Post by: ahh.liz on February 18, 2012, 04:50:12 pm
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 :)
Post by: brightsky on February 18, 2012, 05:12:11 pm
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)
Post by: ahh.liz on February 19, 2012, 05:42:20 pm
State the (i) implied domain and (ii) the range of each of the following.
(a) y = arcos(x^2- 1)
Post by: b^3 on February 19, 2012, 05:49:48 pm
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.
Post by: ahh.liz on February 19, 2012, 06:01:40 pm
Post by: b^3 on February 19, 2012, 06:09:03 pm
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).
(https://s3.amazonaws.com/grapher/exports/gsjmdo2cyu.png)
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.
Post by: ahh.liz on February 19, 2012, 06:23:37 pm
Post by: b^3 on February 19, 2012, 06:25:15 pm
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 :)
Post by: ahh.liz on February 19, 2012, 07:45:54 pm
y = tan(2arcsin(x)) :-[
Post by: brightsky on February 19, 2012, 08:56:02 pm
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
Post by: ahh.liz on February 25, 2012, 07:56:58 pm
Errrm, how would I go about doing something like this?
Show that tan(pi/8) = sqrt(2) - 1.
Post by: b^3 on February 25, 2012, 08:01:22 pm
and
Now you can use
So
Then try solving the equation for
Post by: aznxD on February 25, 2012, 08:05:25 pm
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. :P
Post by: ahh.liz on March 04, 2012, 06:53:33 pm
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.
Post by: b^3 on March 04, 2012, 07:03:54 pm
now
so
Equate real and imaginary parts
Also
So to satisfy both
(as if you check
EDIT: Removed tildes on i's, they are not vectors (thanks Paul), take note of that (my mistake)
Post by: rife168 on March 04, 2012, 08:01:28 pm
Doing so, I found that
From this, you can see that a has to be
Post by: paulsterio on March 04, 2012, 08:11:36 pm
Post by: b^3 on March 04, 2012, 08:27:15 pm
what's with the tildes under the i, it's not a vector? :PForce of habbit (a couple of months after yr 12 and I'm already rusty :P), I'll fix it up now. Thanks Paul, also note guys doing that on the exam will lose you marks, so take note of that.
Post by: ahh.liz on March 08, 2012, 06:57:37 pm
Solve for x in sqrt(2)sin^2(x) = cos(x), x is an element of [0, 2pi]
Post by: nubs on March 08, 2012, 07:44:10 pm
sqrt(2) - sqrt(2)(cos^2(x)) - cos(x) = 0
Let cos(x) = a
-sqrt(2)a^2 - a + sqrt(2) = 0
go on from there :)
Post by: ahh.liz on March 08, 2012, 09:09:53 pm
Post by: ahh.liz on April 06, 2012, 12:10:00 pm
Hey, so we're been learning about vectors recently and we hit dot product, although I understand the mathematical arithmetic, I'm not sure what it is that I'm working out. All I see is a number and I'm not sure what it's trying to tell me :P. Haha. Would anyone be able to shed some light on my little dilemma? :)
Post by: rife168 on April 06, 2012, 04:30:45 pm
It may seem quite circular a definition, but when you think about projections of one vector on another, i.e.
edit: The wikipedia page has some helpful and interesting diagrams and whatnot on it.
Post by: Bhootnike on April 06, 2012, 10:31:55 pm
its called dot product cause we use a dot to show the multiplication btw!
In essence, I think what you work out, most of the time is the angle between two vectors.
This comes into play when considering if vectors are parallel or perpendicular to each other, and thus this can help in proving something like a quadrilateral, square, rhombus, trapezium...
also in a physical aspect, it can be used in cases such as:
REFER TO ATTACHMENT HERE
im gonna show some applied physics here so skip on if you wish!
the stickfigure is pulling the object with a constant force (vector quantity)
Consider the following:
work done = our dot product in this example
the work done in physics, is work done = force x displacement - 2 vector quantities
the component of the force being applied, as in the right diagram, in the direction of motion =
so the work done, through a distance of
this then gives us our:
in a physics sense, to use the least possible force, we would need to pull horizontally, so that we are pulling in the same direction as we want the object to move. Then we would have θ = 0 and cos θ= 1 so that the work done
So...! don't dwell too much cause i dont think we need to go that far, however keep in mind the usefulness of the dotproduct and the applications of it, since who knows.. a question , maybe similarly described above might happen to appear on the exam. AND I HOPE IT DOES SO I GET 10/10. :D haha nah. yer hope that helped...
Post by: ahh.liz on May 16, 2012, 04:20:49 pm
sec^2(x)cos^n(x) where n is an element of Z+
Any suggestions? :P
Post by: pi on May 16, 2012, 04:41:17 pm
Just a couple of questions:
1.
This is from MathsQuest (:o), p. 264
Evaluate:
I got upto, which is only one step from the question, and am now stuck :(
For question 1, its actually a typo. I spent half an hour trying to figure it out as well and realised it was a typo after reading the worked solutions. Cos is meant to be Tan. Then its simple enough to anti diff.
This is why I hate mathquest. Hope this helped.
:)
Post by: Mao on May 17, 2012, 11:11:14 pm
where
Post by: TrueTears on May 17, 2012, 11:22:40 pm
Post by: ahh.liz on May 21, 2012, 08:15:29 pm
Post by: Sinanator on May 24, 2012, 12:21:21 pm