Use Acos@+Bsin@ = Rcos(@-c)

where

R^2 = A^2 + B^2

and

tanc = B/A
RichE
Use Acos@+Bsin@ = Rcos(@-c)

where

R^2 = A^2 + B^2

and

tanc = B/A

did that - my answer doesnt agrree with book - i realy cant se where im going wrong

hold on i got it. Why did u use - c and not + c
ryan750
solve:

5cosx - 12sinx = 6.5
for 0<x<360

thanks

R^2 = A^2 + B^2 = 25 + 12^2 = 13^2 -> R =13

tanc = -12/5

c = -arctan(12/5) = -1.176

13cos(@+1.176) = 6.5

cos(@+1.176) = 0.5

@+1.176 = 5.2359 or 7.3303

@ = 4.06 or 6.154
no its stil comin out rong - can sum1 see what they get
ok - i get same answers as u. But i dont see why u have put - c instead of + c.
i thought that when u have a positive cosine and negative sine u use
cos(x + c) because when all terms positive u use cos(x - c).

But the answers we have differ to the book. It has 20.3 and 260.3 degrees?

i still dont get to this if i use cos(x+c)? maybe book is wrong.
ryan750
ok - i get same answers as u. But i dont see why u have put - c instead of + c.
i thought that when u have a positive cosine and negative sine u use
cos(x + c) because when all terms positive u use cos(x - c).

But the answers we have differ to the book. It has 20.3 and 260.3 degrees?

i still dont get to this if i use cos(x+c)? maybe book is wrong.

It doesn't matter which you use as long as you are careful in equating the coefficients of cos@ and sin@ when expanding either
ryan750

But the answers we have differ to the book. It has 20.3 and 260.3 degrees?

i still dont get to this if i use cos(x+c)? maybe book is wrong.

you can check whether the book is wrong by putting its answers back into the equation
book is rong - nit all