answer：Alright, so I have this problem where I need to find the value of (frac{p(3)}{q(3)}) given the graph of (y = frac{p(x)}{q(x)}). Both (p(x)) and (q(x)) are quadratic polynomials. The graph shows a rational function with a vertical asymptote at (x = -3) and a horizontal asymptote at (y = 3). There's also a hole at (x = 5), which means that both the numerator and the denominator have a common factor of ((x - 5)).First, let me recall what asymptotes tell me about a rational function. A vertical asymptote occurs where the denominator is zero but the numerator isn't, so in this case, (q(x)) must have a factor of ((x + 3)) because the vertical asymptote is at (x = -3). For the horizontal asymptote, since both (p(x)) and (q(x)) are quadratic, the degrees are the same. The horizontal asymptote is the ratio of the leading coefficients. Here, it's given as (y = 3), so the leading coefficient of (p(x)) must be three times that of (q(x)).Now, considering the hole at (x = 5), both (p(x)) and (q(x)) must have a factor of ((x - 5)). So, I can express (p(x)) and (q(x)) as:[p(x) = k(x - 5)(x - a)][q(x) = (x - 5)(x + 3)]Here, (k) is a constant that I need to determine, and (a) is another root of the numerator. Looking at the graph, I notice that it passes through the point ((2, 0)). This means that when (x = 2), (y = 0). Plugging this into the function:[0 = frac{p(2)}{q(2)} = frac{k(2 - 5)(2 - a)}{(2 - 5)(2 + 3)}]Simplifying, the ((2 - 5)) terms cancel out, leaving:[0 = frac{k(2 - a)}{5}]Since the denominator isn't zero, the numerator must be zero. Therefore, (2 - a = 0), which implies (a = 2). So, the numerator becomes:[p(x) = k(x - 5)(x - 2)]Now, the function simplifies to:[frac{p(x)}{q(x)} = frac{k(x - 5)(x - 2)}{(x - 5)(x + 3)} = frac{k(x - 2)}{x + 3}]This is valid for (x neq 5), which accounts for the hole at (x = 5).Next, I need to determine the value of (k). I know that the horizontal asymptote is (y = 3). For a rational function where the degrees of the numerator and denominator are equal, the horizontal asymptote is the ratio of the leading coefficients. Expanding the numerator and denominator:Numerator: (k(x - 2) = kx - 2k)Denominator: (x + 3)The leading coefficients are (k) (from the numerator) and (1) (from the denominator). Therefore, the horizontal asymptote is (y = frac{k}{1} = k). Since the horizontal asymptote is (y = 3), it follows that (k = 3).So, the function is:[frac{p(x)}{q(x)} = frac{3(x - 2)}{x + 3}]Now, I need to find (frac{p(3)}{q(3)}). Let's compute both (p(3)) and (q(3)):First, (p(3)):[p(3) = 3(3 - 2) = 3(1) = 3]Next, (q(3)):[q(3) = (3 + 3) = 6]Therefore, (frac{p(3)}{q(3)} = frac{3}{6} = frac{1}{2}).Wait, let me double-check this. If I plug (x = 3) into the simplified function:[frac{p(3)}{q(3)} = frac{3(3 - 2)}{3 + 3} = frac{3(1)}{6} = frac{3}{6} = frac{1}{2}]Yes, that seems correct. So, the value is (frac{1}{2}).I think I've covered all the steps: identifying the factors from asymptotes and the hole, using the point on the graph to find the remaining root, determining the constant (k) from the horizontal asymptote, and then computing the required value. It all seems consistent.**Final Answer**boxed{dfrac{1}{2}}

answer：Alright, so I have this problem here about triangle ABC. It says that two side lengths are 2 and 3, and the cosine of the included angle is 1/3. I need to find the radius of the circumscribed circle. Hmm, okay, let me think about how to approach this.First, I remember that in a triangle, if I know two sides and the included angle, I can find the third side using the Law of Cosines. The Law of Cosines formula is ( c^2 = a^2 + b^2 - 2abcos(C) ), where C is the included angle between sides a and b. So in this case, I have two sides, let's say 2 and 3, and the included angle has a cosine of 1/3. I need to find the third side.Let me denote the sides as follows: let’s say side a is 2, side b is 3, and the included angle is C, with ( cos(C) = frac{1}{3} ). So, applying the Law of Cosines, the third side, which I'll call side c, should be:( c^2 = 2^2 + 3^2 - 2 times 2 times 3 times frac{1}{3} )Calculating that step by step:First, ( 2^2 = 4 ) and ( 3^2 = 9 ). So, adding those together gives 4 + 9 = 13.Next, the last term is ( 2 times 2 times 3 times frac{1}{3} ). Let me compute that: 2 times 2 is 4, times 3 is 12, times 1/3 is 4. So, subtracting that from 13 gives 13 - 4 = 9.Therefore, ( c^2 = 9 ), so c is the square root of 9, which is 3. Okay, so the third side is 3 units long.Now, I need to find the radius of the circumscribed circle. I recall that the formula for the radius R of the circumscribed circle of a triangle is given by ( R = frac{a}{2sin(A)} ), where a is a side and A is the angle opposite that side. Alternatively, it can also be written using the Law of Sines as ( frac{a}{sin(A)} = 2R ).But in this case, I don't know the sine of the included angle directly. I only know its cosine. So, I need to find ( sin(C) ) first. Since ( cos(C) = frac{1}{3} ), I can use the Pythagorean identity ( sin^2(C) + cos^2(C) = 1 ) to find ( sin(C) ).Let me compute ( sin(C) ):( sin^2(C) = 1 - cos^2(C) = 1 - left(frac{1}{3}right)^2 = 1 - frac{1}{9} = frac{8}{9} )So, ( sin(C) = sqrt{frac{8}{9}} = frac{2sqrt{2}}{3} ). Since angles in a triangle are between 0 and 180 degrees, sine is positive, so we don't need to worry about the negative root.Now, I can use the Law of Sines to find the radius R. The Law of Sines states that ( frac{a}{sin(A)} = frac{b}{sin(B)} = frac{c}{sin(C)} = 2R ). Since I have side c and angle C, I can use that pair to find R.So, plugging in the values I have:( frac{c}{sin(C)} = 2R )( frac{3}{frac{2sqrt{2}}{3}} = 2R )Let me compute the left side:( frac{3}{frac{2sqrt{2}}{3}} = 3 times frac{3}{2sqrt{2}} = frac{9}{2sqrt{2}} )So, ( frac{9}{2sqrt{2}} = 2R )To solve for R, divide both sides by 2:( R = frac{9}{4sqrt{2}} )But it's often preferred to rationalize the denominator. So, multiplying numerator and denominator by ( sqrt{2} ):( R = frac{9sqrt{2}}{4 times 2} = frac{9sqrt{2}}{8} )So, the radius of the circumscribed circle is ( frac{9sqrt{2}}{8} ).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the Law of Cosines, I correctly calculated the third side as 3. Then, using the cosine value, I found the sine correctly as ( frac{2sqrt{2}}{3} ). Then, applying the Law of Sines, I set up the equation correctly and solved for R, rationalizing the denominator appropriately. It all seems to check out.I think that's the correct answer. The radius of the circumscribed circle is ( frac{9sqrt{2}}{8} ).**Final Answer**The radius of the circumscribed circle is boxed{dfrac{9sqrt{2}}{8}}.

answer：Okay, so I have this problem here that I need to prove. It's about a sum involving terms like 1 times (a-1), 2 times (a-2), all the way up to n times (a-n), and it's supposed to equal this fraction: n times (n+1) times (3a - 2n - 1) all over 6. And then, specifically when a is equal to n+1, the sum simplifies to n times (n+1) times (n+2) all over 6. Hmm, that seems a bit complicated, but let me try to break it down.First, I think I should understand what the sum actually looks like. So, for example, if n is 3 and a is, say, 5, then the sum would be 1*(5-1) + 2*(5-2) + 3*(5-3), which is 1*4 + 2*3 + 3*2 = 4 + 6 + 6 = 16. Let me check if that matches the formula. Plugging into the formula: n=3, a=5, so it's 3*4*(3*5 - 2*3 -1)/6. That's 12*(15 -6 -1)/6 = 12*8/6 = 16. Okay, that works. Good, so the formula seems correct for n=3, a=5.Now, I need to prove this formula in general. I think mathematical induction might be a good approach here. Induction is often used for proving statements that involve integers like n. So, let me recall how induction works. There are two main steps: the base case and the inductive step.**Base Case:** I need to show that the formula holds for the smallest value of n, which is usually n=1. Let's test n=1. The sum would be just 1*(a-1). According to the formula, it should be 1*2*(3a - 2*1 -1)/6. Simplifying that, it's 2*(3a - 2 -1)/6 = 2*(3a -3)/6 = (6a -6)/6 = a -1. Which matches the sum. So, the base case holds.**Inductive Step:** Assume that the formula holds for some arbitrary positive integer k. That is, assume that:1*(a-1) + 2*(a-2) + ... + k*(a-k) = k*(k+1)*(3a - 2k -1)/6.Now, I need to show that the formula also holds for k+1. That is, I need to show that:1*(a-1) + 2*(a-2) + ... + k*(a-k) + (k+1)*(a - (k+1)) = (k+1)*(k+2)*(3a - 2(k+1) -1)/6.Starting from the left-hand side, using the induction hypothesis:k*(k+1)*(3a - 2k -1)/6 + (k+1)*(a - k -1).I can factor out (k+1) from both terms:(k+1)*[k*(3a - 2k -1)/6 + (a - k -1)].Now, let's simplify the expression inside the brackets:First, let's write everything over a common denominator, which is 6:k*(3a - 2k -1)/6 + (a - k -1)*6/6.So, combining these:[k*(3a - 2k -1) + 6*(a - k -1)] / 6.Expanding both terms in the numerator:3a*k - 2k^2 - k + 6a - 6k -6.Now, let's combine like terms:3a*k + 6a - 2k^2 - k -6k -6.Combine the k terms:3a*k + 6a - 2k^2 -7k -6.Now, factor out common terms if possible. I notice that 3a*k + 6a can be factored as 3a(k + 2). Similarly, -2k^2 -7k -6 can be factored, but it's a quadratic in k. Let me see:-2k^2 -7k -6 = -(2k^2 +7k +6). Let's try to factor 2k^2 +7k +6.Looking for two numbers that multiply to 12 (2*6) and add up to 7. Those numbers are 3 and 4. So, 2k^2 +7k +6 = 2k^2 +3k +4k +6 = k(2k +3) +2(2k +3) = (k +2)(2k +3).So, -2k^2 -7k -6 = -(k +2)(2k +3).Putting it all together:3a(k +2) - (k +2)(2k +3).Factor out (k +2):(k +2)(3a - (2k +3)).So, the entire expression inside the brackets becomes:(k +2)(3a -2k -3)/6.Therefore, the entire left-hand side is:(k+1)*(k +2)*(3a -2k -3)/6.But notice that 3a -2k -3 can be written as 3a -2(k +1) -1, because 2(k +1) is 2k +2, so 3a -2k -2 -1 = 3a -2k -3.So, we have:(k+1)*(k +2)*(3a -2(k +1) -1)/6.Which is exactly the formula we wanted to prove for n = k +1.Therefore, by mathematical induction, the formula holds for all positive integers n.Now, for the specific case when a = n +1, let's substitute a with n +1 in the general formula:n*(n +1)*(3(n +1) -2n -1)/6.Simplify the numerator inside the parentheses:3n +3 -2n -1 = n +2.So, the formula becomes:n*(n +1)*(n +2)/6.Which is the desired result.I think that covers the proof. I started with the base case, assumed it holds for k, then showed it holds for k +1, and finally substituted a specific value to get the second identity. It all seems to fit together logically. I didn't encounter any contradictions or errors in my steps, so I feel confident about this proof.**Final Answer**The identity is proven, and the specific case when ( a = n+1 ) is:[boxed{frac{n cdot (n+1) cdot (n+2)}{6}}]

answer：First, I need to compute the expression (frac{tan^2 30^circ - cos^2 30^circ}{tan^2 30^circ cos^2 30^circ}).I know that (tan 30^circ = frac{sqrt{3}}{3}) and (cos 30^circ = frac{sqrt{3}}{2}).Calculating (tan^2 30^circ):[tan^2 30^circ = left(frac{sqrt{3}}{3}right)^2 = frac{3}{9} = frac{1}{3}]Calculating (cos^2 30^circ):[cos^2 30^circ = left(frac{sqrt{3}}{2}right)^2 = frac{3}{4}]Now, substitute these values into the expression:[frac{frac{1}{3} - frac{3}{4}}{frac{1}{3} times frac{3}{4}} = frac{frac{4}{12} - frac{9}{12}}{frac{3}{12}} = frac{-frac{5}{12}}{frac{3}{12}} = -frac{5}{3}]Therefore, the final answer is (-frac{5}{3}).

answer：Okay, so I have this geometry problem here. Let me try to understand it step by step. We have a circle with points A, B, C, and M. M is the midpoint of the minor arc AC. That means if I draw the circle and mark points A and C, M is exactly halfway along the smaller arc connecting them. Then, B is any point on the arc AM. So, B can be anywhere from A to M along that arc. Next, we're told to construct MD perpendicular to BC at point D. So, from point M, we draw a line MD that is perpendicular to BC, and they intersect at point D. The goal is to prove that BD equals half of (BC minus AB). In mathematical terms, BD = 1/2 (BC - AB).Alright, let me visualize this. I imagine a circle with points A and C on the circumference. M is the midpoint of the minor arc AC, so it's directly opposite the midpoint of the chord AC if we consider the circle. Then, B is somewhere on the arc from A to M. Drawing BC, which is a chord from B to C, and then from M, drawing MD perpendicular to BC, hitting it at D.I need to find a relationship between BD, BC, and AB. The equation given is BD = 1/2 (BC - AB). So, BD is somehow related to the difference between BC and AB, scaled by half.Let me think about the properties of circles and arcs. Since M is the midpoint of arc AC, it should be equidistant from A and C. That might mean that MA = MC. Also, because M is the midpoint, the angles subtended by arcs AM and MC should be equal. Since B is on arc AM, the position of B affects the lengths of AB and BC. Maybe I can use some triangle properties or congruence here. MD is perpendicular to BC, so triangle MDC is a right triangle. Similarly, if I can find another right triangle involving M, maybe I can establish some congruency.Wait, the problem mentions constructing MD perpendicular to BC. Maybe I should also construct a perpendicular from M to AB? Let's say ME perpendicular to AB at point E. Then, I have two right triangles: MAE and MCD. If I can show these triangles are congruent, that might help.Since M is the midpoint of arc AC, MA = MC. Also, ME and MD are both perpendiculars from M to AB and BC respectively. If ME = MD, then by the hypotenuse-leg theorem, triangles MAE and MCD would be congruent. That would imply AE = CD.But how does that help me with BD? Hmm. Maybe I need to look at another pair of triangles. What about triangles MBE and MBD? If I can show those are congruent, then BE = BD.Wait, if I consider triangle MBE and MBD, they are both right triangles. If MB is common to both, and if ME = MD, then by hypotenuse-leg, they would be congruent. That would mean BE = BD.But how does that tie into the lengths of BC and AB? Let me think. If AE = CD, and E is on AB, then AE is part of AB. So, AE = AB - BE. But earlier, I thought AE = CD. So, CD = AB - BE.But CD is also part of BC. Since BC = BD + DC, substituting CD gives BC = BD + (AB - BE). But since BE = BD, this becomes BC = BD + AB - BD, which simplifies to BC = AB. But that can't be right because BC and AB are different lengths unless B is at a specific point.Hmm, maybe I made a mistake in my reasoning. Let me go back.I have triangles MAE and MCD congruent, so AE = CD. Also, triangles MBE and MBD congruent, so BE = BD. So, AE = CD and BE = BD.Looking at AB, which is AE + EB. So, AB = AE + EB. But AE = CD and EB = BD, so AB = CD + BD.But BC is BD + DC. Since DC = CD, BC = BD + CD. But from AB = CD + BD, that means BC = BD + (AB - BD). Wait, that would mean BC = AB. That can't be right because BC and AB are different unless B is at a specific point.I must have messed up somewhere. Maybe my assumption that ME = MD is incorrect. ME and MD are both perpendiculars from M to AB and BC, but unless AB and BC are equidistant from M, ME might not equal MD.Wait, but since M is the midpoint of arc AC, maybe the distances from M to AB and BC are equal. Is that true? Let me think.In a circle, the distance from a point on the circumference to a chord is related to the angle subtended by the chord. Since M is the midpoint of arc AC, the angles from M to AB and BC might be equal, making the distances equal. So, ME = MD.If that's the case, then triangles MAE and MCD are congruent, and triangles MBE and MBD are congruent. So, AE = CD and BE = BD.Then, AB = AE + EB = CD + BD.But BC = BD + DC = BD + CD.So, BC = BD + CD, and AB = CD + BD. So, AB = BC? That can't be unless AB = BC, which isn't necessarily true.Wait, maybe I need to consider the lengths differently. Let me denote BD as x. Then, BE is also x because triangles MBE and MBD are congruent. So, EB = x.Then, AB = AE + EB = AE + x.But AE = CD, so CD = AB - x.Also, BC = BD + DC = x + DC.But DC = CD, so DC = AB - x.Therefore, BC = x + (AB - x) = AB. So, BC = AB. But that's not possible unless B is equidistant from A and C, which isn't necessarily the case.Hmm, I'm clearly missing something here. Maybe I need to approach this differently.Let me consider the power of a point or some other circle theorems. Since MD is perpendicular to BC, D is the foot of the perpendicular from M to BC. Similarly, if I draw ME perpendicular to AB, E is the foot from M to AB.Since M is the midpoint of arc AC, MA = MC, and angles at M are equal. Maybe I can use some angle chasing here.Looking at angles, since M is the midpoint of arc AC, angle AMC is equal to angle AME. Wait, not sure.Alternatively, maybe I can use coordinates. Let me assign coordinates to the points to make it easier.Let me place the circle with center at the origin (0,0) for simplicity. Let’s assume the circle has radius 1. Let me assign coordinates:Let’s set point A at (1,0). Since M is the midpoint of arc AC, let me place point C at (-1,0). Therefore, M, being the midpoint of the minor arc AC, would be at (0,1).Now, point B is any point on arc AM. Since arc AM is from (1,0) to (0,1), let me parameterize point B. Let’s say point B is at (cos θ, sin θ), where θ is between 0 and 90 degrees (π/2 radians).Now, we need to construct MD perpendicular to BC at D. So, first, let's find the equation of line BC.Point B is (cos θ, sin θ), and point C is (-1,0). The slope of BC is (0 - sin θ)/(-1 - cos θ) = (-sin θ)/(-1 - cos θ) = sin θ / (1 + cos θ).Therefore, the slope of BC is sin θ / (1 + cos θ). The equation of BC is:y - 0 = [sin θ / (1 + cos θ)](x + 1)So, y = [sin θ / (1 + cos θ)](x + 1)Now, MD is perpendicular to BC, so its slope is the negative reciprocal of the slope of BC. Therefore, the slope of MD is -(1 + cos θ)/sin θ.Point M is at (0,1). So, the equation of MD is:y - 1 = [-(1 + cos θ)/sin θ](x - 0)So, y = [-(1 + cos θ)/sin θ]x + 1Now, we need to find point D, which is the intersection of BC and MD.So, set the two equations equal:[sin θ / (1 + cos θ)](x + 1) = [-(1 + cos θ)/sin θ]x + 1Let me solve for x.Multiply both sides by (1 + cos θ) sin θ to eliminate denominators:sin² θ (x + 1) = -(1 + cos θ)² x + (1 + cos θ) sin θExpand both sides:sin² θ x + sin² θ = -(1 + 2 cos θ + cos² θ) x + (1 + cos θ) sin θBring all terms to the left side:sin² θ x + sin² θ + (1 + 2 cos θ + cos² θ) x - (1 + cos θ) sin θ = 0Combine like terms:[sin² θ + 1 + 2 cos θ + cos² θ] x + [sin² θ - (1 + cos θ) sin θ] = 0Note that sin² θ + cos² θ = 1, so:[1 + 1 + 2 cos θ] x + [sin² θ - sin θ - sin θ cos θ] = 0Simplify:(2 + 2 cos θ) x + [sin² θ - sin θ - sin θ cos θ] = 0Factor out 2(1 + cos θ) x:2(1 + cos θ) x + [sin² θ - sin θ(1 + cos θ)] = 0Now, let's factor sin θ from the second term:2(1 + cos θ) x + sin θ (sin θ - (1 + cos θ)) = 0Let me compute sin θ - (1 + cos θ):sin θ - 1 - cos θHmm, not sure if that helps. Maybe express sin² θ as 1 - cos² θ:So, sin² θ = 1 - cos² θTherefore, the equation becomes:2(1 + cos θ) x + (1 - cos² θ - sin θ - sin θ cos θ) = 0Hmm, this is getting complicated. Maybe there's a better way.Alternatively, let's use vector methods or parametric equations. But maybe it's too involved.Wait, perhaps instead of coordinates, I can use properties of circles and triangles.Since M is the midpoint of arc AC, it's the center of the circle through A and C. Wait, no, M is just a point on the circumference.Alternatively, since MD is perpendicular to BC, and ME is perpendicular to AB, and MA = MC, maybe triangles MDC and MEA are congruent.Wait, earlier I thought that, but that led to BC = AB, which isn't necessarily true. Maybe I need to adjust that.Alternatively, perhaps using the fact that in circle, the perpendicular from M to BC meets BC at D, and similarly from M to AB meets at E, and then using some properties of midpoints or similar triangles.Wait, another approach: Since M is the midpoint of arc AC, the angle bisector of angle ABC passes through M. Is that true? Or maybe the angle at M relates to the angles at B.Alternatively, maybe using power of a point. The power of point D with respect to the circle is DB * DC = DM^2 - r^2, but not sure.Wait, perhaps using coordinate geometry was a good idea, but maybe I should proceed step by step.So, earlier, I had:2(1 + cos θ) x + sin θ (sin θ - (1 + cos θ)) = 0Let me compute sin θ - (1 + cos θ):sin θ - 1 - cos θLet me factor out a negative sign:- (1 + cos θ - sin θ)So, the equation becomes:2(1 + cos θ) x - sin θ (1 + cos θ - sin θ) = 0Factor out (1 + cos θ):(1 + cos θ)(2x - sin θ) + sin² θ = 0Wait, not sure. Alternatively, let's solve for x:2(1 + cos θ) x = sin θ (1 + cos θ - sin θ)Therefore,x = [sin θ (1 + cos θ - sin θ)] / [2(1 + cos θ)]Simplify numerator:sin θ (1 + cos θ) - sin² θSo,x = [sin θ (1 + cos θ) - sin² θ] / [2(1 + cos θ)]Split the fraction:x = [sin θ (1 + cos θ)] / [2(1 + cos θ)] - [sin² θ] / [2(1 + cos θ)]Simplify first term:x = sin θ / 2 - [sin² θ] / [2(1 + cos θ)]Now, for the second term, use the identity sin² θ = 1 - cos² θ:x = sin θ / 2 - [1 - cos² θ] / [2(1 + cos θ)]Factor numerator of second term:1 - cos² θ = (1 - cos θ)(1 + cos θ)So,x = sin θ / 2 - [(1 - cos θ)(1 + cos θ)] / [2(1 + cos θ)]Cancel (1 + cos θ):x = sin θ / 2 - (1 - cos θ)/2Combine terms:x = [sin θ - (1 - cos θ)] / 2So,x = (sin θ - 1 + cos θ) / 2Now, let's find y coordinate of D using the equation of MD:y = [-(1 + cos θ)/sin θ]x + 1Substitute x:y = [-(1 + cos θ)/sin θ] * [(sin θ - 1 + cos θ)/2] + 1Simplify:y = [-(1 + cos θ)(sin θ - 1 + cos θ)] / (2 sin θ) + 1Let me expand the numerator:(1 + cos θ)(sin θ - 1 + cos θ) = (1)(sin θ - 1 + cos θ) + cos θ (sin θ - 1 + cos θ)= sin θ - 1 + cos θ + sin θ cos θ - cos θ + cos² θSimplify:sin θ - 1 + cos θ + sin θ cos θ - cos θ + cos² θ= sin θ - 1 + sin θ cos θ + cos² θSo, numerator is -(sin θ - 1 + sin θ cos θ + cos² θ)Therefore,y = [ - (sin θ - 1 + sin θ cos θ + cos² θ) ] / (2 sin θ) + 1= [ - sin θ + 1 - sin θ cos θ - cos² θ ] / (2 sin θ) + 1Split the fraction:= [ - sin θ / (2 sin θ) ] + [1 / (2 sin θ)] + [ - sin θ cos θ / (2 sin θ) ] + [ - cos² θ / (2 sin θ) ] + 1Simplify each term:= -1/2 + 1/(2 sin θ) - cos θ / 2 - cos² θ / (2 sin θ) + 1Combine constants:-1/2 + 1 = 1/2So,= 1/2 + 1/(2 sin θ) - cos θ / 2 - cos² θ / (2 sin θ)Combine the terms with 1/(2 sin θ):= 1/2 - cos θ / 2 + [1 - cos² θ] / (2 sin θ)Note that 1 - cos² θ = sin² θ, so:= 1/2 - cos θ / 2 + sin² θ / (2 sin θ)Simplify sin² θ / sin θ = sin θ:= 1/2 - cos θ / 2 + sin θ / 2Combine terms:= (1 - cos θ + sin θ)/2So, point D has coordinates:x = (sin θ - 1 + cos θ)/2y = (1 - cos θ + sin θ)/2Now, let's find BD. Point B is (cos θ, sin θ), and point D is ((sin θ - 1 + cos θ)/2, (1 - cos θ + sin θ)/2)Compute the distance BD:BD = sqrt[ (x_D - x_B)^2 + (y_D - y_B)^2 ]Compute x_D - x_B:[(sin θ - 1 + cos θ)/2 - cos θ] = [sin θ - 1 + cos θ - 2 cos θ]/2 = [sin θ - 1 - cos θ]/2Similarly, y_D - y_B:[(1 - cos θ + sin θ)/2 - sin θ] = [1 - cos θ + sin θ - 2 sin θ]/2 = [1 - cos θ - sin θ]/2So, BD = sqrt[ ( (sin θ - 1 - cos θ)/2 )^2 + ( (1 - cos θ - sin θ)/2 )^2 ]Notice that both terms inside the sqrt are the same except for a negative sign in the first term. Let me compute:= sqrt[ ( (sin θ - 1 - cos θ)^2 + (1 - cos θ - sin θ)^2 ) / 4 ]But (sin θ - 1 - cos θ)^2 = ( - (1 + cos θ - sin θ) )^2 = (1 + cos θ - sin θ)^2Similarly, (1 - cos θ - sin θ)^2 is the same as (1 + cos θ - sin θ)^2 with cos θ and sin θ signs changed, but squared, so same value.Wait, actually, (1 - cos θ - sin θ)^2 = (1 - (cos θ + sin θ))^2 = 1 - 2(cos θ + sin θ) + (cos θ + sin θ)^2Similarly, (sin θ - 1 - cos θ)^2 = ( -1 + sin θ - cos θ)^2 = 1 - 2(sin θ - cos θ) + (sin θ - cos θ)^2But this might not be helpful. Alternatively, notice that both terms are (1 + cos θ - sin θ)^2 and (1 - cos θ - sin θ)^2.Wait, no, actually, let me compute:(sin θ - 1 - cos θ)^2 = ( -1 + sin θ - cos θ )^2 = 1 - 2 sin θ + 2 cos θ + sin² θ - 2 sin θ cos θ + cos² θSimilarly, (1 - cos θ - sin θ)^2 = 1 - 2 cos θ - 2 sin θ + cos² θ + 2 sin θ cos θ + sin² θAdding both:1 - 2 sin θ + 2 cos θ + sin² θ - 2 sin θ cos θ + cos² θ + 1 - 2 cos θ - 2 sin θ + cos² θ + 2 sin θ cos θ + sin² θCombine like terms:1 + 1 = 2-2 sin θ - 2 sin θ = -4 sin θ2 cos θ - 2 cos θ = 0sin² θ + sin² θ = 2 sin² θcos² θ + cos² θ = 2 cos² θ-2 sin θ cos θ + 2 sin θ cos θ = 0So, total:2 - 4 sin θ + 2 sin² θ + 2 cos² θFactor 2:2(1 - 2 sin θ + sin² θ + cos² θ)But sin² θ + cos² θ = 1, so:2(1 - 2 sin θ + 1) = 2(2 - 2 sin θ) = 4(1 - sin θ)Therefore, BD = sqrt[ 4(1 - sin θ) / 4 ] = sqrt(1 - sin θ)Wait, that seems off. Wait, let me check.Wait, the numerator inside the sqrt was [ (sin θ - 1 - cos θ)^2 + (1 - cos θ - sin θ)^2 ] = 2(1 - sin θ)Wait, no, earlier I had:After adding both squares, I got 2 - 4 sin θ + 2 sin² θ + 2 cos² θ, which simplifies to 2(1 - 2 sin θ + sin² θ + cos² θ) = 2(2 - 2 sin θ) = 4(1 - sin θ)So, BD = sqrt[4(1 - sin θ)/4] = sqrt(1 - sin θ)Wait, sqrt(4(1 - sin θ)/4) is sqrt(1 - sin θ). So, BD = sqrt(1 - sin θ)But I need to express BD in terms of BC and AB.Let me compute BC and AB.Point B is (cos θ, sin θ), point C is (-1,0).Distance BC:sqrt[ (cos θ + 1)^2 + (sin θ - 0)^2 ] = sqrt[ (cos θ + 1)^2 + sin² θ ]Expand:= sqrt[ cos² θ + 2 cos θ + 1 + sin² θ ] = sqrt[ (cos² θ + sin² θ) + 2 cos θ + 1 ] = sqrt[1 + 2 cos θ + 1] = sqrt[2 + 2 cos θ] = sqrt[2(1 + cos θ)] = 2 cos(θ/2)Similarly, distance AB:Point A is (1,0), point B is (cos θ, sin θ).Distance AB:sqrt[ (cos θ - 1)^2 + (sin θ - 0)^2 ] = sqrt[ (cos θ - 1)^2 + sin² θ ]Expand:= sqrt[ cos² θ - 2 cos θ + 1 + sin² θ ] = sqrt[ (cos² θ + sin² θ) - 2 cos θ + 1 ] = sqrt[1 - 2 cos θ + 1] = sqrt[2 - 2 cos θ] = 2 sin(θ/2)So, BC = 2 cos(θ/2), AB = 2 sin(θ/2)Therefore, BC - AB = 2 cos(θ/2) - 2 sin(θ/2) = 2( cos(θ/2) - sin(θ/2) )Then, 1/2 (BC - AB) = ( cos(θ/2) - sin(θ/2) )Now, BD = sqrt(1 - sin θ). Let me express sqrt(1 - sin θ) in terms of θ/2.Recall that 1 - sin θ = 1 - 2 sin(θ/2) cos(θ/2)But also, 1 - sin θ = [sin(θ/2) - cos(θ/2)]^2Because:[sin(θ/2) - cos(θ/2)]^2 = sin²(θ/2) - 2 sin(θ/2) cos(θ/2) + cos²(θ/2) = 1 - sin θTherefore, sqrt(1 - sin θ) = |sin(θ/2) - cos(θ/2)|Since θ is between 0 and π/2, θ/2 is between 0 and π/4, so cos(θ/2) > sin(θ/2), so sqrt(1 - sin θ) = cos(θ/2) - sin(θ/2)Therefore, BD = cos(θ/2) - sin(θ/2) = 1/2 (BC - AB)Hence, BD = 1/2 (BC - AB)So, the proof is complete.

answer：First, I need to understand the operation defined by ( a star b star c = (a + b + c)^2 ). This means that when I see the star operation applied to three variables, it represents the square of their sum.Next, I'll substitute the given values into the expression. So, replacing ( a ) with 3, ( b ) with -5, and ( c ) with 2, the expression becomes:[(3 + (-5) + 2)^2]Now, I'll perform the addition inside the parentheses:[3 - 5 + 2 = 0]Finally, I'll square the result:[0^2 = 0]Therefore, the value of ( a star b star c ) when ( a = 3 ), ( b = -5 ), and ( c = 2 ) is 0.