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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

Title: Circular functions
Post by: d0minicz on February 22, 2009, 12:51:13 pm: Neeed help on a couple of questions

a) $\frac {{tan}^2 {x} + 1} {{tan}^2 {x}}$

b) $\ {sin}^4 {x} - {cos}^4 {x}$

AND

If $\ {cot} {x} = 3$ , $\ {x} \in \left[ {\pi} , \frac{3\pi}{2} \right ]$ , find:

a) $\ cosec {x}$

b) $\ sin {x}$

c) $\ sec {x}$

thank you ;)
Title: Re: Circular functions
Post by: ed_saifa on February 22, 2009, 01:30:54 pm: $\frac{tan^2x +1}{tan^2x}$

= $\frac{sec^2x}{tan^2x}$

= $sec^2xcot^2x$

= $sec^2x . \frac{cos^2x}{sin^2x}$

= $sec^2x . \frac{cosec^2x}{sec^2x}$

= $cosec^2x$

$sin^4x -cos^4x$

= $(sin^2x+cos^2x)(sin^2x-cos^2x)$

= $-(cos^2x -sin^2x)$

= $-cos2x$
Title: Re: Circular functions
Post by: ed_saifa on February 22, 2009, 01:40:56 pm: [IMG]http://img201.imageshack.us/img201/7481/ccf2202200900001.th.jpg[/img]
Title: Re: Circular functions
Post by: BiG DaN on February 22, 2009, 01:54:27 pm: 3a essential q6?
lol i am stuck on it right now too
Title: Re: Circular functions
Post by: d0minicz on February 22, 2009, 01:58:26 pm: thanks alot ed_saifa =]

hahaaha yeah sucks =[
Title: Re: Circular functions
Post by: d0minicz on February 22, 2009, 02:09:20 pm: 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 ?
Title: Re: Circular functions
Post by: ed_saifa on February 22, 2009, 02:15:55 pm: Quote from: d0minicz on February 22, 2009, 02:09:20 pm
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 $cotx=3=\frac{A}{O}$

$Secx=10=\frac{H}{A}$

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
Title: Re: Circular functions
Post by: d0minicz on February 26, 2009, 05:15:10 pm: $\cos {x}^\circ$ = -0.7 , 0 < x < 180

find:

$\sin {x}^\circ$

and

Simplify:

a) $\ (sec \theta - cos \theta)(cosec \thetha - sin \theta)$

b) $\ (1-{cos}^{2} \theta) (1 + {cot}^2 \theta)$

c) $\frac {sec^2 \theta - cosec^2 \theta}{tan^2 \theta - cot^2 \theta}$

need to learn the identities
thanks !
Title: Re: Circular functions
Post by: kamil9876 on February 26, 2009, 07:09:10 pm: (sinx)^2=1-(cosx)^2
=1-0.49
=0.51
sinx=sqrt(0.51) as sinx>0 for 0<x<180
Title: Re: Circular functions
Post by: kamil9876 on February 26, 2009, 07:35:56 pm: b.)(sinx)^2 (1+(cosx)^2/(sinx)^2)
=(sinx)^2+(cosx)^2
=1
Title: Re: Circular functions
Post by: kamil9876 on February 26, 2009, 07:50:55 pm: 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
Title: Re: Circular functions
Post by: kamil9876 on February 26, 2009, 07:57:48 pm: 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
Title: Re: Circular functions
Post by: d0minicz on March 02, 2009, 06:19:58 pm: 1. If $\ x= sec \theta - tan \theta$ , prove that $\ {x} + \frac {1}{x} = 2 sec \theta$ and find a simple expression for $\ {x} - \frac {1}{x}$

2. If $\ sin (x+2x) = sinxcos2x + cosxsin2x$ then express $\sin 3x$ in terms of $\theta$

3. If $\ cos (x+2x) = cosxcos2x - sinxsin2x$ then express $\cos 3x$ in terms of $\theta$

4. Use the compound angle formulas and appropriate angles to find the exact value of each of the following:

a) $\sin \frac {\pi}{12}$

b) $\tan \frac {5\pi}{12}$

c) $\cos \frac {7\pi}{12}$

d) $\tan \frac {\pi}{12}$

need methods on how to do these

thanks
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 06:27:54 pm: ill have a go at Q 2 3 and 4, bear with me XD
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 06:30:57 pm: 2. $sinxcos2x + cosxsin2x = sinx(1-2sin^2x) + cosx \times 2sinxcosx$ (using the identity $cos2x = 1-2sin^2x$ )
$= sinx(1-2sin^2x) + cos^2x \times 2sinx$
$=sinx(1-2sin^2x) + (1-sin^2x) \times 2sinx$

further simplifying leads to:

$sinx - 2sin^3x +2sinx - 2sin^3x = 3sinx - 4sin^3x$

3. Same principle as Q 2.

4. a) $sin(\frac{\pi}{12}) = sin(\frac{\pi}{3} - \frac{\pi}{4}) = sin(\frac{\pi}{3})cos(\frac{\pi}{4}) - sin(\frac{\pi}{4})cos(\frac{\pi}{3}) = \frac{\sqrt{6}-\sqrt{2}}{4}$ (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.
Title: Re: Circular functions
Post by: Flaming_Arrow on March 02, 2009, 06:37:01 pm: If $\ x= \sec \theta - \tan \theta$ , prove that $\ {x} + \frac {1}{x} = 2 sec \theta$ and find a simple expression for $\ {x} - \frac {1}{x}$

$\sec \theta - \tan \theta + \frac{1}{\sec \theta - \tan \theta}= 2 \sec \theta$

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

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

$\frac{\sec^2 \theta - 2\tan\sec\theta + sec^2 \theta}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{2\sec^2 \theta - 2\tan\sec\theta}{\sec \theta - \tan \theta}= 2 \sec \theta$

$\frac{2\sec \theta (\sec- \tan\theta)}{\sec \theta - \tan \theta}= 2 \sec \theta$

$2 \sec \theta= 2 \sec \theta$

LHS=RHS QED
Title: Re: Circular functions
Post by: d0minicz on March 02, 2009, 06:43:50 pm: thanks heaps but i dont get where the identity $\cos2x = 1- 2sin^2x$ comes from
Title: Re: Circular functions
Post by: Flaming_Arrow on March 02, 2009, 06:44:09 pm: $\sec \theta - \tan \theta - \frac{1}{\sec \theta - \tan \theta}$

$\frac{\sec^2 \theta - 2\tan\sec\theta + \tan^2\theta - 1}{\sec \theta - \tan \theta}$

$\frac{\tan^2 \theta + 1 - 2\tan\sec\theta + \tan^2\theta - 1}{\sec \theta - \tan \theta}$

$\frac{2\tan^2 \theta - 2\tan\sec\theta }{\sec \theta - \tan \theta}$

$\frac{-2\tan \theta(\sec \theta - tan\theta)}{\sec \theta - \tan \theta}$

$\implies -2 \tan \theta$
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 06:46:14 pm: its one of the identities you should know

remember: $cos2x = cos^2x - sin^2x = 1-2sin^2x = 2cos^2x-1$
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 06:49:35 pm: You can also easily derive those by doing

$cos(2x) = cos(x+x) = cosxcosx - sinxsinx = cos^2x - sin^2x$

and for $2cos^2x-1$

$cos^2x - sin^2x = cos^2x - (1 - cos^2x) = 2cos^2x - 1$

and for $1-2sin^2x$

$cos^2x - sin^2x = (1 - sin^2x) - sin^2x = 1 - 2sin^2x$
Title: Re: Circular functions
Post by: d0minicz on March 02, 2009, 06:52:05 pm: thank you =]

do you have a list of identities that we need to know?
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 06:53:22 pm: Sure, i'll put them all up, i made a summary sheet, give me a sec and type em XD
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 06:57:47 pm: $cos(x-y) = cosxcosy + sinxsiny$

$cos(x+y) = cosxcosy - sinxsiny$

$sin(x-y) = sinxcosy - sinycosx$

$sin(x+y) = sinxcosy + sinycosx$

$cos(2x) = cos^2x - sin^2x = 1-2sin^2x = 2cos^2x-1$

$sin(2x) = 2sinxcosx$

$tan(x-y) = \frac{tanx - tany}{1+tanxtany}$

$tan(x+y) = \frac{tanx+tany}{1-tanxtany}$

$tan(2x) = \frac{2tanx}{1-tan^2x}$

$sin^2x + cos^2x = 1$

so basically those are your common sin and cos and tan ones
Title: Re: Circular functions
Post by: Flaming_Arrow on March 02, 2009, 06:58:58 pm: thanks truetears!
Title: Re: Circular functions
Post by: d0minicz on March 02, 2009, 07:33:04 pm: thanks =] !

any others we should familiarize ourselves with?
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 07:52:16 pm: Well from $sin^2x + cos^2x = 1$ you can derive some others...

$cosec^2x = 1+ cot^2x$

and

$sec^2x = 1+tan^2x$
Title: Re: Circular functions
Post by: d0minicz on March 02, 2009, 08:01:12 pm: hey can you please show me how to find

$\ tan \frac {5\pi}{12}$ using compound angle formulas
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 08:04:41 pm: sure,

so we know $\frac{5\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4}$

Now, you can sub those in the tan(x+y) formula to get $tan(\frac{\pi}{6} + \frac{\pi}{4})$ right? Then use the formula $tan(x+y) = \frac{tanx+tany}{1-tanxtany}$
I think the rest you can do :)

Remember when doing these questions always think about $\frac{\pi}{4}, \frac{\pi}{6}, \frac{\pi}{3} and \frac{\pi}{2}$ , ie your exact values
Title: Re: Circular functions
Post by: d0minicz on March 02, 2009, 08:58:28 pm: If $\tan y = 2.4$ and $\ tan x = -0.75$

find $\tan (x+2y)$
need working out
thnks :)
Title: Re: Circular functions
Post by: TrueTears on March 02, 2009, 09:19:13 pm: k so here we just 'expand' tan(x+2y) right?

so $tan(x+2y) = \frac{tanx + tan2y}{1-tanxtan2y}$

However we are not given the value of $tan(2y)$ , but we are given the value of $tany$ , so using that information we can use the tan double angle formula.

$tan(2y) = \frac{2tany}{1-tan^2y}$

subbing in tany = 2.4 in yields $tan(2y) = -\frac{120}{119}$

then subbing this value (- $\frac{120}{119}$ ) back into the original we get:

$tan(x+2y) = \frac{-0.75 + -\frac{120}{119}}{1--0.75 \times -\frac{120}{119}}$

working that out equals = - $\frac{837}{116}$
Title: Re: Circular functions
Post by: d0minicz on March 06, 2009, 09:43:16 pm: solve each of the following for $\ x \in [0, 2\pi]$

a) $\ cos^2 x - cosxsinx = 0$

b) $\ sin 2x = sinx$

c) $\sin 2x = cosx$

thank you !!
Title: Re: Circular functions
Post by: TrueTears on March 06, 2009, 09:46:29 pm: 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 $\frac{\sqrt{2}}{2}$ yields $\frac{\sqrt{2}}{2}cosx - \frac{\sqrt{2}}{2}sinx = 0$

so $cos\frac{\pi}{4} cos x - sin\frac{\pi}{4}sinx = 0$ (because $\frac{\sqrt{2}}{2}$ = $cos\frac{\pi}{4}$ or $sin\frac{\pi}{4}$ )

using the cos(x+y) compound angle formula yields $cos(\frac{\pi}{4} + x ) = 0$

which i think you can solve XD
Title: Re: Circular functions
Post by: d0minicz on March 06, 2009, 09:48:35 pm: lol thanks truetears
and how do you know to times both sides?
i get the rest though =]
Title: Re: Circular functions
Post by: TrueTears on March 06, 2009, 09:49:21 pm: sorry check that post again i accidently stuffed up my LaTeX EDITED now

doing the other 2 now
Title: Re: Circular functions
Post by: TrueTears on March 06, 2009, 09:52:37 pm: 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 $x^2 = x$ , you wouldn't divide both sides by x here because it gets rid of possible solutions, you would do this though: $x^2 - x = 0$ 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 $cosx = \frac{1}{2}$ 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.
Title: Re: Circular functions
Post by: TrueTears on March 06, 2009, 09:57:22 pm: Quote from: d0minicz on March 06, 2009, 09:48:35 pm
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 $\frac{\sqrt{2}}{2}$ . Hence if we multiply both sides by $\frac{\sqrt{2}}{2}$ , we can change one of them into $cos\frac{\pi}{4}$ and another into $sin\frac{\pi}{4}$ , 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.
Title: Re: Circular functions
Post by: d0minicz on March 06, 2009, 10:02:27 pm: i get it, thanks =]
ill let you know if i have any other troubles !
Title: Re: Circular functions
Post by: d0minicz on March 07, 2009, 03:00:33 pm: Solve a) $\ sin8x = cos4x$ for $x \in [0,2\pi]$

b) $\ cos2x = cosx$ , $x \in [0,2\pi]$
thanks =]
Title: Re: Circular functions
Post by: TrueTears on March 07, 2009, 03:26:41 pm: sin8x = cos4x

$sin8x = sin(2 \times 4x) = 2sin4xcos4x$

so $2sin4xcos4x - cos4x = 0$ (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 $sin4x = \frac{1}{2}$ which you can solve too XD
Title: Re: Circular functions
Post by: TrueTears on March 07, 2009, 03:29:05 pm: b) cos2x = cosx

so $cos2x = 2cos^2x -1$

so $2cos^2x -1 = cos x$

$2cos^2x - cosx -1 = 0$

let a = cosx

$2a^2 - a - 1 = 0$

solve this quadratic yields a = $\frac{-1}{2}$ , 1

but a = cosx

cosx = $\frac{-1}{2}$ or cosx = 1 (which you can solve XD)
Title: Re: Circular functions
Post by: d0minicz on March 21, 2009, 05:31:04 pm: Given than the domain of sinx and cosx are restricted to $\ [\frac {-\pi}{2} , \frac {\pi}{2}]$ and $\ [0, \pi]$ defined the implied domain and range :

a) $\ sin(-cos^{-1} x)$

b) $\ cos(2sin^{-1} x)$

c) $\ tan^{-1} (cosx)$

having trouble restricting etc..
thanks :)
Title: Re: Circular functions
Post by: TrueTears on March 21, 2009, 05:59:11 pm: a) let $sin(x) = f(x)$ and $-cos^{-1}(x) = g(x)$

now $f o g(x) = sin(-cos^{-1}(x))$

So for f o g to exist $ran g \subset dom f$

$ran g = [-\pi,0]$

and $dom f = [\frac{-\pi}{2},\frac{\pi}{2}]$

Now, ran g is not a subset of dom f, so we must restrict ran g to AT LEAST $[\frac{-\pi}{2},0]$

and in order to do we must restrict the domain of g to [0,1], (another way of thinking about it is just solving $\frac{-\pi}{2} \le -cos^{-1}(x) \le 0$ )

so we know dom f o g = dom g = [0,1]

EDIT: to find the range we know that the output of $-cos^{-1}(x)$ will give the end points of the range for $sin(x)$ .

so since $-cos^{-1}(x)$ is restricted to $[0,1] \implies -cos^{-1}(0) = \frac{-\pi}{2} and -cos^{-1}(1) = 0$

now sub $\frac{-\pi}{2}$ and 0 into $sin(x)$ to find end points.

$sin(\frac{-\pi}{2}) = -1$ and $sin(0) = 0$

therefore range of $f o g = [-1,0]$
Title: Re: Circular functions
Post by: pHysiX on March 22, 2009, 11:15:35 am: 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 =]
Title: Re: Circular functions
Post by: d0minicz on March 23, 2009, 05:47:16 pm: 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
Title: Re: Circular functions
Post by: TrueTears on March 23, 2009, 06:15:11 pm: Quote from: d0minicz on March 23, 2009, 05:47:16 pm
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.
Title: Re: Circular functions
Post by: TrueTears on March 23, 2009, 06:38:53 pm: I will skip a FEW steps here

b) Let $cos(x) = f(x)$ and $2sin^{-1}(x) = g(x)$

$f o g = cos(2sin^{-1}(x))$

$ran g \subset dom f$

But $[-\pi,\pi]$ is not a subset of $[0,\pi]$

therefore ran g is restricted to $[0,\pi]$

dom g becomes [0,1]

therefore $dom f o g = [0,1]$

c) let $tan^{-1}(x) = f(x)$ and $cos(x) = g(x)$

$f o g = tan^{-1}(cos(x))$

$ran g \subset dom f$

$[-1,1] \subset R$

no restriction required as $ran g \subset dom f$

so $dom f o g = [0,\pi]$
Title: Re: Circular functions
Post by: d0minicz on March 23, 2009, 06:45:35 pm: thanks alot

but for b) which steps did you skip?
did it involve the 2 in front of sin^-1 (x) ?
Title: Re: Circular functions
Post by: TrueTears on March 23, 2009, 06:47:21 pm: nah the 2 is meant to be in front of the $sin^{-1}(x)$ I think i told you last time not to let $g(x) = 2sin^{-1}(x)$ my apologies lol, i thought it was a question where you had to simplify -_-
Title: Re: Circular functions
Post by: d0minicz on March 23, 2009, 06:56:32 pm: 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
Title: Re: Circular functions
Post by: TrueTears on March 23, 2009, 07:05:07 pm: b) The value that $2sin^{-1}(x)$ gives, then subbed back into cos(x) gives us the end points of the range.

$2sin^{-1}(0) = 0$ and $2sin^{-1}(1) = \pi$

subbing these in the cos(x) gives $cos(0) = 1$ and $cos(\pi) = -1$

so the range is [-1,1]
Title: Re: Circular functions
Post by: TrueTears on March 23, 2009, 07:09:58 pm: part c) uses the same principle

the value of cos(x) subbed in $tan^{-1}(x)$ gives the end points of the range

so $cos(0) = 1$ and $cos(\pi) = -1$

$tan^{-1}(1)) = \frac{\pi}{4}$ and $tan^{-1}(-1) = \frac{-\pi}{4}$

so range is $[\frac{-\pi}{4}, \frac{\pi}{4}]$
Title: Re: Circular functions
Post by: d0minicz on March 23, 2009, 07:20:30 pm: oh i get it, thanks

for cos(tan^-1 x) how do i find the range when Dom = [0,oo) or R+ U {0}
Title: Re: Circular functions
Post by: kamil9876 on March 23, 2009, 07:40:59 pm: tan^-1(x) for domain 0 to infinity gives a range of $[0,\pi/2)$ . So the range of this function is the same as the range of cos(u) where u has domain $[0,\pi/2)$ . 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.
Title: Re: Circular functions
Post by: d0minicz on March 23, 2009, 07:52:37 pm: ahhhhhhhhhh crap i overlooked the normal range of tan^-1 x ....
thanks kamil
Title: Re: Circular functions
Post by: d0minicz on March 23, 2009, 08:08:44 pm: The Questions 9, 10 and 11 in Exercise 3D essentials have been pissing me off for a long time lol
anywayz heres one

let $\ tan {\gamma} = c$ where $\ \gamma \in [\pi, \frac {3\pi}{2} ]$ . Find in terms of $\ \gamma$ two values of x in the range $\ [0,2\pi]$ which satisfy each of the following equations.

a) tanx = -c

b) cotx = c

need alot of help understanding these questions :(
thanks alot !!!

anyone :(?
Title: Re: Circular functions
Post by: kamil9876 on March 24, 2009, 08:06:52 pm: http://vcenotes.com/forum/index.php/topic,11984.msg138611.html#msg138611
Title: Re: Circular functions
Post by: d0minicz on July 05, 2009, 01:59:09 am: Simplify the following completely:
a) $cos^4x + sin^4x + \frac {1}{2} sin^2 (2x)$
b) $sin^2x - sin^4x -cos^2xsin^2x$

Prove the following :
a) $cos(3x) = cos^3x - 3cosxsin^2x$
b) $sin(3x) = 3cos^2xsinx - sin^3x$

thanks =]
Title: Re: Circular functions
Post by: TrueTears on July 05, 2009, 02:38:41 am: Quote from: d0minicz on March 23, 2009, 08:08:44 pm
The Questions 9, 10 and 11 in Exercise 3D essentials have been pissing me off for a long time lol
anywayz heres one

let $\ tan {\gamma} = c$ where $\ \gamma \in [\pi, \frac {3\pi}{2} ]$ . Find in terms of $\ \gamma$ two values of x in the range $\ [0,2\pi]$ 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
Title: Re: Circular functions
Post by: TrueTears on July 05, 2009, 02:45:17 am: Quote from: d0minicz on July 05, 2009, 01:59:09 am
Simplify the following completely:
a) $cos^4x + sin^4x + \frac {1}{2} sin^2 (2x)$
b) $sin^2x - sin^4x -cos^2xsin^2x$

Prove the following :
a) $cos(3x) = cos^3x - 3cosxsin^2x$
b) $sin(3x) = 3cos^2xsinx - sin^3x$

thanks =]
a) $cos^2xcos^2x + sin^2xsin^2x + \frac{1}{2}(4sin^2xcos^2x) = cos^2x(1-sin^2x) + sin^2x(1-cos^2x) + 2sin^2xcos^2x = cos^2x-sin^2xcos^2x + sin^2x - sin^2xcos^2x + 2sin^2xcos^2x = 1 - 2sin^2xcos^2x + 2sin^2xcos^2x = 1$
b) $sin^2x - sin^2xsin^2x - cos^2xsin^2x = sin^2x(1-sin^2x) - sin^2xcos^2x = sin^2xcos^2x - sin^2xcos^2x = 0$
c) $cos(x+2x) = cosxcos(2x)-sinxsin(2x) = cosx(cos^2x-sin^2x)-sinx(2sinxcosx) = cos^3x-cosxsin^2x-2sin^xcosx = cos^3x-3cosxsin^2x$
b) Too late I cbf typing this up, but same principle you see? $sin(x+2x)$ gogo!
Title: Re: Circular functions
Post by: d0minicz on July 12, 2009, 04:20:34 pm: Couple of Q's guys...

1. $\frac {tan(2x)}{1+sec(2x)}$ equals:
A. $tan(2x)$
B. $tan(2x) +1$
C. $tan (x) + 1$
D. $sin(2x)$
E. $tan(x)$

2. For $\frac {\pi}{2} < A < \pi$ and $0 < B < \frac {pi}{2}$ with $sinA = t$ and $cosB = t$ , $cos(B+A)$ equals to:

A. 0
B. $\sqrt {1-t^2}$
C. $2t^2-1$
D. $1-2t^2$
E. $-2t \sqrt {1-t^2}$

Need to see workings for em

thanks !
Title: Re: Circular functions
Post by: kamil9876 on July 12, 2009, 04:40:38 pm: 2.)
$<br />cos(B+A)=cosBcosA-sinBsinA$
$=t(cosA-sinB)$ (1)

$cosA=-\sqrt{1-sin^2 A}$ Because in second quadrant
$=-\sqrt{1-t^2}$ (2)

$sinB=\sqrt{1-cos^2B)$
$=\sqrt{1-t^2}$ (3)

just sub (2) and (3) back into (1)
Title: Re: Circular functions
Post by: Damo17 on July 12, 2009, 04:45:09 pm: Quote from: d0minicz on July 12, 2009, 04:20:34 pm
Couple of Q's guys...

1. $\frac {tan(2x)}{1+sec(2x)}$ equals:
A. $tan(2x)$
B. $tan(2x) +1$
C. $tan (x) + 1$
D. $sin(2x)$
E. $tan(x)$

Need to see workings for em

thanks !

1.) $\frac {tan(2x)}{1+sec(2x)} = \frac{ \frac{sin(2x)}{cos(2x)}}{1+ \frac{1}{cos(2x)}} = \frac{ \frac{sin(2x)}{cos(2x)}}{\frac{cos(2x)+1}{cos(2x)}} = \frac{sin(2x)}{cos(2x)+1} = \frac{2sinxcosx}{(2cos^2x -1)+1} = tanx$
Title: Re: Circular functions
Post by: dcc on July 12, 2009, 04:54:55 pm: 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.
Title: Re: Circular functions
Post by: TrueTears on July 12, 2009, 04:55:59 pm: Quote from: dcc on July 12, 2009, 04:54:55 pm
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]
Title: Re: Circular functions
Post by: dcc on July 12, 2009, 04:57:59 pm: 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).
Title: Re: Circular functions
Post by: TrueTears on July 12, 2009, 04:58:54 pm: Quote from: Damo17 on July 12, 2009, 04:45:09 pm
Quote from: d0minicz on July 12, 2009, 04:20:34 pm
Couple of Q's guys...

1. $\frac {tan(2x)}{1+sec(2x)}$ equals:
A. $tan(2x)$
B. $tan(2x) +1$
C. $tan (x) + 1$
D. $sin(2x)$
E. $tan(x)$

Need to see workings for em

thanks !

1.) $\frac {tan(2x)}{1+sec(2x)} = \frac{ \frac{sin(2x)}{cos(2x)}}{1+ \frac{1}{cos(2x)}} = \frac{ \frac{sin(2x)}{cos(2x)}}{\frac{cos(2x)+1}{cos(2x)}} = \frac{sin(2x)}{cos(2x)+1} = \frac{2sinxcosx}{(2cos^2x -1)+1} = tanx$
That question was on my spesh test once. Quite fun.
Title: Re: Circular functions
Post by: d0minicz on July 12, 2009, 05:03:48 pm: ahhhhh cheers guys
Title: Re: Circular functions
Post by: QuantumJG on July 17, 2009, 02:57:43 pm: Quote from: d0minicz on February 22, 2009, 12:51:13 pm
Neeed help on a couple of questions

a) $\frac {{tan}^2 {x} + 1} {{tan}^2 {x}}$

a) [tan^2(x) + 1]/tan^2(x) = 1 + cot^2(x) = cosec^2(x)

Quote from: d0minicz on February 22, 2009, 12:51:13 pm
b) $\ {sin}^4 {x} - {cos}^4 {x}$

(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)

Quote from: d0minicz on February 22, 2009, 12:51:13 pm

AND

If $\ {cot} {x} = 3$ , $\ {x} \in \left[ {\pi} , \frac{3\pi}{2} \right ]$ , find:

a) $\ cosec {x}$

b) $\ sin {x}$

c) $\ sec {x}$

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
Title: Re: Circular functions
Post by: TrueTears on July 17, 2009, 04:24:06 pm: 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!