三角形ABC中,bcosC=ccosB,若tanA=-2倍根号2,求sin（4B+π/3）的值
1个回答
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/cosC
sinBcosC = cosBsinC
sinBcosC - cosBsinC = 0
sin(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√2
tan(180° - 2B) = -2√2
-tan2B = -2√2
tan2B = 2√2
利用万能公式,
得:sin4B = 2tan2B/(1 + tan²2B)
= 2 * 2√2/[1 + (-2√2)²]
= 4√2/9
cos4B = (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
相关问题
