Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 114

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 127

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 136

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 137

Warning: Cannot modify header information - headers already sent by (output started at /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php:909) in /home/hsn/public_html/forum/admin/sources/classes/output/formats/html/htmlOutput.php on line 141
Maths Problems/Questions - HSN forum - Page 2

Jump to content


Maths Problems/Questions


32 replies to this topic

#21 Allan

    Top of the Class

  • Moderators
  • PipPipPipPipPip
  • 253 posts
  • Gender:Male

Posted 17 April 2004 - 03:19 PM

Working that out I get for part a

pi/6 and 5pi/6

For part b I get

pi/3 and 2pi/3

#22 sparky

    Fully Fledged Genius

  • Moderators
  • PipPipPipPipPipPipPip
  • 1,323 posts
  • Location:Airdrie

Posted 17 April 2004 - 03:27 PM

Thanks, just noticed made a mistake in the first line of each one rolleyes.gif stupid!!

I am revising trig equ's just now! smile.gif So they'll prob be more q's later!

Thanks
Mark

#23 Ally

    HSN Legend

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 2,912 posts
  • Interests:Just finished 1st year at uni studying medicine.
  • Gender:Male

Posted 18 April 2004 - 12:37 AM

More Questions....

The list seems endless for me. lol!


Anyway, back to the point. Here is the question:


Solve the equation 2sinx°-3cosx° = 2.5 in the interval 0≤x<360.

Help would be much appreciated.

Ally

#24 Ally

    HSN Legend

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 2,912 posts
  • Interests:Just finished 1st year at uni studying medicine.
  • Gender:Male

Posted 18 April 2004 - 12:53 AM

And another 1...

The line y = -1 is a tangent to a circle which passes through (0,0) and (6,0). Find the equation of this circle.

It's always the circle or trigonometry questions that stump me...

...Or maybe there just meant to be the hardest...



#25 Steve

    Top of the Class

  • Admin
  • PipPipPipPipPip
  • 435 posts
  • Location:Edinburgh
  • Gender:Male

Posted 18 April 2004 - 11:18 AM

QUOTE
Solve the equation 2sinx°-3cosx° = 2.5 in the interval 0<x<360


Right, this is quite a trcky one; you have to use the Wave Function first (sorry, our notes on that aren't out yet tongue.gif).

Bearing that in mind, you should have another go at the question, if you still can't get the answer, my working is here (not any more, it was wrong!) (don't know that it's right though biggrin.gif, please post to tell me if it is)

Don't listen to me, listen to Allan - I always make stupid mistakes mad.gif
HSN contribute: Help the site grow!

Looking for a Maths tutor in West Lothian? Just PM me!

#26 Allan

    Top of the Class

  • Moderators
  • PipPipPipPipPip
  • 253 posts
  • Gender:Male

Posted 18 April 2004 - 11:22 AM

QUOTE (ally @ Apr 18 2004, 01:37 AM)
More Questions....

The list seems endless for me. lol!


Anyway, back to the point. Here is the question:


Solve the equation 2sinx°-3cosx° = 2.5 in the interval 0≤x<360.

Help would be much appreciated.

Ally

For that question you need to express 2sin x - 3cos x in the form ksin(x - α) [or kcos(x - α)

I work that out to be √13sin(x - 56.3)

Now that it's in that form you should be able to solve it like any normal trig equation:

√13sin(x - 56.3) = 2.5



#27 Allan

    Top of the Class

  • Moderators
  • PipPipPipPipPip
  • 253 posts
  • Gender:Male

Posted 18 April 2004 - 11:34 AM

QUOTE (Steve @ Apr 18 2004, 12:18 PM)
don't know that it's right though biggrin.gif, please post to tell me if it is

Steve, you've put the 2 and the 3 the wrong way round when doing tanα.

And it's 2.5, not 25 tongue.gif

Got me all confused there, my answer didn't work out the same, but I think that's why

#28 Ally

    HSN Legend

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 2,912 posts
  • Interests:Just finished 1st year at uni studying medicine.
  • Gender:Male

Posted 18 April 2004 - 07:58 PM

Thanks for that.

I get it now. rolleyes.gif

Have you had any luck on the circles question?



#29 sparky

    Fully Fledged Genius

  • Moderators
  • PipPipPipPipPipPipPip
  • 1,323 posts
  • Location:Airdrie

Posted 20 April 2004 - 07:52 PM

Ok I have two questions, from the A/B Exercises in the TJ 3 Booklet incase anyone has it.

1 By writing cos3x as cos(2x + x) show that cos3x = 4cos³x - 3cosx

And if anyone can give me the working for this, as I don't know if I have set mines down properly,

log (to the base 6 ) x + log (to the base 6 ) (2x+1) =2

Thanks
Mark

#30 Allan

    Top of the Class

  • Moderators
  • PipPipPipPipPip
  • 253 posts
  • Gender:Male

Posted 20 April 2004 - 08:20 PM

QUOTE (sparky @ Apr 20 2004, 08:52 PM)
log (to the base 6 ) x + log (to the base 6 ) (2x+1) =2

using the log rule logAB = logA + logB

you get log[base 6]x(2x + 1) = 2
log[base 6]2x² + x = 2

Then that can be re-written using the log rule y = a[power x] if x = log[base a]y

2x² + x = 6²
2x² + x - 36 = 0
(2x + 9)(x - 4) = 0
x = -9/2 and x = 4

Conditions for the rule above state that x > 0 so the answer is x = 4

#31 sparky

    Fully Fledged Genius

  • Moderators
  • PipPipPipPipPipPipPip
  • 1,323 posts
  • Location:Airdrie

Posted 20 April 2004 - 08:27 PM

biggrin.gif Thanks!

(You are great at maths btw! tongue.gif )
Mark

#32 Ally

    HSN Legend

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 2,912 posts
  • Interests:Just finished 1st year at uni studying medicine.
  • Gender:Male

Posted 21 April 2004 - 09:45 PM

QUOTE (sparky @ Apr 20 2004, 07:52 PM)
Ok I have two questions, from the A/B Exercises in the TJ 3 Booklet incase anyone has it.

1 By writing cos3x as cos(2x + x) show that cos3x = 4cos³x - 3cosx


OK Sparky here goes:

cos3x = cos(2x+x)
=> cos2xcosx - sin2xsinx
=> (2cos²x-1)(cosx) - sin2xsinx
=> 2cos³x-cosx - (2sinxcosx)(sinx)
=> 2cos³x-cosx - (2sin²xcosx)
=> 2cos³x-cosx - (2(1-cos²x)(cosx)
=> 2cos³x-cosx - (2(cosx-cos³x)
=> 2cos³x-cosx - (2cosx-2cos³x)
=> 2cos³x-cosx - 2cosx + 2cos³x

= 4cos³-3cosx

Hope that helps!!

#33 Discogirl17

    HSN Legend

  • Moderators
  • PipPipPipPipPipPipPipPip
  • 3,034 posts
  • Location:South Lanarkshire
  • Gender:Female

Posted 24 April 2004 - 11:24 AM

Thanx everyone. I'm printing off the answers to those questions so that I can look at the method more closely and do some examples myself. Thanx ur all geniuses.
I know "geniuses" isn't a word but still! smile.gif
Half ideas,half quality, half a million pound law suit!





1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users