kingcal07
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#1
Report Thread starter 12 years ago
#1
how do you solve this equation ?

3sin6bcosec2b = 4 for 0 <b <90degrees

please state every step so that i know how to do this question
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zKlown
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#2
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Cosec = 1/sin

Bring the 4 over and you get 3sin.1/sin = 4

One of the sin will cancel out the 1/sin so your left with 2sin6b x 2b = 4

That's as far as I can take it without fully working it out which I CBA to do, try that
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kingcal07
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thanks very much i `ll try and carry it on from that
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kingcal07
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i m still stuck....from the second step
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zKlown
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Lol, I ain't too strong on Maths, especially Trig and I'm sure if you find it hard others will too
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Aurel-Aqua
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(Original post by zKlown)
Cosec = 1/sin

Bring the 4 over and you get 3sin.1/sin = 4

One of the sin will cancel out the 1/sin so your left with 2sin6b x 2b = 4

That's as far as I can take it without fully working it out which I CBA to do, try that
Wrong.
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Aurel-Aqua
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(Original post by kingcal07)
how do you solve this equation ?

3sin6bcosec2b = 4 for 0 <b <90degrees

please state every step so that i know how to do this question
It needs a lot of good manipulation (or someone could show if it's actually simpler than I think it is).
3\sin{(6b)}\text{cosec}{(2b)} = 4 \implies 3\sin(6b) = 4\sin(2b)

Now, nothing useful can be seen, until you actually find the common factors. Note that:
\sin(6b) = \sin(4b+2b) = \sin(4b)cos(2b)+\cos(4b)\sin(2b) and \sin(2b) = \sin(4b-2b) = \sin(4b)cos(2b)-\cos(4b)\sin(2b).

I think you could work from there. If you're stuck still, ask.
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kingcal07
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thank you so much for this ...
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kingcal07
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I only reached up to 7cos(4b)sin(2b)-sin(4b)cos(2b)
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Aurel-Aqua
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(Original post by kingcal07)
I only reached up to 7cos(4b)sin(2b)-sin(4b)cos(2b)
Quote me next time.

7cos(4b)sin(2b)-sin(4b)cos(2b)=0 implies 7cos(4b)sin(2b)=sin(4b)cos(2b). Now you can do something to both sides: Assuming that cos4b isn't equal to zero, and cos2b isn't equal to zero, we divide both sides by cos4bcos2b to get tan4b = 7tan2b. You'll have to expand the left hand side using double angle formula (that is, in terms of tan2b), and from that you should get a nice quadratic (don't forget to cancel common factors first).
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kingcal07
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cheers mate
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scaredoftests
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I seem not to understand this question

I have a circle with centre

(1,-2) and radius 4

question:

The point A has coordinates(2,-1). find the equation of the chord of the circle which has A as its mid-point.

how do I this? does the chord cross the centre of the circle?
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zKlown
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#13
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#13
(Original post by zKlown)
Cosec = 1/sin

Bring the 4 over and you get 3sin.1/sin = 4

One of the sin will cancel out the 1/sin so your left with 2sin6b x 2b = 4

That's as far as I can take it without fully working it out which I CBA to do, try that
Why am I getting negged for this?

It was 2 years ago!
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Jay.M04
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What happened to the coefficient on both sides
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Jay.M04
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#15
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(Original post by Aurel-Aqua)
It needs a lot of good manipulation (or someone could show if it's actually simpler than I think it is).
3\sin{(6b)}\text{cosec}{(2b)} = 4 \implies 3\sin(6b) = 4\sin(2b)

Now, nothing useful can be seen, until you actually find the common factors. Note that:
\sin(6b) = \sin(4b+2b) = \sin(4b)cos(2b)+\cos(4b)\sin(2b) and \sin(2b) = \sin(4b-2b) = \sin(4b)cos(2b)-\cos(4b)\sin(2b).

I think you could work from there. If you're stuck still, ask.
What happened to the coefficients on both sides
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