The question is :

Show that f(h(x)) = 1/√2sinx+1/√2cosx

When f(h(x)) = sin(x+π/4)

I know the next step is sinxcosπ/4+cosxsinπ/4 but could someone tell me why this is and if there is some sort of formula used to arrive at this stage? Thanks

Graham

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# Help needed understanding higher 2001 maths question

Started by graham83, Apr 06 2011 02:07 PM

1 reply to this topic

### #1

Posted 06 April 2011 - 02:07 PM

### #2

Posted 06 April 2011 - 08:25 PM

graham83, on 06 April 2011 - 02:07 PM, said:

The question is :

Show that f(h(x)) = 1/√2sinx+1/√2cosx

When f(h(x)) = sin(x+π/4)

I know the next step is sinxcosπ/4+cosxsinπ/4 but could someone tell me why this is and if there is some sort of formula used to arrive at this stage? Thanks

Graham

Show that f(h(x)) = 1/√2sinx+1/√2cosx

When f(h(x)) = sin(x+π/4)

I know the next step is sinxcosπ/4+cosxsinπ/4 but could someone tell me why this is and if there is some sort of formula used to arrive at this stage? Thanks

Graham

it is from the sine expansion, sin(a + b) = sin(a)cos(b) + cos(a)sin(b) where a=x and b= π/4 I'm sure this is one of the formulae given in the exam (check a past paper to check) and from there sin(π/4)=cos(π/4)=1/√2 from your exact value triangles

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