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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

Heres the question i cannot solve:

sin x + 8 cos x represented in the form Asin(x-y)

Thanks,

C25

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi C25

The method for doing this is to 'force' it into the following format:

Asin(x-y) = Asinxcosy -Acosxsiny (use of compound angle formula; slight complication as you have a + sign between the elements))

so

sinx + 8cosx = 1sinx + 8cosx

= A(1/A sinx + 8/A cosx) where A = sqr root (1 + 64)

[This use of Pythagoras means there is a 'right angled triangle' with angle y where cosy = 1/A and siny = -8/A. Strictly no such triangle exists because of the minus 8; more usually there is no sign change when using the compound angle formula; but it won't matter as long as you get a 'y' that's over 180 degrees.]

So, find A by the above Pythag. calculation and y by y = inv tan (8/1); this will be an acute angle; so now find the reflex angle that has cos positive and sin negative; so an angle between 270 and 360 is expected.

Hope that makes sense; I've got some jet lag today; post again if you need more clarification.

Bob

*Last edited by Bob (2010-10-04 21:35:37)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**soroban****Member**- Registered: 2007-03-09
- Posts: 452

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**careless25****Real Member**- Registered: 2008-07-24
- Posts: 560

Thanks a lot both of you ...

Soroban thats a nice trick...took me a few minutes to understand but finally got it

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