This discussion is now closed.

Check out other Related discussions

ocr maths c3 help?

2

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

Reply 1

zKlown

19

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

Reply 2

kingcal07

OP

2

thanks very much i `ll try and carry it on from that

Reply 3

kingcal07

OP

2

i m still stuck....from the second step

Reply 4

zKlown

19

Lol, I ain't too strong on Maths, especially Trig and I'm sure if you find it hard others will too

Reply 5

Aurel-Aqua

5

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.

Reply 6

Aurel-Aqua

5

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.

Reply 7

kingcal07

OP

2

thank you so much for this ...

Reply 8

kingcal07

OP

2

I only reached up to 7cos(4b)sin(2b)-sin(4b)cos(2b)

Reply 9

Aurel-Aqua

5

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).

Reply 10

kingcal07

OP

2

cheers mate

Reply 11

scaredoftests

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?

Reply 12

zKlown

19

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!

Reply 13

Jay.M04

1

What happened to the coefficient on both sides

Reply 14

Jay.M04

1

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

ocr maths c3 help?

Related discussions

Articles for you

How The Student Room is moderated