Trigonometry Excersises Request (1 Viewer)

[rsingh]

Hey guys,

I just started yr 12 and need some help with Further Trig. I have the Fitzpatrick and Coroneous book, and seem to be going fine, however, I would like some more difficult exersises.

If someone could please scan (from textbooks) or post up some exercises, that would be great! Maybe from Cambridge or something?

This would be greatly appreciated!

Thanks.

 

[paganinio]

i recommend you to go to library

 

[grimreaper]

I reckon the stuff in fitz\coroneus will be harder than anything you get in the hsc... trig isnt a very difficult topic in general

 

[paper cup]

try the resources section on the main page.
anyhow the hardcore ppl will be here soon to help.

 

[Sharky]

I have little idea as to exactly what u'd like, but i'll post u a question anyways just for something to do (it has "some" use of trig so im gonna assume u'll like it)

Show that:

cos(3Theta) = 4cos³(Theta) - 3cos(Theta)

Hence, solve 1-6x+8x³=0

 

[ngai]

find exact value of cos(37.5 degrees)*cos(7.5 degrees)

Edit: i better give a hint for that...
cos(x+y)=cosxcosy-sinxsiny
cos(x-y)=cosxcosy+sinxsiny

 

[Slidey]

(Possible) Spoiler for Sharky's post:

cos3@=cos2@cos@-sin2@sin@=cos^3@-sin^2@cos@-2sin^2cos@
i.e.: cos^3@-3sin^2@cos@
sin^2@=1-cos^2@, so
cos^3@-3(1-cos^2@)cos@=4cos^3@-3cos@

let x=cos@,
1-6x+8x^3=0
1-6cos@+8cos^3@=0
1+2(cos3@)=0
cos3@=-1/2
3@=arccos(-1/2)
3@=120 and 240
@=40, 80
x=cos@
x=cos40, cos80

2 solutions... there should be 3. Unless there is a repeated root... and all roots are real. Maybe I am not wrong after all.

 

[mojako]

here we go again.. how did you get the 2 solutions? can you type it up?

anyway there are three, by graphing program
well I showed this Q to my teacher and he told me that you can have 3 values of 3@ which give 3 unique cos@ values, even though the 3@ values aren't really unique. but then when I asked how that can possibly happen, he refused to answer and said that it's one of the grey areas of mathematics which only God knows.. guess he was lying...

uh.. trig is soo confusing..
just think about it..
how does this cos(3Theta) = 4cos³(Theta) - 3cos(Theta) relate to 1-6x+8x³=0??
okay, it uses the word "hence" but.. I cant see any link!!

 

[Estel]

Slide:
cos3@=-1/2
3@ = 2pi/3, 4pi/3, 5pi/3
:. roots are cos(2pi/9), cos(4pi/9), cos(5pi/9)
i.e. you neglected the root -cos80

 

[Sharky]

That's some clever shorthand notation from Slide Rule there, i might just borrow it.
(let @ be a Theta)

I love the hidding text, never thought to do that.

So straight out, here's how i did it:

cos3@
=cos(2@+@)
=cos2@cos@-sin2@sin@
=(cos²@-sin²@)cos@-(2sin@cos@)sin@
=cos³@-sin²@cos@-2sin²@cos@
=cos³@-(1-cos²@)cos@-2(1-cos²@)cos@
=cos³@-cos@+cos³@-2cos@+2cos³@
=4cos³@-3cos@



1-6x+8x³=0
Let x=cos@
1-6cos@+8cos³@=0
-6cos@+8cos³@=-1
2(4cos³@-3cos@)=-1
2cos3@=-1
cos3@=-1/2
3@=120, 240, 480

(This third value has been considered because we want the values of @ for 0<@<180 and not 0<@<360 to avoid repetition, so the values for 3@ for 0<3@<540 need to be considered)

@=40, 80, 160
Hence
x=cos40, cos80, cos160 (or -cos20)

These are the three, and only 3, real roots.

 

[Slidey]

*scratches his head about the third root*

I never did understand the domain stuff for trig.

cos160 = cos(180-20) and 2nd quad, so -cos20. OK. That's fine. So I just need you to elaborate on how you restrict and expand the domain and such?

Now, I concede that if you DO have a domain such as 0<=@<=540, that cos480 would also be a solution, but... how do you get that domain? =/

 

[mojako]

its explained before...

(This third value has been considered because we want the values of @ for 0<@<180 and not 0<@<360 to avoid repetition, so the values for 3@ for 0<3@<540 need to be considered)

you know cos@ will be unique for 0<=@<=180 (first and second quadrants) and for other domains the values of cos@ will just repeat.

but I wont bother doing this domain thing.. since its cubic its meant to have 3 solutions so just find 3 values of 3@ and find their cos@ values.

 

[Sharky]

but... how do you get that domain? =/

well, it's just that we wanted a domain of @ being 0<@<180

This is so that we don't get repeated solutions, as the cos values in the 3rd and 4th quadrants are the same as they are in the 1st and 2nd.

This means that we if we want values of 0<@<180, then by tripling the inequality we get 0<3@<540.

i.e. we want values of 3@ that range from 0 to 540.

Now, in terms of cos3@=(-1/2):

The general solution to this for all values of 3@ will be:
3@ = 360n±120
(for any positive and negative integer value for n)

So, the values for 3@ that lie within our domain of 0<3@<540 would be 120,240,480.

 

[Slidey]

Thanks Sharky.