ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: d0minicz on February 22, 2009, 12:51:13 pm
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Neeed help on a couple of questions
a) 
b) 
AND
If
,
, find:
a) 
b) 
c) 
thank you ;)
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=
=
=
=
=

=
= )
=
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[IMG]http://img201.imageshack.us/img201/7481/ccf2202200900001.th.jpg[/img]
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3a essential q6?
lol i am stuck on it right now too
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thanks alot ed_saifa =]
hahaaha yeah sucks =[
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How do you know how to draw the triangle
ie... which side has a length of 3
what if it was Sec x = 10
how would you draw the triangle to represent it ?
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How do you know how to draw the triangle
ie... which side has a length of 3
what if it was Sec x = 10
how would you draw the triangle to represent it ?
It says

Just draw a right angled triangle and put the angle x wherever. This then tells you with the the A or O side so you can label the length. If you are missing 1 side, use pythag
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= -0.7 , 0 < x < 180
find:

and
Simplify:
a) (cosec \thetha - sin \theta))
b)  (1 + {cot}^2 \theta))
c) 
need to learn the identities
thanks !
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(sinx)^2=1-(cosx)^2
=1-0.49
=0.51
sinx=sqrt(0.51) as sinx>0 for 0<x<180
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b.)(sinx)^2 (1+(cosx)^2/(sinx)^2)
=(sinx)^2+(cosx)^2
=1
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a.)(1/cosx - cosx)(1/sinx - sinx)
=((1-cos^2(x))/cosx)((1-sin^2(x))/sinx)
=(sin^2(x)/cosx)(cos^2(x)/sinx)
=sin^2(x)cos^2(x)/(cosxsinx)
=sinxcosx
=sin(2x)/2
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c.)
[(1/cos^2(x))-(1/sin^2(x)]/[((sin^2(x)/cos^2(x))-((cos^2(x))/((sin^2(x))]
multiply the numerator and denominator by sin^2(x)cos^2(x) to get:
[sin^2(x)-cos^2(x)]/[sin^4(x)-cos^4(x)]
factorise the denominator using difference of squares
[sin^2(x)-cos^2(x)]/[(sin^2(x)-cos^2(x))(sin^2(x)+cos^2(x))
=1
Because of cancelling the (sin^2(x)-cos^2(x)) found in the denominator and using the pythagoras theorem identity
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1. If
, prove that
and find a simple expression for 
2. If
then express
in terms of 
3. If
then express
in terms of 
4. Use the compound angle formulas and appropriate angles to find the exact value of each of the following:
a) 
b) 
c) 
d) 
need methods on how to do these
thanks
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ill have a go at Q 2 3 and 4, bear with me XD
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2.
(using the identity
)
 + cos^2x \times 2sinx)
 + (1-sin^2x) \times 2sinx)
further simplifying leads to:

3. Same principle as Q 2.
4. a)
(after subbing in the exact values and doing some arithmetic, i left that steps out)
the others are the same principle, if you are still stuck, let me know and i'll do them.
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If
, prove that
and find a simple expression for 

^2 + 1}{\sec \theta - \tan \theta}= 2 \sec \theta)



}{\sec \theta - \tan \theta}= 2 \sec \theta)

LHS=RHS QED
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thanks heaps but i dont get where the identity
comes from
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}{\sec \theta - \tan \theta})

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its one of the identities you should know
remember:
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You can also easily derive those by doing
 = cos(x+x) = cosxcosx - sinxsinx = cos^2x - sin^2x)
and for 
 = 2cos^2x - 1)
and for 
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thank you =]
do you have a list of identities that we need to know?
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Sure, i'll put them all up, i made a summary sheet, give me a sec and type em XD
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 = cosxcosy + sinxsiny)
 = cosxcosy - sinxsiny)
 = sinxcosy - sinycosx)
 = sinxcosy + sinycosx)
 = cos^2x - sin^2x = 1-2sin^2x = 2cos^2x-1)
 = 2sinxcosx)
 = \frac{tanx - tany}{1+tanxtany})
 = \frac{tanx+tany}{1-tanxtany})
 = \frac{2tanx}{1-tan^2x})

so basically those are your common sin and cos and tan ones
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thanks truetears!
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thanks =] !
any others we should familiarize ourselves with?
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Well from
you can derive some others...

and
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hey can you please show me how to find
using compound angle formulas
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sure,
so we know 
Now, you can sub those in the tan(x+y) formula to get
right? Then use the formula  = \frac{tanx+tany}{1-tanxtany})
I think the rest you can do :)
Remember when doing these questions always think about
, ie your exact values
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If
and 
find )
need working out
thnks :)
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k so here we just 'expand' tan(x+2y) right?
so  = \frac{tanx + tan2y}{1-tanxtan2y})
However we are not given the value of
, but we are given the value of
, so using that information we can use the tan double angle formula.
 = \frac{2tany}{1-tan^2y})
subbing in tany = 2.4 in yields  = -\frac{120}{119})
then subbing this value (-
) back into the original we get:
 = \frac{-0.75 + -\frac{120}{119}}{1--0.75 \times -\frac{120}{119}})
working that out equals = -
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solve each of the following for

a) 
b) 
c) 
thank you !!
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a) cosx(cosx - sinx) = 0
therefore cosx = 0 and cosx - sinx = 0
first one cosx = 0 you can solve
but the other one cosx - sinx = 0 needs further simplifying
times both sides by
yields 
so
(because
=
or
)
using the cos(x+y) compound angle formula yields  = 0)
which i think you can solve XD
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lol thanks truetears
and how do you know to times both sides?
i get the rest though =]
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sorry check that post again i accidently stuffed up my LaTeX EDITED now
doing the other 2 now
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b) sin2x = sinx
2sinxcosx = sinx
2sinxcosx - sinx = 0 (notice here we don't divide both sides by sinx to get rid of it because we will be losing possible solutions. Consider
, you wouldn't divide both sides by x here because it gets rid of possible solutions, you would do this though:
then factor and solve. Same principle applies here)
so factor out a sinx leaves
sinx(2cosx -1 ) = 0
now using the null factor law
sinx = 0, i think you can solve
and 2cosx - 1 = 0, rearranging leaves
which you can solve as well :)
c) This is the same principle to b) , do not divide both sides by cosx because it gets rid of possible solutions. Instead bring it over to the other side of the equation and factor it out.
basically these questions just always try to get it to =0 and then use the null factor law.
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lol thanks truetears
and how do you know to times both sides?
i get the rest though =]
Basically whenever you have a situation like
cosx + sinx = 1
or cosx - sinx = 1
or sinx - cosx = 1
People will often try to square both sides and then solve. Obviously you can, but remember squaring leaves us with REDUNDANT roots. (meaning answers which will not actually work for the equation if you sub it back in)
so in this case, always try to aim to get it into the compound angle formula form, we know sinx and cosx share a same value, that is
. Hence if we multiply both sides by
, we can change one of them into
and another into
, that way we can change the whole thing into a compound angle formula, which in this case is cos(x+y)
Hope that explains it, but if you don't understand, feel free to ask.
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i get it, thanks =]
ill let you know if i have any other troubles !
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Solve a)
for 
b)
, 
thanks =]
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sin8x = cos4x
 = 2sin4xcos4x)
so
(notice you don't divide both sides by cos4x as it gets rid of possible solutions)
now factor out a cos4x yields
cos4x(2sin4x -1 ) =0
so cos4x = 0 You can solve this
and
which you can solve too XD
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b) cos2x = cosx
so 
so 

let a = cosx

solve this quadratic yields a =
, 1
but a = cosx
cosx =
or cosx = 1 (which you can solve XD)
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Given than the domain of sinx and cosx are restricted to
and
defined the implied domain and range :
a) )
b) )
c) )
having trouble restricting etc..
thanks :)
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a) let
and  = g(x) )
now  = sin(-cos^{-1}(x)))
So for f o g to exist 

and 
Now, ran g is not a subset of dom f, so we must restrict ran g to AT LEAST 
and in order to do we must restrict the domain of g to [0,1], (another way of thinking about it is just solving
)
so we know dom f o g = dom g = [0,1]
EDIT: to find the range we know that the output of
will give the end points of the range for
.
so since
is restricted to  = \frac{-\pi}{2} and -cos^{-1}(1) = 0)
now sub
and 0 into
to find end points.
and  = 0)
therefore range of
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for this one:
\ cos^2 x - cosxsinx = 0, factor out cos(x) and u'll get cos(x)(cos(x)-sin(x))=0.
so cos(x)=0 and cos(x)-sin(x)=0. Now, for me, i prefer the brute force way, so instead of thinking about pi/4, i'd go:
cos(x)-sin(x)=0
cos(x)=sin(x)
therefore, tan(x)=1
just my 2 cents =]
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thanks guys
Truetears can you pls show me ur working out for b) and c)
thanks a bunch
and i need help with this question
sinx + cosx = 1
thanks
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thanks guys
Truetears can you pls show me ur working out for b) and c)
thanks a bunch
and i need help with this question
sinx + cosx = 1
thanks
for sinx + cosx = 1
look at the 2nd post on page 3 :) its a minus in the middle, but it's the same principle, in the meanwhile I shall answer b) and c) for you.
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I will skip a FEW steps here
b) Let
and  = g(x) )
))

But
is not a subset of 
therefore ran g is restricted to 
dom g becomes [0,1]
therefore 
c) let
and  = g(x) )
))


no restriction required as 
so 
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thanks alot
but for b) which steps did you skip?
did it involve the 2 in front of sin^-1 (x) ?
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nah the 2 is meant to be in front of the
I think i told you last time not to let
my apologies lol, i thought it was a question where you had to simplify -_-
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thanks man
can you pls show me the steps u take for solving the range
i tried subbing the domain into the equation but need some clarification through ur steps
thank you
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b) The value that
gives, then subbed back into cos(x) gives us the end points of the range.
and  = \pi)
subbing these in the cos(x) gives
and  = -1)
so the range is [-1,1]
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part c) uses the same principle
the value of cos(x) subbed in
gives the end points of the range
so
and  = -1)
and  = \frac{-\pi}{4})
so range is
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oh i get it, thanks
for cos(tan^-1 x) how do i find the range when Dom = [0,oo) or R+ U {0}
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tan^-1(x) for domain 0 to infinity gives a range of
. So the range of this function is the same as the range of cos(u) where u has domain
. THis gives a range of (0,1].
I see many people having problems with domain/range of composite functions, especially trig ones. Think of it as a computer/machine taking in numbers and spitting them out. So basically f(g(x)) takes some number x, then takes it through the machine g to produce a new number. Then takes that number and puts it through f. So in our case we're applying tan^-1 to numbers from 0 to infinity... producing numbers from 0 to pi/2 (but not including pi/2). Now all these numbers that belonged to that range are now inputed into the cosine function. And this gives numbers between 0 and 1. But not including 0 because cos(pi/2)=0 coz we didn't put pi/2 into our functions since it wasnt produced by tan^-1(x) earlier, just numbers very close to it.
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ahhhhhhhhhh crap i overlooked the normal range of tan^-1 x ....
thanks kamil
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The Questions 9, 10 and 11 in Exercise 3D essentials have been pissing me off for a long time lol
anywayz heres one
let
where
. Find in terms of
two values of x in the range
which satisfy each of the following equations.
a) tanx = -c
b) cotx = c
need alot of help understanding these questions :(
thanks alot !!!
anyone :(?
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http://vcenotes.com/forum/index.php/topic,11984.msg138611.html#msg138611
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Simplify the following completely:
a)  )
b) 
Prove the following :
a)  = cos^3x - 3cosxsin^2x )
b)  = 3cos^2xsinx - sin^3x )
thanks =]
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The Questions 9, 10 and 11 in Exercise 3D essentials have been pissing me off for a long time lol
anywayz heres one
let
where
. Find in terms of
two values of x in the range
which satisfy each of the following equations.
a) tanx = -c
b) cotx = c
need alot of help understanding these questions :(
thanks alot !!!
anyone :(?
Even better: http://vcenotes.com/forum/index.php/topic,9192.msg134547.html#msg134547
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Simplify the following completely:
a)  )
b) 
Prove the following :
a)  = cos^3x - 3cosxsin^2x )
b)  = 3cos^2xsinx - sin^3x )
thanks =]
a)  = cos^2x(1-sin^2x) + sin^2x(1-cos^2x) + 2sin^2xcos^2x = cos^2x-sin^2xcos^2x + sin^2x - sin^2xcos^2x + 2sin^2xcos^2x = 1 - 2sin^2xcos^2x + 2sin^2xcos^2x = 1)
b)  - sin^2xcos^2x = sin^2xcos^2x - sin^2xcos^2x = 0)
c)  = cosxcos(2x)-sinxsin(2x) = cosx(cos^2x-sin^2x)-sinx(2sinxcosx) = cos^3x-cosxsin^2x-2sin^xcosx = cos^3x-3cosxsin^2x)
b) Too late I cbf typing this up, but same principle you see?
gogo!
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Couple of Q's guys...
1.
equals:
A.  )
B.  +1 )
C.  + 1 )
D.  )
E.  )
2. For
and
with
and
,
equals to:
A. 0
B. 
C. 
D. 
E. 
Need to see workings for em
thanks !
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2.)
=cosBcosA-sinBsinA)
(1)
Because in second quadrant
(2)
)
(3)
just sub (2) and (3) back into (1)
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Couple of Q's guys...
1.
equals:
A.  )
B.  +1 )
C.  + 1 )
D.  )
E.  )
Need to see workings for em
thanks !
1.)
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If that was a multiple choice question on a VCE MM exam (non-CAS), what I would use to do is input each option into the Y menu, and then graph the functions. If the two functions overlapped everywhere (except perhaps some naughty discontinuities), then they were obviously equal.
Only do this if you are quick at using your calculator though. I rationalised that it was quicker for me to type these in then do the algebra, but you might not be the same. Also don't check 'silly' answers which are obviously not correct, as that is a waste of time.
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If that was a multiple choice question on a VCE MM exam (non-CAS), what I would use to do is input each option into the Y menu, and then graph the functions. If the two functions overlapped everywhere (except perhaps some naughty discontinuities), then they were obviously equal.
Only do this if you are quick at using your calculator though. I rationalised that it was quicker for me to type these in then do the algebra, but you might not be the same. Also don't check 'silly' answers which are obviously not correct, as that is a waste of time.
Yeah true, I do another thing, if you can not figure out the answer just type in a random value for x and test each x to see if you get the same answer. [if non calc just sub in an exact value]
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I wouldn't use any method similar to this without a calculator. Exact values will (with an infinitely larger probability) overlap more often (i.e. cos/sin at pi/4).
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Couple of Q's guys...
1.
equals:
A.  )
B.  +1 )
C.  + 1 )
D.  )
E.  )
Need to see workings for em
thanks !
1.) }{1+sec(2x)} = \frac{ \frac{sin(2x)}{cos(2x)}}{1+ \frac{1}{cos(2x)}} = \frac{ \frac{sin(2x)}{cos(2x)}}{\frac{cos(2x)+1}{cos(2x)}} = \frac{sin(2x)}{cos(2x)+1} = \frac{2sinxcosx}{(2cos^2x -1)+1} = tanx)
That question was on my spesh test once. Quite fun.
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ahhhhh cheers guys
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Neeed help on a couple of questions
a) 
a) [tan^2(x) + 1]/tan^2(x) = 1 + cot^2(x) = cosec^2(x)
b) 
(sin^2(x) + cos^2(x))(sin^2(x) - cos^2(x)) = (sin^2(x) - cos^2(x)) = (1 - cos^2(x) - cos^2(x)) = 1 - 2cos^2(x) = -cos(2x)
AND
If
,
, find:
a) 
b) 
c) 
thank you ;)
x = arccot(3)
a) cosecx = cosec(arccot(3)), where, cosec(x)<0
cosec(arccot(3)) = -sqrt(1 + cot^2(arccot(3)) = -sqrt(10)
b) sin(x) = 1/cosec(x) = -1/sqrt(10) = sqrt(10)/10
c) sec(x) = -sqrt(1 + tan^2(arccot3)), since, sec(x)<0
= -sqrt(1 + [1/(cot^2(arccot3))])
= -sqrt(10/9)
= -sqrt(10)/3
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Eh won't those answered ages ago?
EDIT* There is never a good reason to quote the preceding post in its entirety. Especially if its large!