Bryn Posted June 18, 2004 Share 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] Link to comment Share on other sites More sharing options...
NSX Posted June 18, 2004 Share 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 Link to comment Share on other sites More sharing options...
Bryn Posted June 18, 2004 Author Share 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? Link to comment Share on other sites More sharing options...
Dave Posted June 19, 2004 Share 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. Link to comment Share on other sites More sharing options...
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