TrueTears question thread

[TrueTears]

ahh thanks again Damo17.

[TrueTears]

ah i see, so how do i know when to leave the answers as they are or to change them to be within the interval of

eg, was simplified to

however, was left as (according to book's answers)

so, how do i know when to keep simplifying or stop?

yeah so is there a generalization which i can follow?

[Damo17]

ah i see, so how do i know when to leave the answers as they are or to change them to be within the interval of

eg, was simplified to

however, was left as (according to book's answers)

so, how do i know when to keep simplifying or stop?

yeah so is there a generalization which i can follow?

The does not change the point of . Adding or subtracting from any angle will yield the same base angle/ same point on the graph. That is why that have taken it out.

As to the , that does change the point on the graph so they have kept it in.

Unfortunately I am limited in my knowledge of this so if you still don't understand i'm sure someone else could explain it much better.

[TrueTears]

ah i seem to understand now, but isn't the same as ? so why can't be simplified to ?

[Damo17]

ah i seem to understand now, but isn't the same as ? so why can't be simplified to ?

I thought that myself and to be honest I have no idea why.

[TrueTears]

ah i seem to understand now, but isn't the same as ? so why can't be simplified to ?

hmmm, could anyone else see if they can find a reason why  ?

in the meanwhile....

Factorise the first expression into linear factors over C, given that the second expression is one of the linear factors.

,

[Matt The Rat]

The expressions are equal.

For the factorisation,



is the way I'd approach it. Then just equate co-efficients and then factor the quadratic bracket again to have it as its linear factors.

[TrueTears]

ahhh, thanks very much matt the rat

[Damo17]

The expressions are equal.

I did know that, but I am not sure why they would leave the there instead of putting it to just .

[shinny]

Was the value or a possible range of alpha given? The value inside the cis should be within the principal domain, so that may cause it to be left as -3pi. If it isn't, then without having an idea of what alpha is, there is no way to determine what value should be used to put it within the principal domain and anything is acceptable really.

[TrueTears]

no range was given it was just like that, so im guessing both answers are correct.

Also, this question:

If is a zero of the polynomial , find the values of and given that they are real.

Well this is a fairly easy question, since the coefficients are real, so 2 of the roots would be and . (conjugate pairs.)

So u sub both those in the former equation and solve the 2 simultaneous equations, however i keep getting and , even after solving on calc. But, the question says they are real numbers. Can anyone see what's wrong?

[shinny]

The conjugate root theorem ONLY applies if all coefficients are real, which isn't the case here. I might have a go at it later in latex if I get the time.

[TrueTears]

but isn't all the coefficients of all real?

[ell]

Imagine it like this:

[TrueTears]

Imagine it like this:

ahhhh i see, now i get it thanks ell and shinny