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question:3. In an equilateral triangle ABC, P is a point on side AB, Q is a point on side AC, and AP = CQ. It is measured that the distance between point A and the midpoint M of line segment PQ is 19 mathrm{~cm}. Then the distance from point P to point C is qquad mathrm{cm}.
answer:3. 38 .
question:4. Given the three interior angles of triangle A B C are angle A, angle B, and angle C. Iffrac{sin A+sqrt{3} cos A}{cos A-sqrt{3} sin A}=tan frac{7 pi}{12},then the maximum value of sin 2 B+2 cos C is ( ).(A) frac{1}{2}(B) 1(C) frac{3}{2}(D) 2
answer:4. C.begin{array}{l}text { Given } frac{sin A+sqrt{3} cos A}{cos A-sqrt{3} sin A}=frac{tan A+tan frac{pi}{3}}{1-tan A cdot tan frac{pi}{3}} =tan left(A+frac{pi}{3}right) Rightarrow angle A+frac{pi}{3}=frac{7 pi}{12} Rightarrow angle A=frac{pi}{4} .end{array}Since 2 angle B+2 angle C=frac{3 pi}{2}, we have,begin{array}{l}sin 2 B+2 cos C=sin left(frac{3 pi}{2}-2 Cright)+2 cos =-cos 2 C+2 cos C =1-2 cos ^{2} C+2 cos C =-2left(cos C-frac{1}{2}right)^{2}+frac{3}{2} leqslant frac{3}{2} .end{array}When angle C=frac{pi}{3}, the equality holds.Thus, the maximum value of sin 2 B+2 cos C is frac{3}{2}.
question:Problem III - 1The non-zero real numbers a and b satisfy the equationfrac{a^{2} b^{2}}{a^{4}-2 b^{4}}=1Find, with justification, all the values taken by the expressionfrac{a^{2}-b^{2}}{a^{2}+b^{2}}
answer:SolutionThe first equality in the statement we write asa^{2} b^{2}=a^{4}-2 b^{4} Leftrightarrow a^{4}-b^{2} a^{2}-2 b^{4}=0 Leftrightarrowleft(a^{2}+b^{2}right)left(a^{2}-2 b^{2}right)=0From this, either a^{2}=-b^{2} (which is clearly impossible), or a^{2}=2 b^{2}.In the latter case, substituting into the second expression, we obtainfrac{2 b^{2}-b^{2}}{2 b^{2}+b^{2}}=frac{1}{3}
question:3.3. For what least a is the inequality frac{sqrt[3]{operatorname{ctg}^{2} x}-sqrt[3]{operatorname{tg}^{2} x}}{sqrt[3]{sin ^{2} x}-sqrt[3]{cos ^{2} x}}<a satisfied for all permissible x inleft(-frac{3 pi}{2} ;-piright)? Round the answer to the nearest hundredth if necessary.
answer:Answer: -2.52 (exact value: -2 sqrt[3]{2} ).
question:therefore 、 m distinct positive even numbers and n distinct positive odd numbers have a total sum of 1987, for all such m and n, what is the maximum value of 3 m+4 n? Please prove your conclusion.The sum of m distinct positive even numbers and n distinct positive odd numbers is 1987, for all such m and n, what is the maximum value of 3 m+4 n? Please prove your conclusion.
answer:The maximum value of 3 m+4 n is 221. The proof is as follows: Let a_{1}+cdots+a_{n}+b_{1}+cdots+b_{n}=1987. Here, a_{i} (1 leqslant i leqslant m) are distinct positive even numbers, and b_{j} (1 leqslant j leqslant n) are distinct positive odd numbers. Clearly, u is always an odd number, andbegin{array}{l}a_{1}+cdots+a_{mathrm{m}} geqslant 2+4+cdots+2 m=m(m+1), b_{1}+cdots+b_{n} geqslant 1+3+cdots+(2 n-1)=n^{2} . therefore m^{2}+m+n^{2} leqslant 1987 text {. (where } n text { is odd) }end{array}This inequality is equivalent to left(m+frac{1}{2}right)^{2}+n^{2} leqslant 1987+frac{1}{4}. By the Cauchy-Schwarz inequality,begin{array}{c}3left(m+frac{1}{2}right)+4 n leqslant sqrt{3^{2}+4^{2}} cdot sqrt{left(m+frac{1}{2}right)^{2}+n^{2}} leqslant 5 sqrt{1987+frac{1}{4}} therefore quad 3 m+4 n leqslantleft[left(5 cdot sqrt{1987+frac{1}{4}}-frac{3}{2}right)right] .end{array}where [x] denotes the integer part of x.Thus, 3 m+4 n leqslant 221.On the other hand, when m=27, n=35, m^{2}+m+n^{2} leqslant 1987 and 3 m+4 n=221. Therefore, for m and n satisfying the conditions of the proposition, the maximum value of 3 m+4 n is 221.
question:Given sin x + cos x = -1, find the value of sin^{2005}x + cos^{2005}x.
answer:Since sin x + cos x = -1, according to the basic relationship of trigonometric functions of the same angle, either sin x or cos x equals 0, and the other equals -1. Therefore, the value of sin^{2005}x + cos^{2005}x is -1. Hence, the answer is boxed{-1}.