 \sin x/2} = \frac{\cos x/2}{\sin x/2}

It's one of those weird things nobody uses or notices. :tongue:
(edited 8 years ago)
Original post by jordanwu
Lol really? XD But you're studying maths at uni right?

Yes :colondollar: but there isn't a huge focus on manipulating trig. The stuff you see at A-level form small parts of the course.
Original post by Zacken
csc⁡x+cot⁡x=1sin⁡x+cos⁡xsin⁡x=1+cos⁡xsin⁡x=1+2cos⁡2x/2−12cos⁡x/2sin⁡x/2=cos⁡x/2sin⁡x/2\displaystyle \csc x + \cot x = \frac{1}{\sin x} + \frac{\cos x}{\sin x} = \frac{1+ \cos x}{\sin x} = \frac{1 + 2 \cos^2 x/2 - 1}{2 \cos x/2 \sin x/2} = \frac{\cos x/2}{\sin x/2}

It's one of those weird things nobody uses or notices. :tongue:


I see, thanks :tongue:

Latex can be funny in that things disappear.
(edited 8 years ago)
Reply 33
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Zacken
22
Original post by SeanFM

Latex can be funny in that things disappear.


It's one of those weird quirks of latex, it doesn't have \cosec x, just \csc x. Thanks for reminding me!
Original post by Zacken
It's one of those weird quirks of latex, it doesn't have \cosec x, just \csc x. Thanks for reminding me!


Thanks for editing my post too :wink:
Reply 35
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Zacken
22
Original post by SeanFM
Thanks for editing my post too :wink:


See your VM's, I'm so sorry!
Reply 36
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jordanwu
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13
Original post by SeanFM
Thanks for editing my post too :wink:


So I've ended up with this: image.jpg, please could you check through it and tell me whether I've made any mistakes? Thanks!
Original post by jordanwu
So I've ended up with this: image.jpg, please could you check through it and tell me whether I've made any mistakes? Thanks!


Your values of tan(pi/3) and tan(pi/6) aren't quite right :redface:

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