三角形ABC中,bcosC=ccosB,若tanA=-2倍根号2,求sin（4B+π/3）的值

跟据正弦定理,b/sinB = c/sinC得:b/c = sinB/sinC已知:bcosC=ccosB得:b/c = cosB/cosC两式相等,得:sinB/sinC = cosB/cosCsinBcosC = cosBsinCsinBcosC - cosBsinC = 0sin(B - C) = 0所以B - C = 0 (A,B,C不超於180度)得B = C且A + B + C = 180°所以A + B + B = 180°A = 180° - 2B代入tanA = -2√2tan(180° - 2B) = -2√2-tan2B = -2√2tan2B = 2√2利用万能公式,得:sin4B = 2tan2B/(1 + tan²2B)= 2 * 2√2/[1 + (-2√2)²]= 4√2/9cos4B = (1 - tan²2B)/(1 + tan²2B)= [1 - (2√2)²]/[1 + (2√2)²]= -7/9所求:sin(4B + π/3)= sin4Bcosπ/3 + cos4Bsinπ/3= (4√2/9)*(1/2) + (-7/9)/(√3/2)= (4√2 - 7√3)/18