answer:So first we need to know what arccos( tan (arcsin x) ) ) is. Then we have sin y = x. The range of sin is [-1, 1], so -1 le x le 1. But the range of arcsin is [-frac{pi}{2}, frac{pi}{2}], so -frac{pi}{2} le y le frac{pi}{2}. That means 0 le cos y le 1. But tan(cdot) has a range of (-infty, infty), so cos y ne 0. Therefore, 0 < cos y le 1. Note that if y = frac{pi}{2}, then cos y = 0 and tan(arcsin x) is undefined. So we must have -frac{pi}{2} le y < frac{pi}{2}. That means 0 < cos y < 1. When y = arcsin x, we have tan y = tan (arcsin x) Then we have tan y = frac{sin y}{cos y} = frac{x}{cos y} So tan y = frac{x}{cos y} = frac{x}{+sqrt{1-x^2}} Then we have arccos (tan (arcsin x) ) = arccos (frac{x}{+sqrt{1-x^2}}) The range of arccos is [0,pi], so 0 leq arccos (frac{x}{+sqrt{1-x^2}}) leq pi Therefore, 0 leq sin (arccos (frac{x}{+sqrt{1-x^2}})) leq 1 So 0 leq sin (arccos (tan (arcsin x))) leq 1 Then we have sin u = sin (arccos (tan (arcsin x))) We know that tan (arcsin x) = frac{x}{+sqrt{1-x^2}}, so cos u = frac{x}{+sqrt{1-x^2}} This is possible because sin^2 u + cos^2 u = 1. We have sin^2 u + left(frac{x}{sqrt{1-x^2}}right)^2 = 1 We know that 0 leq sin u leq 1, so we must have sin u = sqrt{frac{1-2x^2}{1-x^2}} Now we have sin (arccos (tan (arcsin x))) = sin u = sqrt{frac{1-2x^2}{1-x^2}} We want to find the number of solutions to sin (arccos (tan (arcsin x))) = x. That's equivalent to finding the number of solutions to sqrt{frac{1-2x^2}{1-x^2}} = x Squaring both sides, we have frac{1-2x^2}{1-x^2} = x^2 implies 1-2x^2 = (1-x^2)x^2 implies 1-2x^2 = x^2-x^4 That is equivalent to x^4-3x^2+1=0 So we want to find the number of solutions to x^4-3x^2+1=0 That is equivalent to finding the number of solutions to (x^2)^2-3(x^2)+1=0 We can use the quadratic formula to solve for x^2 That means x^2 = frac{3 + sqrt{5}}{2} or x^2 = frac{3 - sqrt{5}}{2} But we must also check that they are in the range of [0,1] We know that 0 le x le 1, so we must have 0 le x^2 le 1. Therefore, we must have x^2 = frac{3 - sqrt{5}}{2} So we have x = sqrt{frac{3 - sqrt{5}}{2}}. There is only one positive solution to sin (arccos (tan (arcsin x))) = x, which is x = sqrt{frac{3 - sqrt{5}}{2}}.

answer:After (1,1,1) is rotated 180^circ about the y-axis, it goes to (-1,1,-1). After (-1,1,-1) is reflected through the yz-plane, it goes to (1,1,-1). After (1,1,-1) is reflected through the xz-plane, it goes to (1,-1,-1). After (1,-1,-1) is rotated 180^circ about the y-axis, it goes to (-1,-1,1). Finally, after (-1,-1,1) is reflected through the xz-plane, it goes to (-1,1,1). [asy] import three; size(250); currentprojection = perspective(6,3,2); triple I = (1,0,0), J = (0,1,0), K = (0,0,1), O = (0,0,0); triple P = (1,1,1), Q = (-1,1,-1), R = (1,1,-1), S = (1,-1,-1), T = (-1,-1,1), U = (-1,1,1); draw(O--2*I, Arrow3(6)); draw((-2)*J--2*J, Arrow3(6)); draw(O--2*K, Arrow3(6)); draw(O--P); draw(O--Q); draw(O--R); draw(O--S); draw(O--T); draw(O--U); draw(P--Q--R--S--T--U,dashed); label("x", 2.2*I); label("y", 2.2*J); label("z", 2.2*K); dot("(1,1,1)", P, N); dot("(-1,1,-1)", Q, SE); dot("(1,1,-1)", R, dir(270)); dot("(1,-1,-1)", S, W); dot("(-1,-1,1)", T, NW); dot("(-1,1,1)", U, NE); [/asy]

answer:Let O be the origin, and let A, B, C, D be points in space so that overrightarrow{OA} = mathbf{a}, overrightarrow{OB} = mathbf{b}, overrightarrow{OC} = mathbf{c}, and overrightarrow{OD} = mathbf{d}. [asy] import three; size(180); currentprojection = perspective(6,3,2); triple A, B, C, D, O; A = (-1/sqrt(55),0,3*sqrt(6)/sqrt(55)); B = (sqrt(5/11), -sqrt(6/11), 0); C = (sqrt(5/11), sqrt(6/11), 0); D = (-1/sqrt(55),0,-3*sqrt(6)/sqrt(55)); O = (0,0,0); draw(O--A,Arrow3(6)); draw(O--B,Arrow3(6)); draw(O--C,Arrow3(6)); draw(O--D,Arrow3(6)); draw(A--B--D--C--cycle,dashed); draw(B--C,dashed); label("A", A, N); label("B", B, W); label("C", C, SE); label("D", D, S); label("O", O, NW); label("mathbf{a}", A/2, W); label("mathbf{b}", B/2, N); label("mathbf{c}", C/2, NE); label("mathbf{d}", D/2, W); [/asy] Note that cos angle AOB = -frac{1}{11}, so by the Law of Cosines on triangle AOB, [AB = sqrt{1 + 1 - 2(1)(1) left( -frac{1}{11} right)} = sqrt{frac{24}{11}} = 2 sqrt{frac{6}{11}}.]Similarly, AC = BC = BD = CD = 2 sqrt{frac{6}{11}}. Let M be the midpoint of overline{BC}. Since triangle ABC is equilateral with side length 2 sqrt{frac{6}{11}}, BM = CM = sqrt{frac{6}{11}}, and AM = sqrt{3} cdot sqrt{frac{6}{11}} = sqrt{frac{18}{11}}. [asy] import three; size(180); currentprojection = perspective(6,3,2); triple A, B, C, D, M, O; A = (-1/sqrt(55),0,3*sqrt(6)/sqrt(55)); B = (sqrt(5/11), -sqrt(6/11), 0); C = (sqrt(5/11), sqrt(6/11), 0); D = (-1/sqrt(55),0,-3*sqrt(6)/sqrt(55)); O = (0,0,0); M = (B + C)/2; draw(O--A,dashed); draw(O--B,dashed); draw(O--C,dashed); draw(O--D,dashed); draw(A--B--D--C--cycle); draw(B--C); draw(A--M); draw(M--O,dashed); label("A", A, N); label("B", B, W); label("C", C, SE); label("D", D, S); label("M", M, S); label("O", O, NW); [/asy] Then by Pythagoras on right triangle BMO, [MO = sqrt{BO^2 - BM^2} = sqrt{1 - frac{6}{11}} = sqrt{frac{5}{11}}.]By the Law of Cosines on triangle AMO, [cos angle AOM = frac{AO^2 + MO^2 - AM^2}{2 cdot AO cdot MO} = frac{1 + frac{5}{11} - frac{18}{11}}{2 cdot 1 cdot sqrt{frac{5}{11}}} = -frac{1}{sqrt{55}}.]Then begin{align*} mathbf{a} cdot mathbf{d} &= cos angle AOD &= cos (2 angle AOM) &= 2 cos^2 angle AOM - 1 &= 2 left( -frac{1}{sqrt{55}} right)^2 - 1 &= -frac{53}{55}. end{align*}

answer:Note that begin{align*}(sin t+icos t)^n &=left[cosleft({{pi}over2}-tright) +isinleft({{pi}over2}-tright)right]^n &=cos nleft({{pi}over2}-tright)+ isin nleft({{pi}over2}-tright) &=cosleft({{npi}over2}-ntright)+ isinleft({{npi}over2}-ntright),end{align*}and that displaystyle sin nt+icos nt =cosleft({{pi}over2}-ntright) +isinleft({{pi}over2}-ntright). Thus the given condition is equivalent to cosleft({{npi}over2}-ntright) = cosleft({{pi}over2}-ntright) quad {rm and} quad sinleft({{npi}over2}-ntright) = sinleft({{pi}over2}-ntright).In general, cosalpha=cosbeta and sinalpha=sinbeta if and only if alpha -beta=2pi k. Thus {{npi}over2}-nt-{{pi}over2}+nt=2pi k,which yields n=4k+1. Because 1le nle1000, conclude that 0le kle 249, so there are 250 values of n that satisfy the given conditions.