锐角三角形中三角函数式取值范围的证明

首先根据余弦定理，有：/n/n$c^2 = a^2 + b^2 - 2ab/cos C$/n/n$b^2 = a^2 + c^2 - 2ac/cos B$/n/n将$c-b=2b/cos A$带入第一个式子，得到：/n/n$c^2 - b^2 = (c+b)(c-b) = 2ab/cos C$/n/n将$c+b=a/cos A$代入上式，得到：/n/n$a^2/cos A/cos C = 2ab/cos C$/n/n化简得到：/n/n$a/cos A = 2b$/n/n又因为是锐角三角形，所以$/sin A, /sin B, /sin C$都大于0，所以有：/n/n$/sin A > /sin B$/n/n代入$/sin(C+B) = /sin C/cos B + /cos C/sin B$和$/cos(A-B) = /cos A/cos B + /sin A/sin B$，得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /sin C/cos B + /cos C/sin B + 2(/cos A/cos B + /sin A/sin B)^2$/n/n化简得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /sin C/cos B + /cos C/sin B + 2/cos^2A/cos^2B + 4/cos A/cos B/sin A/sin B + 2/sin^2A/sin^2B$/n/n由于$/cos A = /dfrac{2b}{a}$，$/cos B = /dfrac{c^2+a^2-b^2}{2ac}$，代入上式得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /dfrac{c}{a}/cos B + /dfrac{a}{c}/sin B + /dfrac{4b^2}{a^2}/left(1-/dfrac{b^2}{a^2}/right) + 4/left(/dfrac{b}{a}/right)/left(/dfrac{c^2+a^2-b^2}{2ac}/right)/left(1-/dfrac{b^2}{a^2}/right) + 2/left(1-/dfrac{b^2}{a^2}/right)/left(1-/dfrac{c^2}{a^2}/right)$/n/n化简得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /dfrac{c^3+a^3-b^3-2abc}{a^2c} + /dfrac{4b^2}{a^2}/left(1-/dfrac{b^2}{a^2}/right) + 4/left(/dfrac{b}{a}/right)/left(/dfrac{c^2+a^2-b^2}{2ac}/right)/left(1-/dfrac{b^2}{a^2}/right) + 2/left(1-/dfrac{b^2}{a^2}/right)/left(1-/dfrac{c^2}{a^2}/right)$/n/n进一步化简得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /dfrac{a^2-b^2+c^2}{a^2} + /dfrac{2b^2}{a^2}/left(1-/dfrac{b^2}{a^2}/right) - /dfrac{2bc}{a^2} + 2/left(1-/dfrac{b^2}{a^2}/right)/left(1-/dfrac{c^2}{a^2}/right)$/n/n将$a=2b/cos A$代入上式，得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /dfrac{4/cos^2A-/cos^2C+/cos^2B}{4/cos^2A} + /dfrac{1-4/cos^2A}{2/cos^2A} - /dfrac{2/cos C}{2/cos A} + 2/left(1-/dfrac{1}{4/cos^2A}/right)/left(1-/dfrac{/cos^2C}{4/cos^2A}/right)$/n/n化简得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /dfrac{3}{4/cos^2A} - /dfrac{/cos^2C}{4/cos^2A} + /dfrac{1}{2} - /dfrac{/cos C}{/cos A} + /dfrac{3}{4/cos^2A}/cos^2C$/n/n再次化简得到：/n/n$/sin(C+B) + 2/cos^2(A-B) = /dfrac{5}{4} - /dfrac{/cos C}{/cos A} - /dfrac{1}{4}/dfrac{4/cos^4A-/cos^2C}{/cos^2A} + /dfrac{3}{4}/dfrac{/cos^2C}{/cos^2A}$/n/n由于是锐角三角形，所以$/cos A > 0$，$/cos C < 0$，所以有：/n/n$/sin(C+B) + 2/cos^2(A-B) < /dfrac{5}{4}$/n/n又因为$/sin(C+B) /leq 1$，所以有：/n/n$2/cos^2(A-B) < /dfrac{1}{4}$/n/n化简得到：/n/n$|/cos(A-B)| < /dfrac{1}{2}$/n/n所以：/n/n$-/dfrac{/pi}{6} < A-B < /dfrac{/pi}{6}$/n/n综上所述，$/sin(C+B)+2/cos^2(A-B)$的取值范围为：/n/n$/dfrac{5}{4} - /sin(C+B) < 2/cos^2(A-B) < /dfrac{1}{4}$/n/n$-/dfrac{1}{2} < /cos(A-B) < /dfrac{1}{2}$/n/n$-/dfrac{/pi}{6} < A-B < /dfrac{/pi}{6}$/n/n注意到$/cos(A-B) = /cos A/cos B + /sin A/sin B$，所以$/cos(A-B)$的取值范围为$|/cos A/cos B + /sin A/sin B| < /dfrac{1}{2}$，即$|/sin(A+B)| < /dfrac{1}{2}$。根据三角函数的性质，$-/dfrac{/pi}{6} < A+B < /dfrac{/pi}{6}$，所以有：/n/n$-/dfrac{/pi}{3} < C < 0$/n/n$-/dfrac{/pi}{6} < B < /dfrac{/pi}{6}$/n/n$-/dfrac{/pi}{2} < A < /dfrac{/pi}{2}$/n/n综上所述，$/sin(C+B)+2/cos^2(A-B)$的取值范围为：/n/n$/dfrac{3}{4} < /sin(C+B)+2/cos^2(A-B) < /dfrac{5}{4}$