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標題:

AMA題....求救

發問:

It is given that sinA = 3/5 and cos(A+B) = -7/25 , where A and B are acute angles. Without solving for A and B, find the value of sinB. 求救....麻煩列出steps,另外,只可用compound angle formula...又或者其他中四應該學到既野(代數等等...).....= =麻煩各位了

最佳解答:

cos(A+B) =cosAcosB - sinAsinB we have cosA = 4/5 and cos^2 B+sin^2 B=1 (4/5)cosB - (3/5)sinB = -7/25 20cosB - 15sinB = -7 let y=sinB 20 sqrt(1-y^2) - 15y= -7 400(1-y^2) = (15y-7)^2 = 225y^2 - 210y + 49 625y^2 - 210y - 351 = 0 by quadratic formular y = [210 + 960]/1250 or [210-960]/1250] (rejected since sinB > 0) so we have done. y = 117/125

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002做得好!!!就係依個quadratic formula!!!!!我同學都係咁做ga! ncy0916唔係唔好,不過一般人會好難先諗到sinB = sin(A+B-A) 依一步既......所以就...唔好意思......我唯有俾002分啦...因為佢個解法近我既程度多d7638E7481407D16B