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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

**0**

# 2001 Paper 1, Q 7b)(ii)

Started by kr88, Apr 18 2005 08:32 AM

1 reply to this topic

### #1

Posted 18 April 2005 - 08:32 AM

### #2

Posted 18 April 2005 - 09:47 AM

Hi, welcome to HSN.

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

a)

i)

= sin (x + /4)

ii)

= cos (x + /4)

b)

i) sin (x + /4)

You have to use the addition formula here which is

sin (x + /4) = sin x cos /4 + cos x sin /4

= sin x * (1/√2) + cos x * (1/√2)

= 1/√2 sinx + 1 /√2 cos x

ii)

cos (x + /4)

You have to use the addition formula which is

cos (x + /4) = cos x cos ( /4) - sin x sin ( /4)

= 1/√2 cosx - 1 /√2 sin x

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 = /4, 3 / 4

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

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

a)

i)

*f(h(x))*= f(x + /4)= sin (x + /4)

ii)

*g(h(x))*= g(x + /4)= cos (x + /4)

b)

i) sin (x + /4)

You have to use the addition formula here which is

**sin(A + B) = sin A cos B + cos A sin B**sin (x + /4) = sin x cos /4 + cos x sin /4

= sin x * (1/√2) + cos x * (1/√2)

= 1/√2 sinx + 1 /√2 cos x

ii)

cos (x + /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 differentcos (x + /4) = cos x cos ( /4) - sin x sin ( /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))*= 11/√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 +

2/√2 sinx = 1

sin x = √2/2

sin x = √2/ √2 . √2

sin x = 1/√2

x = /4, 3 / 4

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

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