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June 09, 2026, 02:07:56 am
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Author Topic: Circular functions  (Read 14532 times)  Share 

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d0minicz

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Circular functions
« 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 ;)
« Last Edit: February 22, 2009, 01:08:01 pm by d0minicz »
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ed_saifa

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Re: Circular functions
« Reply #1 on: February 22, 2009, 01:30:54 pm »
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=

=

=

=

=





=

=

=
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ed_saifa

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Re: Circular functions
« Reply #2 on: February 22, 2009, 01:40:56 pm »
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BiG DaN

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Re: Circular functions
« Reply #3 on: February 22, 2009, 01:54:27 pm »
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3a essential q6?
lol i am stuck on it right now too
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d0minicz

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Re: Circular functions
« Reply #4 on: February 22, 2009, 01:58:26 pm »
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thanks alot ed_saifa =]

hahaaha yeah sucks =[
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d0minicz

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Re: Circular functions
« Reply #5 on: February 22, 2009, 02:09:20 pm »
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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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ed_saifa

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Re: Circular functions
« Reply #6 on: February 22, 2009, 02:15:55 pm »
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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



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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d0minicz

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Re: Circular functions
« Reply #7 on: February 26, 2009, 05:15:10 pm »
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= -0.7 ,            0 < x < 180

find:





and


Simplify:

a)

b)

c)

need to learn the identities
thanks !
« Last Edit: February 26, 2009, 06:01:01 pm by d0minicz »
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kamil9876

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Re: Circular functions
« Reply #8 on: February 26, 2009, 07:09:10 pm »
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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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kamil9876

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Re: Circular functions
« Reply #9 on: February 26, 2009, 07:35:56 pm »
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b.)(sinx)^2 (1+(cosx)^2/(sinx)^2)
  =(sinx)^2+(cosx)^2
  =1
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kamil9876

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Re: Circular functions
« Reply #10 on: February 26, 2009, 07:50:55 pm »
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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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kamil9876

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Re: Circular functions
« Reply #11 on: February 26, 2009, 07:57:48 pm »
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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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d0minicz

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Re: Circular functions
« Reply #12 on: March 02, 2009, 06:19:58 pm »
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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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TrueTears

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Re: Circular functions
« Reply #13 on: March 02, 2009, 06:27:54 pm »
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ill have a go at Q 2 3 and 4, bear with me XD
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TrueTears

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Re: Circular functions
« Reply #14 on: March 02, 2009, 06:30:57 pm »
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2. (using the identity )




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.
« Last Edit: March 02, 2009, 06:39:55 pm by TrueTears »
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