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2002 - Paper 1 - Question 3(b) - HSN forum

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2002 - Paper 1 - Question 3(b)


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#1 AmAnDa

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Posted 07 April 2005 - 04:34 PM

2002 - Question 3 - Paper 1

I've done part (a):
f(g(x)) = sin2x , g(f(x)) = 2sinx

But I'm stuck on part (b)

Can anyone help?!

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#2 oddbins

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Posted 07 April 2005 - 04:51 PM

You should get:
2sin2x=2sinx
Then you take the 2sinx over to the other side to get:
2sin2x-2sinx=o

Then use this formula on your formula sheet:
sin2A = 2sinA cosA

So then you should get:
4sinx cosx - 2sinx=0

Then you can solve it. Does this help, or do you need me to explain anymore?


#3 Dave

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Posted 07 April 2005 - 04:53 PM

so you have

2sin2x=2sinx

the thing that is tested is do you know sin 2x = 2sinxcosx

if you do then sub it in to get

2(2sinxcosx)=2sinx

then its just solving the equation

4sinxcosx = 2sinx

4sinxcosx-2sinx = 0

factorise

2sinx(2cosx-1) = 0

2sinx=0 or 2cosx-1 = 0
x= 0,180,360 cosx= 1/2
x = 60,300

If i am not here i am somewhere else



#4 AmAnDa

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Posted 07 April 2005 - 05:11 PM

Thanx!! biggrin.gif

Yip thats helped alot!! I totally forget bout the formula! Oops! blink.gif silly me!

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#5 Steve

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Posted 07 April 2005 - 07:24 PM

Always remember in the exam to look at the Formulae List at the front when you are doing a trig question. smile.gif
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