ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: jasopan on April 17, 2010, 05:56:26 pm
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Evalaluate cos(pi/8) using half angle Formulae leaving answer in exact values
Thanks!
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 = \cos\left(\frac{\frac{\pi}{4}}{2}\right) )
Use half-angle formula:  = \frac{1}{2} \left(1 + \cos(x)\right) )
In this case: 
So  = \frac{1}{2} \left(1 + \cos(\frac{\pi}{4}\right) )
 = \sqrt{\frac{1}{2} \left(1 + \cos(\frac{\pi}{4}\right)} )
So:
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^^ Yup, got the sin cos ones, but now I'm stuck on the tan
I can't seem to use the formula, or should i do
tanx=sinx/cosx and then simplify?
And also is there a way to solve the sin(pi/8) using sinX=2sin(x/2)cos(X/2)
Thanks again
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there is another way to evaluate sin(pi/8) and cos(pi/8) using complex numbers if anyone is interested. In fact, it make a a rather good question. It's in the essentials book, pg 168, Q8
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 = \cos\left(\frac{\frac{\pi}{4}}{2}\right) )
Use half-angle formula:  = \frac{1}{2} \left(1 + \cos(x)\right) )
In this case: 
So  = \frac{1}{2} \left(1 + \cos(\frac{\pi}{4}\right) )
 = \sqrt{\frac{1}{2} \left(1 + \cos(\frac{\pi}{4}\right)} )
So:

sucky uni maths stuff :( its so hard D:
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lol, uni maths stull is AWESOME, and that not uni maths, it's spech
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lol, uni maths stull is AWESOME, and that not uni maths, it's spech
we're doing this at uni maths D: i dont do spesh so i not know haha
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wtf, i just realised that. How are you doing uni maths without spech? not only how but why???
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wtf, i just realised that. How are you doing uni maths without spech? not only how but why???
how because they let me..i dont know y every1 says u have to do spesh b4 uni maths :S but they let me
WHY? so i can drop a subject and profit in free hours haha
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there is another way to evaluate sin(pi/8) and cos(pi/8) using complex numbers if anyone is interested. In fact, it make a a rather good question. It's in the essentials book, pg 168, Q8
Can you post up the proof? :)
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reli? dammit i hate latex have never used it cos i ceebes
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Quick outline? Just to quench my curiosity? :P
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latex is the bomb, once u get into uni latex is ur best friend for maths xD might as well learn the skill now
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i'll take you thu the logic you'll learn more by doing the numbers youself(and it'll save me time lol)
ok
1)It's the same logic as questions 3 and 4. Let z^2 be 1+i and let z be a+bi
2) square (a+bi) and let it equal 1+i, solve for a and b
part ii) you should be able to do, it's pretty standard, just basic using de moirves
part b)
1) expand the and from part aii into carterian form, not symplyfing cos(pi/8) and sin(pi/8)
2) equate coefficents
Hope that helps:) Ask if you need more help
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Lol I don't have the book. :(
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dammit, TT you have e-books of everything, do you have essentials???
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Hey can someone re-explain the
)
And one more;
Show that: 
tA :)
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And one more;
Show that: 
tA :)
 = sin(2x+x))
 = sin(2x)cos(x) + cos(2x)sin(x))
cos(x))(cos(x)) + (1 - 2sin^2(x))(sin(x)) )
cos^2(x) + sin(x) - 2sin^2(x)sin(x))
)(1-sin^2(x)) + sin(x) - 2sin^2(x)sin(x))
 - 2sin^3(x) + sin(x) - 2sin^3(x))
as required.
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 = \tan \left(\frac{\frac{\pi}{4}}{2}\right) )
Use the identity
.
The rest is trivial.
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^ Whoa, my book doesn't even have that identity ... >:(
Gotta google some now :-\
@Yitzi_K yes! got it! thanks alotalotalotalot
Another one since everyone's so helpful :D
Simplify
divide 
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You can derive the formula from this image:
http://upload.wikimedia.org/wikipedia/commons/2/21/Weierstrass_substitution.png
:D
Look at
, then use
as normal.
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dammit, TT you have e-books of everything, do you have essentials???
yeah i do...
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^2 - \sqrt{2}}{(\sqrt{2})^2 + \sqrt{2}}} )
Let
.

}{x(x + 1)} )

So:


^2}{(\sqrt{2})^2 - 1^2}} )
^2}{2 - 1}} )
^2} )

:)
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^ Holy crap. Is it that hard?
I tried to use tanx=sinx/cosx to get tanx as the book didn't have the right formula and thats what i got stuck on :S
Thanks again brightsky, this is gold
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or u could use the double angle formula
=\frac{2\tan \frac{\pi}{8}}{1-\tan ^{2}(\frac{\pi}{8})})
let )
and note )
rearrange to get a quadratic
solve and u get 
as > 0)
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What does it mean if they go 'if x is an acute angle' ? does that mean its only in the first quadrant ? so all sin cos tan are positive?
And just to confirm if domain is pi > x > pi/2, then domain of x/2 is (pi/4, pi/2)
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Yep. Acute means
for angle
, hence in the first quadrant.
Yes for your second question as well (I'm pretty sure).
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for 
Is the domain written;
or
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The second one.
is two statements combined.
and 
So
and
.
Or
.
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Yup, all clear now, thanks for clearing that up
Edit sorry another one, if x is an acute angle, and then we are asked to find sin(x/2) are all tanx/2 sinx/2 cosx/2 still positive? do we have to state the new domain is pi/4>x>0?
EDIT2 nvm, i understand it now ;D