標題:
maths 中四5題
發問:
圖片參考：http://imgcld.yimg.com/8/n/HA07492551/o/701106070066513873456421.jpg 想問 9,12,13,14,54 題 特別係恆等式 最唔識
最佳解答:
9) L.H.S. = (1 + sinA + cosA)^2 = (1 + sinA + cosA)(1 + sinA + cosA) = 1 + sinA + cosA + sinA + sin^2A + sinAcosA + cosA + sinAcosA + cos^2A = 1 + 1 + 2sinA + 2cosA + 2sinAcosA = 2(1 + sinA + cosA + sinAcosA) = 2(1 + sinA)(1 + cosA) = R.H.S. 11) tanθ = -8/15 sinθ = -8/17 cosθ = 15/17 (cosθ在第四象限為正數) sinθ + 3cosθ / cosθ = -8/17 + 45/17 / 15/17 = 35/17 * 17/15 = 37/15 12) L.H.S. = sinA + cosA / sinA - cosA = (sinA + cosA)(sinA + cosA) / (sinA - cosA)(sinA + cosA) = sin^2A + 2sinAcosA + cos^2A / sin^2A - cos^2A = 1 + 2sinAcosA / (1 - cos^2A) - cos^2A = 1 + 2sinAcosA / 1 - 2cos^2A = R.H.S. 13) sina - cosa / sina + cosa = tana - 1 / tana + 1 = 3 - 1 / 3 + 1 = 2/4 = 1/2 14) x^2 + 2^2 = sqrt13^2 x^2 + 4 = 13 x = 3(捨去) or -3 cos(180 - φ) = 3 / sqrt13 -cosφ = 3 / sqrt13 cosφ = - 3 / sqrt13 sin^2φ = 1 - cos^2φ = 1 - 9/13 = 4/13 sinφ = 2 / sqrt13 54) 2tanx / 1 + tan^2x = 1 2tanx = 1 + tan^2x tan^2x + 2tanx + 1 = 0 (tanx + 1)^2 = 0 tanx = -1 x = 225
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