Bryn Posted June 18, 2004 Posted June 18, 2004 won't be too many more question, got my last maths exam monday but you help would be much appreciated. Solve for [math]0 \leq x <360[/math] [math]2cos(x + 50)^o = sin(x + 40)^o[/math] and [math]2cosx+sinx = 5[/math]
NSX Posted June 18, 2004 Posted June 18, 2004 Try the sum / difference formulaes: sin(A+B)=sin A cos B + cos A sin B sin(A-B)=sin A cos B - cos A sin B cos(A+B)=cos A cos B - sin A sin B cos(A-B)=cos A cos B + sin A sin B hm...i'm not sure anymore; it gets kind of messy
Bryn Posted June 18, 2004 Author Posted June 18, 2004 i did, what do i do with it once i got it in the form 2(cos x cos 50-sin x sin 50) = sin x cos 40 + sin 40 cos x?
Dave Posted June 19, 2004 Posted June 19, 2004 Remember that cos(50) is a constant, rearrange the second equation in terms of either sin(x) or cos(x) and then substitute it into that equation.
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