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Maths Problems/Questions - HSN forum - Page 2

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Maths Problems/Questions

Started by Ally, Apr 14 2004 06:13 PM

32 replies to this topic

#21 Allan

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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

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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

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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
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#24 Ally

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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...


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#25 Steve

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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
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#26 Allan

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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

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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

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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?


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#29 sparky

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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

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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

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Posted 20 April 2004 - 08:27 PM

biggrin.gif Thanks!

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

#32 Ally

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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!!
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#33 Discogirl17

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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
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