Ahh.liz's Specialist Mathematics Queries

[ahh.liz]

It seems as if I'm going to be dropping in quite a lot so I decided upon saving some room and making my own little page. So let us begin 

1. Using the double angle formulas, find the exact value of:
(a) sin(pi/8)

I do have another question but I can't express it properly without making it look like a bunch of gobbledygook. Would anyone be able to also tell me how to make my mathematical expression look more mathematical?

Thank you very muchly

[brightsky]

1. a)
cos(pi/4) = 1- 2sin^2(pi/8)
sqrt(2)/2 = 1 - 2sin^2(pi/8)
(2-sqrt(2))/2 = 2sin^2(pi/8)
sin^2(pi/8) = (2-sqrt(2))/4
sin(pi/8) = sqrt(2-sqrt(2))/2 (reject negative solution since pi/8 is in the first quadrant)

[ahh.liz]

Ooh thank you so much. How about this one?

State the (i) implied domain and (ii) the range of each of the following.

(a) y = arcos(x^2- 1)

[b^3]

ai) For arcos(x), the domain is [-1,1] and range is [0,pi]
Now for arcos(x²-1) that means that x₂-1 has to give the same [-1,1]
x₂-1=-1
x²=0
x=0

and
x²-1=1
x²=2
x=+-root(2)
So the domain is [-root(2), root(2)]

Now the domain will be the same, [0,pi]

basically just remember that what ever is going into the arcos() has to be -1=<x=<1 so that the arcos() will not come out undefined.

[ahh.liz]

So what exactly is the significance of the x=0?

[b^3]

I was just testing the endpoints.

Really you can approach this/look at it from a fog/gof approach, range of x^2-1 must fit into the domain of arcos(x).

Red and green lines is the domain of arcos(x) ([-1,1]) and the range of x^2-1 must fit into those, so you find when they get cut off by solving the equation x^2-1=1. Sor eally the x=0 wasnt needed, as I didn't quite visualise it properly.

[ahh.liz]

That makes a lot more sense now! Thank you very much!

[b^3]

That makes a lot more sense now! Thank you very much!

Yeh, sorry, I should have just explained it that way in the first place

[ahh.liz]

That's perfectly alright I'm still trying to get around this inverse function thing How would you work out the domain and range of this function?

y = tan(2arcsin(x))

[brightsky]

work it out as per any composite function
in this case, we have f(g(x)) where f(x) = tan(x) and g(x) = 2arcsin(x)

or we can simply do:
x E [-1,1]
2arcsin(x) E [-pi, -pi/2) U (-pi/2,pi/2) U (pi/2, pi]
arcsin(x) E [-pi/2, -pi/4) U (-pi/4, pi/4) U (pi/4, pi/2]
x E [-1, -sqrt(2)/2)) U (-sqrt(2)/2, sqrt(2)/2) U (sqrt(2)/2, 1]
hence that is your domain
to find the range, just sub this back into your equation

[ahh.liz]

Thanks

Errrm, how would I go about doing something like this?

Show that tan(pi/8) = sqrt(2) - 1.

[b^3]

You know that
and (double angle formula)
Now you can use , so .
So
Then try solving the equation for and keep in mind that wheny ou use the quadratic formula to take the postivie solution as will be in the first quadrant, i.e. positive.

[aznxD]

Using the double angle formulas:
tan(pi/4) = (2tan(pi/8))/1-tan^2(pi/8)
let t = tan(pi/8)
1 = (2t)/1-t^2
1 - t^2 = 2t
t^2 + 2t - 1 = 0
t = -1 +- sqrt(2)
tan(pi/8) = -1 +- sqrt(2)
Since pi/8 is in the first quadrant reject the negative solution
Hence tan(pi/8) = -1 + sqrt(2)

Edit: Beaten.

[ahh.liz]

Thanks again guys!

Find values for a (a is an element of R) for which ai is a solution to:
(a) P(z) = z^3 + 3z^2 + 36z + 108

I tried doing P(ai) = 0 and got a little confused with my answer. So I shall present this to the genius that is the atarnotes community. Any ideas?

The solutions say that the answer's actually +/- 6.

[b^3]

now
so (where a few people get stuck)


Equate real and imaginary parts


or
Also


So to satisfy both
(as if you check , then you get which does not equal )

EDIT: Removed tildes on i's, they are not vectors (thanks Paul), take note of that (my mistake)