#1

Alright, I'm not in the mood for any 'lolol divide by zero' crap or 'do your own homework crap'. This is

This one part of the chapter is ****ing me over big time and it's really annoying. If you could solve these for me, leaving methods so I can practice more, I'd be such a help.

(0)= theta.

6sin(0)cos(0)-3cos(0)+2sin(0)-1

3cos^2(0)-13cos(0)=3sin^2(0)-9

tan^2(0)-4tan(0)= 0

I need all the help I can get.

**revision**for a college exam and I need help.This one part of the chapter is ****ing me over big time and it's really annoying. If you could solve these for me, leaving methods so I can practice more, I'd be such a help.

(0)= theta.

__For -(pie) < (0) < (pie), solve;__6sin(0)cos(0)-3cos(0)+2sin(0)-1

For 0 < (0) 2(pie), solve;For 0 < (0) 2(pie), solve;

3cos^2(0)-13cos(0)=3sin^2(0)-9

__For 180 < (0) < 180, solve;__tan^2(0)-4tan(0)= 0

I need all the help I can get.

#2

Yay, radians!

Use sin^2 O+cos^2 O =1

so you end up getting something like 6sin^2 O = 3

then cancel it, square root it, inverse sine it and find the solutions for -n<0<n

I couldn't be bothered doing it for you.

Just remember that the suare root will give you +ve and -ve numbers, I lost 3 marks in my mock because I forgot

Use sin^2 O+cos^2 O =1

so you end up getting something like 6sin^2 O = 3

then cancel it, square root it, inverse sine it and find the solutions for -n<0<n

I couldn't be bothered doing it for you.

Just remember that the suare root will give you +ve and -ve numbers, I lost 3 marks in my mock because I forgot

#3

Yay, radians!

Use sin^2 O+cos^2 O =1

so you end up getting something like 6sin^2 O = 3

then cancel it, square root it, inverse sin it and find the solutions for -n<0<n

I'm alright with that actually. It's pretty much everything else.

Cheers anyway!

#4

Do you have log tables?

They're pretty standard here...

They're pretty standard here...

#5

Do you have log tables?

They're pretty standard here...

We do log, but I've never approached tables.

#6

Isn't that trig identities?

Tan(o) = sin(o)/cos(o)

And sin^2 (o) + cos^2 (o) = 1

The basic trick is seeing how it wants you to divide up the bits, rearranging the squared one and tan can be expanded out etc; i'll see if i can work some of hte examples you gave

Tan(o) = sin(o)/cos(o)

And sin^2 (o) + cos^2 (o) = 1

The basic trick is seeing how it wants you to divide up the bits, rearranging the squared one and tan can be expanded out etc; i'll see if i can work some of hte examples you gave

#7

Isn't that trig identities?

Tan(o) = sin(o)/cos(o)

Forgot to mention that one

Edit:

If you're struggling with ones like:

For 180 < (0) < 180, solve;

tan^2(0)-4tan(0)= 0

take tanO=y, so y^2-4y=0 and see if that helps...

*Last edited by bequickorbedead at May 27, 2008,*

#8

Alright, I'm not in the mood for any 'lolol divide by zero' crap or 'do your own homework crap'. This is

This one part of the chapter is ****ing me over big time and it's really annoying. If you could solve these for me, leaving methods so I can practice more, I'd be such a help.

(0)= theta.

For 0 < (0) 2(pie), solve;

3cos^2(0)-13cos(0)=3sin^2(0)-9

tan^2(0)-4tan(0)= 0

I need all the help I can get.

how can you solve something with no '='?

im doing this exam too

but im **** at it

**revision**for a college exam and I need help.This one part of the chapter is ****ing me over big time and it's really annoying. If you could solve these for me, leaving methods so I can practice more, I'd be such a help.

(0)= theta.

6sin(0)cos(0)-3cos(0)+2sin(0)-1

__For -(pie) < (0) < (pie), solve;__6sin(0)cos(0)-3cos(0)+2sin(0)-1

For 0 < (0) 2(pie), solve;

3cos^2(0)-13cos(0)=3sin^2(0)-9

__For 180 < (0) < 180, solve;__tan^2(0)-4tan(0)= 0

I need all the help I can get.

how can you solve something with no '='?

im doing this exam too

but im **** at it

#9

Really guys, I had my core 2 test 2 weeks ago, and i revised like a mong, the trcik with trig identities is just to do loads of worked examples from tests with the mark scheme to help and eventually you see what its directing you towards and how to use the standard squared and tan rules

#10

how can you solve something with no '='?

im doing this exam too

but im **** at it

Sorry, yeah it equals 0.

#11

Really guys, I had my core 2 test 2 weeks ago, and i revised like a mong, the trcik with trig identities is just to do loads of worked examples from tests with the mark scheme to help and eventually you see what its directing you towards and how to use the standard squared and tan rules

I'm doing that. I'm also doing extra practice from the book. Anything wrong with that?

#12

Can't you scan the questions? I assume it's a previous exam paper right?

If I could see it on paper I could do it easily.

If I could see it on paper I could do it easily.

#13

Can't you scan the questions? I assume it's a previous exam paper right?

If I could see it on paper I could do it easily.

Wait there, I'll do it in paint...

#14

Wait there, I'll do it in paint...

This ought to be good lol

#15

This ought to be good lol

Sorry! Been working with a mate on MSN doing these.

#16

Is that 2 different questions?

EDIT: And do you want me to scan you some old papers I've got upstairs?

EDIT: And do you want me to scan you some old papers I've got upstairs?

*Last edited by BlackLuster at May 27, 2008,*

#17

I'm doing that. I'm also doing extra practice from the book. Anything wrong with that?

No there is nothing wrong with that, the only thing that would be wrong is if you weren't doing the work, which you are

Which exam board is it?

#18

Is that 2 different questions?

Yep. For the context, refer to the OP.

I've got one sorted, so I didn't write it down.

#19

No there is nothing wrong with that, the only thing that would be wrong is if you weren't doing the work, which you are

Which exam board is it?

Fair do.

#20

It is just trig identities. you need to reduce it to either an identity (e.g. sin theta = x) or to a quadratic in a single trig function (sin, cor or tan). Substitute in y (e.g. sine theta = y), solve, and then find what theta is.

#21

Damn, that is really, really difficult.

Give me a while on this one.

Give me a while on this one.

#22

Cheers dude.

#23

tan^2$ - 4tan$ = 0

tan$(tan$ - 4) = 0

Therefore either tan$ = 0

or tan$ - 4 = 0

Tan $ = 0

$ = -180, 0, 180 (degrees)

tan$ = 4

$ = tan-1($) (i haven't got a calculator on me)

EDIT: $ = 75.96 degrees

So for -180 to 180 you would have theta = -104.04 and 75.96 degrees

tan$(tan$ - 4) = 0

Therefore either tan$ = 0

or tan$ - 4 = 0

Tan $ = 0

$ = -180, 0, 180 (degrees)

tan$ = 4

$ = tan-1($) (i haven't got a calculator on me)

EDIT: $ = 75.96 degrees

So for -180 to 180 you would have theta = -104.04 and 75.96 degrees

*Last edited by BlackLuster at May 27, 2008,*

#24

I'm trying to do the one with 6sinthetha... it's quite hard...

The shorter tan one is easy.

The shorter tan one is easy.

#25

I'm trying to do the one with 6sinthetha... it's quite hard...

The shorter tan one is easy.

Same here. I've tried factorising and that doesn't really work. I've tried dividing through by cos theta and sin theta (which you're not even supposed to do) and that didn't really work, just got a load of cosecs and secs.

But i'm still working on it.

#26

mmmm trig identities, I don't mind them,

Although, i've done the 3cos^2theta-13costheta=3sin^2theta-9

Using the identity sin^2theta + cos^2theta = 1

I've got as far as costheta(3costheta-10)=6

Not quite sure what to do with that. Normally if it equaled 0, then either costheta = 0 or 3costheta-10=0

Hmmm...

This is harder than the trig identity question I got in my C2 exam a few weeks ago. Maybe thats a good thing :P

Although, i've done the 3cos^2theta-13costheta=3sin^2theta-9

Using the identity sin^2theta + cos^2theta = 1

I've got as far as costheta(3costheta-10)=6

Not quite sure what to do with that. Normally if it equaled 0, then either costheta = 0 or 3costheta-10=0

Hmmm...

This is harder than the trig identity question I got in my C2 exam a few weeks ago. Maybe thats a good thing :P

*Last edited by OddOneOut at May 27, 2008,*

#27

mmmm trig identities, I don't mind them,

Although, i've done the 3cos^2theta-13costheta=3sin^2theta-9

Using the identity sin^2theta + cos^2theta = 1

I've got as far as costheta(3costheta-10)=6

Not quite sure what to do with that. Normally if it equaled 0, then either costheta = 0 or 3costheta-10=0

Hmmm...

This is harder than the trig identity question I got in my C2 exam a few weeks ago. Maybe thats a good thing :P

That was a good exam though, nice and easy, very forgiving questions.

And i got basically the same factorisation you got, but none of them would make possible roots

#28

Arrrg I hate trig.

</useless post>

</useless post>

#29

That was a good exam though, nice and easy, very forgiving questions.

And i got basically the same factorisation you got, but none of them would make possible roots

Yeah. What exam board? I did OCR MEI, I taught my mate how to convert from degrees to radians about 10 minutes before the exam, luckily the first question asked to convert something like 7pi/6 into degrees xD

Hmmm I think I've seen where I couldv'e gone wrong....am using small square note paper atm ><

actually no, I jsut came out with the same thing

edit: no, I've seen where I've gone wrong...

*Last edited by OddOneOut at May 27, 2008,*

#30

Yeah. What exam board? I did OCR MEI, I taught my mate how to convert from degrees to radians about 10 minutes before the exam, luckily the first question asked to convert something like 7pi/6 into degrees xD

Hmmm I think I've seen where I couldv'e gone wrong....am using small square note paper atm ><

Yep, OCR for me too.

Yeah I remember that question too, nice and easy. I remember the trapezium rule question (number 10 I think) threw me a bit and I think i lost all marks for it, but I know i've got 36 marks for section A, it really was a very nice exam paper.

Where might you have gone wrong?

#31

Yep, OCR for me too.

Yeah I remember that question too, nice and easy. I remember the trapezium rule question (number 10 I think) threw me a bit and I think i lost all marks for it, but I know i've got 36 marks for section A, it really was a very nice exam paper.

Where might you have gone wrong?

Ah, I don't mind the trapezium rule. I wasn't too sure about the equation of the straight line for the exponential curve question at the end. But it looked about right haha. There was some other things I weren't too sure about but I can't remember them now.

well,

Ok for ease of notation theta=x

3cos^2x-13cosx=-3cos^2x-6

I tried adding 3cos^2x with -13cosx haha.

But now I'm left with cosx(6cosx-13)=-6

:S

edit: ooooh. Maybe you're not supposed to factorise, it IS possible to work out 6cos^2x-13cosx=6=0 right?

Then again, that might not have any real roots...

*Last edited by OddOneOut at May 27, 2008,*

#32

Where did you get this line from?

3cos^2x-13cosx=-3cos^2x-6

Because if you have that part correct then the rest is easy, the 3cos^2x will cancel and x will be easy to find.

3cos^2x-13cosx=-3cos^2x-6

Because if you have that part correct then the rest is easy, the 3cos^2x will cancel and x will be easy to find.

#33

^ thats -3cos^2x

3sin^2x-9 = 3(1-cos^2x) - 9 = 3-3cos^2x-9 => -3cos^2x-6

I've got a quadratic now

6cos^2x-13cosx+6=0

It has real roots

cosx= 18/12 or 2/3

cannot equal 18/12

I think I could've just solved that

I might just go get my calculator and write it out properly on some paper

3sin^2x-9 = 3(1-cos^2x) - 9 = 3-3cos^2x-9 => -3cos^2x-6

I've got a quadratic now

6cos^2x-13cosx+6=0

It has real roots

cosx= 18/12 or 2/3

cannot equal 18/12

I think I could've just solved that

I might just go get my calculator and write it out properly on some paper

#34

SOLVED!!!

I wrote it all out, checked it, and I'm confident it's correct.

I'll take a picture of it (no scanner xD) and upload it if you want.

I wrote it all out, checked it, and I'm confident it's correct.

I'll take a picture of it (no scanner xD) and upload it if you want.

#35

SOLVED!!!

I wrote it all out, checked it, and I'm confident it's correct.

I'll take a picture of it (no scanner xD) and upload it if you want.

Cheers! Big help.

At least it wasn't just me who was struggling

#36

Cheers! Big help.

At least it wasn't just me who was struggling

Yeah, might have solved it sooner if I hadn't tried adding cos to cos^2

Totally forgot you could make it into a quadratic then solve it.

http://i11.photobucket.com/albums/a184/LilMissOddOneOut/IMG_3650.jpg

You'll have to read it sideways though :P

#37

It's "pi". Not "pie".

...

Just so you know.

...

Just so you know.

#38

It's "pi". Not "pie".

...

Just so you know.

Anything wrong with phonetics?

#39

Yeah, might have solved it sooner if I hadn't tried adding cos to cos^2

Totally forgot you could make it into a quadratic then solve it.

http://i11.photobucket.com/albums/a184/LilMissOddOneOut/IMG_3650.jpg

You'll have to read it sideways though :P

Very good, and you have good maths etiquette as well.

#40

Yeah, might have solved it sooner if I hadn't tried adding cos to cos^2

Totally forgot you could make it into a quadratic then solve it.

http://i11.photobucket.com/albums/a184/LilMissOddOneOut/IMG_3650.jpg

You'll have to read it sideways though :P

Cheers!