In triangle ABC , Summation a^3 cos(B-C)
Answers
Answer:
sin2x+sin2y=2sin(x+y)cos(x−y)
sin2x=2sinxcosx
cos(x−y)=sin2x+sin2y2sin(x+y)=sinxcosx+sinycosysin(x+y)
So now consider the first term from the question:
a3cos(B−C)=a3sin2B+sin2C2sin(B+C)=a3sinBcosB+sinCcosCsin(180∘−A)=a3bkcosB+ckcosCak=a2bcosB+a2ccosC
We’ll get the corresponding result for the other terms, so
a3cos(B−C)+b3cos(C−A)+c3cos(A−B)
=a2bcosB+a2ccosC+b2acosA+b2ccosC+c2acosA+c2bcosB
=ab(acosB+bcosA)+ac(acosC+ccosA)+bc(bcosC+ccosB)
=ab(c)+ac(b)+bc(a)
=3abc ✓
EDIT: I forgot my usual practice of deriving the identities from Euler’s formula. First the sum angle formulas:
cos(x+y)+isin(x+y)=ei(x+y)=eixeiy=(cosx+isinx)(cosy+isiny)=(cosxcosy−sinxsiny)+i(sinxcosy+cosxsiny)
sin(x+y) is the imaginary part, cos(x−y) is the real part with −y for y, flipping the sign to a +. So
sin(x+y)cos(x−y)=(sinxcosy+cosxsiny)(cosxcosy+sinxsiny)
=cosxsinx(cos2y+sin2y)+cosysiny(sin2x+cos2x)
=cosxsinx+cosysiny
This is the form we need it in, but for the identity we take the irresistable step sin2x=2cosxsinx to turn it into the sum of sines:
2sin(x+y)cos(x−y)=sin2x+sin2y