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2001 Paper 1, Q 7b)(ii) - HSN forum

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2001 Paper 1, Q 7b)(ii)


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#1 kr88

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Posted 18 April 2005 - 08:32 AM

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heyhey, well ive gotten past (i) of this question, the answer to which is
1/square root2 sinx + 1/square root2 cos x
BUT, when you hate to take g(h(x))from f(h(x)), surely these are the same so the answer would be zero? sorry, i know im not explaining it very well but if you have the past papers you should get it! any help much appreciated! xx

#2 Ally

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Posted 18 April 2005 - 09:47 AM

Hi, welcome to HSN. smile.gif

I'll do all of the question, since it'll make it more clearer.

a)
i)
f(h(x)) = f(x + pi.gif /4)
= sin (x + pi.gif /4)

ii)
g(h(x)) = g(x + pi.gif /4)
= cos (x + pi.gif /4)

b)
i) sin (x + pi.gif /4)

You have to use the addition formula here which is sin(A + B) = sin A cos B + cos A sin B

therefore.gif sin (x + pi.gif /4) = sin x cos pi.gif /4 + cos x sin pi.gif /4
= sin x * (1/√2) + cos x * (1/√2)
= 1/√2 sinx + 1 /√2 cos x

ii)
cos (x + pi.gif /4)

You have to use the addition formula which is cos(A + B) = cos A cos B - sin A sin B. Notice that the addition formula used this time is different

cos (x + pi.gif /4) = cos x cos (pi.gif /4) - sin x sin (pi.gif /4)
= 1/√2 cosx - 1 /√2 sin x

f(h(x)) = 1/√2 sinx + 1 /√2 cos x
g(h(x)) = 1/√2 cosx - 1 /√2 sin x

f(h(x)) - g(h(x)) = 1
1/√2 sinx + 1 /√2 cos x - (1/√2 cosx - 1 /√2 sin x) = 1
1/√2 sinx + 1 /√2 cos x - 1/√2 cosx + 1 /√2 sin x = 1
1/√2 sinx + 1 /√2 cos x - 1/√2 cosx + 1 /√2 sin x = 1
2/√2 sinx = 1
sin x = √2/2
sin x = √2/ √2 . √2
sin x = 1/√2

x = pi.gif /4, 3 pi.gif / 4

If you don't understand any of the steps, post again.





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