💡First, I need to determine the side lengths of each square by dividing their perimeters by 4.For Region A:[text{Side length of A} = frac{16}{4} = 4 text{ units}]For Region B:[text{Side length of B} = frac{32}{4} = 8 text{ units}]For Region C:[text{Side length of C} = frac{20}{4} = 5 text{ units}]Next, I'll calculate the areas of Region B and Region C by squaring their respective side lengths.Area of Region B:[8 times 8 = 64 text{ square units}]Area of Region C:[5 times 5 = 25 text{ square units}]Finally, I'll find the ratio of the area of Region B to the area of Region C:[frac{64}{25}]

💡Okay, so I have this geometry problem here about a rectangle ABCD. A point M is selected on the arc AB of the circumcircle, not being any of the vertices A or B. Then, we're supposed to project M onto the sides AD, AB, BC, and CD, getting points P, Q, R, and S respectively. The goal is to prove that lines PQ and RS are perpendicular using analytical methods.Alright, let's break this down. First, since it's a rectangle, all angles are right angles, and opposite sides are equal and parallel. The circumcircle of a rectangle would have its center at the intersection of the diagonals, right? So, the center is the midpoint of both diagonals AC and BD.Now, point M is on the arc AB. Since it's a rectangle, the circumcircle should pass through all four vertices, so arc AB is a quarter-circle, I think. But wait, no, in a rectangle, the circumcircle is actually a circle with diameter equal to the diagonal of the rectangle. So, the arc AB would be a semicircle if we consider the diameter as AB, but since it's a rectangle, the diagonal is longer than the sides.Wait, maybe I should think in terms of coordinates. Let me set up a coordinate system to make this easier. Let's place the rectangle ABCD such that point A is at (0,0), B is at (a,0), C is at (a,b), and D is at (0,b). Then, the circumcircle would have its center at the midpoint of the diagonal AC, which is at (a/2, b/2), and the radius would be half the length of the diagonal, which is (sqrt(a² + b²))/2.So, the equation of the circumcircle is (x - a/2)² + (y - b/2)² = (a² + b²)/4.Point M is on the arc AB, which is the arc from A to B not containing the other two vertices. So, M is somewhere on that arc. Let's parameterize point M. Maybe using an angle parameter θ. Let's say M has coordinates (a/2 + (a/2)cosθ, b/2 + (b/2)sinθ). Wait, is that correct?Actually, the parametric equations for a circle centered at (h,k) with radius r are x = h + r cosθ, y = k + r sinθ. So, in this case, h = a/2, k = b/2, and r = sqrt(a² + b²)/2. So, point M can be represented as:x = a/2 + (sqrt(a² + b²)/2) cosθy = b/2 + (sqrt(a² + b²)/2) sinθBut maybe it's simpler to use a different parameterization. Alternatively, since the rectangle is axis-aligned, perhaps we can use a parameter t such that M moves along the arc AB.But maybe I'm overcomplicating. Let's instead use coordinates for M as (x, y) lying on the circumcircle, so (x - a/2)² + (y - b/2)² = (a² + b²)/4.Now, projections of M onto the sides. Let's recall that projecting a point onto a line gives the closest point on that line to the given point. So, for each side, we need to find the projection of M onto that side.Let's define the sides:- AD: from A(0,0) to D(0,b). This is the line x=0.- AB: from A(0,0) to B(a,0). This is the line y=0.- BC: from B(a,0) to C(a,b). This is the line x=a.- CD: from C(a,b) to D(0,b). This is the line y=b.So, projecting M(x,y) onto AD (x=0): the projection P would be (0, y). Wait, no. The projection of a point onto a vertical line x=0 is indeed (0, y). Similarly, projecting onto AB (y=0) would be (x, 0). Projecting onto BC (x=a) would be (a, y). Projecting onto CD (y=b) would be (x, b).Wait, but in the problem statement, the projections are onto the lines AD, AB, BC, and CD. So, P is the projection onto AD, which is x=0, so P is (0, y). Q is the projection onto AB, which is y=0, so Q is (x, 0). R is the projection onto BC, which is x=a, so R is (a, y). S is the projection onto CD, which is y=b, so S is (x, b).Wait, but the problem says P, Q, R, S are projections onto AD, AB, BC, CD respectively. So, P is projection on AD: (0, y). Q is projection on AB: (x, 0). R is projection on BC: (a, y). S is projection on CD: (x, b).Wait, but that seems too straightforward. Let me confirm. For a vertical line x=0, the projection of (x,y) is indeed (0,y). For horizontal line y=0, projection is (x,0). For vertical line x=a, projection is (a,y). For horizontal line y=b, projection is (x,b). Yes, that's correct.So, we have:P = (0, y)Q = (x, 0)R = (a, y)S = (x, b)Now, we need to find the equations of lines PQ and RS and show that they are perpendicular.First, let's find the coordinates of P, Q, R, S.Given M(x,y), then:P = (0, y)Q = (x, 0)R = (a, y)S = (x, b)So, line PQ connects P(0,y) to Q(x,0). Let's find its slope.Slope of PQ: (0 - y)/(x - 0) = -y/xSimilarly, line RS connects R(a,y) to S(x,b). Let's find its slope.Slope of RS: (b - y)/(x - a)Now, to check if PQ and RS are perpendicular, the product of their slopes should be -1.So, let's compute:Slope_PQ * Slope_RS = (-y/x) * ((b - y)/(x - a)) = (-y(b - y))/(x(x - a))We need this product to be -1.So, (-y(b - y))/(x(x - a)) = -1Simplify:(-y(b - y))/(x(x - a)) = -1Multiply both sides by x(x - a):-y(b - y) = -x(x - a)Simplify signs:y(b - y) = x(x - a)So, y(b - y) = x² - a xNow, let's rearrange:y b - y² = x² - a xBring all terms to one side:x² - a x - y b + y² = 0Hmm, not sure if this is helpful yet. Maybe we can use the fact that M(x,y) lies on the circumcircle of the rectangle.The equation of the circumcircle is (x - a/2)² + (y - b/2)² = (a² + b²)/4Let's expand this:x² - a x + (a²)/4 + y² - b y + (b²)/4 = (a² + b²)/4Simplify:x² - a x + y² - b y + (a² + b²)/4 = (a² + b²)/4Subtract (a² + b²)/4 from both sides:x² - a x + y² - b y = 0So, x² - a x + y² - b y = 0Wait, that's interesting. Earlier, we had from the perpendicular condition:x² - a x - y b + y² = 0Which is the same as x² - a x + y² - b y = 0So, that's exactly the equation of the circumcircle.Therefore, the condition for PQ and RS being perpendicular is equivalent to M lying on the circumcircle, which it does by definition.Therefore, PQ and RS are perpendicular.Wait, so that's the proof? Let me recap.We found that for PQ and RS to be perpendicular, the condition reduces to x² - a x + y² - b y = 0, which is exactly the equation of the circumcircle that M lies on. Therefore, since M is on the circumcircle, the condition holds, hence PQ and RS are perpendicular.That seems to make sense. So, the key was realizing that the condition for perpendicularity leads directly to the equation of the circumcircle, which M satisfies by construction.I think that's a solid approach. Maybe I should write it out more formally.

💡Okay, so I have this question here asking which of the given options is a linear equation in one variable. The options are A, B, C, and D. Let me try to figure this out step by step.First, I need to remember what a linear equation in one variable is. From what I recall, a linear equation is an equation where the highest power of the variable is 1. That means there are no squared terms, cube terms, or any other exponents higher than 1. Also, it should only involve one variable. If there are two variables, then it's a linear equation in two variables, which is different.Let me look at each option one by one.Starting with option A: 2x - 1 = 0. Hmm, this looks straightforward. The variable here is x, and it's only raised to the first power. There are no other variables involved. So, this seems like a linear equation in one variable. That might be the answer, but let me check the others just to be sure.Moving on to option B: 1/x = 4. Okay, this one is a bit different. Here, x is in the denominator. I remember that equations with variables in the denominator are called rational equations or fractional equations. They aren't linear because of the reciprocal nature. So, this isn't a linear equation. Plus, it's still only one variable, but the form isn't linear. So, B is out.Next is option C: 4x² - 2x = 0. Alright, here I see an x squared term. That means the highest power of x is 2, which makes this a quadratic equation, not linear. Even though it's only one variable, the exponent is too high. So, C isn't the right answer either.Finally, option D: 5x - y = 8. This one has two variables, x and y. Even though both are to the first power, the presence of two variables means it's a linear equation in two variables, not just one. So, D doesn't fit the criteria either.Going back, option A is the only one that fits all the requirements: it's a linear equation with only one variable. The others either have variables in the denominator, higher exponents, or multiple variables. So, I think A is the correct answer.Just to make sure I didn't miss anything, let me recap:- A: Linear, one variable. Correct.- B: Fractional equation, one variable. Not linear.- C: Quadratic equation, one variable. Not linear.- D: Linear, two variables. Not one variable.Yep, that seems right. So, I'm confident that A is the correct choice.

💡Okay, so I have this math problem here: [frac{3^{2015} + 3^{2013}}{3^{2015} - 3^{2013}}.] And I need to figure out its value. The options are A) 4/5, B) 5/4, C) 9/7, D) 8/10. Hmm, let's see how I can approach this.First, I notice that both the numerator and the denominator have terms with exponents of 3, specifically 3^{2015} and 3^{2013}. That seems like a big difference in exponents, but maybe there's a way to simplify this expression by factoring out a common term.Let me think. If I look at 3^{2015}, that's the same as 3^{2013 + 2}, which is 3^{2013} multiplied by 3^2. So, 3^{2015} = 3^{2013} * 9. That might be useful.So, let's rewrite both the numerator and the denominator using this idea.Starting with the numerator: 3^{2015} + 3^{2013} = 3^{2013} * 9 + 3^{2013}. I can factor out 3^{2013} from both terms, right? So that becomes 3^{2013} * (9 + 1) = 3^{2013} * 10.Similarly, the denominator: 3^{2015} - 3^{2013} = 3^{2013} * 9 - 3^{2013}. Again, factor out 3^{2013}, so that's 3^{2013} * (9 - 1) = 3^{2013} * 8.Now, putting it all together, the original fraction becomes:[frac{3^{2013} * 10}{3^{2013} * 8}.]Oh, wait a minute! The 3^{2013} terms are in both the numerator and the denominator, so they can cancel each other out. That simplifies things a lot.So now, the fraction is just [frac{10}{8}.] Hmm, 10 divided by 8. I can simplify that fraction by dividing both numerator and denominator by 2. So, 10 ÷ 2 = 5, and 8 ÷ 2 = 4. That gives me [frac{5}{4}.]Looking back at the options, B is 5/4. So, that must be the answer.Wait, let me double-check to make sure I didn't make a mistake. I factored out 3^{2013} correctly, right? Yes, because 3^{2015} is indeed 3^{2013} times 3 squared, which is 9. Then, adding and subtracting 3^{2013} gives me the 10 and 8 in the numerator and denominator. Then, canceling out the common term makes sense. Simplifying 10/8 to 5/4 is correct. Yeah, I think that's right.So, yeah, the answer should be B) 5/4.

💡Okay, so I have this problem where I need to prove that a rational function ( R(x) = frac{P(x)}{Q(x)} ), with ( P ) and ( Q ) being coprime polynomials, can be expressed as a polynomial ( A(x) ) plus a sum of fractions of the form ( frac{c_{ik}}{(x - a_i)^k} ). Hmm, I think this is related to partial fraction decomposition, which I remember from my algebra class. Let me try to recall how that works.First, partial fraction decomposition is a method used to break down a complex rational function into a sum of simpler fractions. The idea is that if the denominator can be factored into linear or irreducible quadratic factors, then the rational function can be expressed as a sum of fractions with these factors as denominators. In this case, since ( P ) and ( Q ) are coprime, ( Q(x) ) can be factored into linear terms, right? Or maybe some higher-degree terms if it's irreducible over the real numbers. But since we're dealing with complex numbers, I think all polynomials can be factored into linear terms.Wait, but the problem mentions ( (x - a_i)^k ), which suggests that the denominator might have repeated roots. So, if ( Q(x) ) has multiple roots, say ( a_i ) with multiplicity ( k ), then the partial fraction decomposition would include terms like ( frac{c_{i1}}{x - a_i} + frac{c_{i2}}{(x - a_i)^2} + dots + frac{c_{ik}}{(x - a_i)^k} ). That makes sense because each repeated root contributes multiple terms in the decomposition.But before I get into the partial fractions, the problem also mentions a polynomial ( A(x) ). I think that comes from polynomial long division. If the degree of ( P(x) ) is greater than or equal to the degree of ( Q(x) ), then we can divide ( P(x) ) by ( Q(x) ) to get a polynomial quotient ( A(x) ) and a remainder ( S(x) ), such that ( P(x) = Q(x) cdot A(x) + S(x) ), where the degree of ( S(x) ) is less than the degree of ( Q(x) ). So, ( R(x) = A(x) + frac{S(x)}{Q(x)} ). That seems right.So, the first step is to perform polynomial division to separate the polynomial part ( A(x) ) from the proper rational function ( frac{S(x)}{Q(x)} ). Then, the next step is to decompose ( frac{S(x)}{Q(x)} ) into partial fractions. Since ( P ) and ( Q ) are coprime, ( S(x) ) and ( Q(x) ) must also be coprime, right? Because if they had a common factor, then ( P(x) ) and ( Q(x) ) would have that common factor as well, which contradicts them being coprime.Now, I need to recall how partial fraction decomposition works for rational functions where the denominator has repeated linear factors. Let's say ( Q(x) ) factors as ( (x - a_1)^{k_1} (x - a_2)^{k_2} dots (x - a_n)^{k_n} ). Then, the partial fraction decomposition of ( frac{S(x)}{Q(x)} ) would be:[frac{S(x)}{Q(x)} = sum_{i=1}^{n} left( frac{c_{i1}}{x - a_i} + frac{c_{i2}}{(x - a_i)^2} + dots + frac{c_{ik_i}}{(x - a_i)^{k_i}} right)]So, combining this with the polynomial ( A(x) ), the entire rational function ( R(x) ) can be written as:[R(x) = A(x) + sum_{i=1}^{n} sum_{k=1}^{k_i} frac{c_{ik}}{(x - a_i)^k}]Which is exactly the form given in the problem. So, I think this is the right approach.But let me double-check if I missed anything. The problem states that ( P ) and ( Q ) are coprime, which is important because it ensures that the partial fraction decomposition doesn't have any common factors in the numerator and denominator, simplifying the decomposition process. Also, the fact that ( Q(x) ) can be factored into linear terms (assuming we're working over the complex numbers) allows us to express the decomposition in terms of ( (x - a_i)^k ) terms.Wait, what if ( Q(x) ) has irreducible quadratic factors? Then, the partial fraction decomposition would include terms with quadratic denominators, right? But the problem specifically mentions terms of the form ( frac{c_{ik}}{(x - a_i)^k} ), which are all linear denominators. So, does that mean we're assuming ( Q(x) ) factors completely into linear terms? Or is there something else?Hmm, maybe the problem is assuming that we're working over the complex numbers, where every polynomial factors into linear terms. So, in that case, the decomposition would indeed only involve linear denominators. If we were working over the real numbers, we might have irreducible quadratic factors, but since the problem doesn't specify, I think it's safe to assume we're working over the complex numbers, where everything factors into linear terms.Another thing to consider is whether the polynomial ( A(x) ) is unique. I think it is, because polynomial division gives a unique quotient and remainder. Similarly, the partial fraction decomposition is unique as well, given the factorization of ( Q(x) ). So, the entire expression ( A(x) + sum frac{c_{ik}}{(x - a_i)^k} ) should be unique.Let me try to outline the steps formally:1. Perform polynomial long division on ( P(x) ) and ( Q(x) ) to write ( R(x) = A(x) + frac{S(x)}{Q(x)} ), where ( deg(S) < deg(Q) ).2. Factor ( Q(x) ) into linear terms: ( Q(x) = prod_{i=1}^{n} (x - a_i)^{k_i} ).3. Express ( frac{S(x)}{Q(x)} ) as a sum of partial fractions:[frac{S(x)}{Q(x)} = sum_{i=1}^{n} sum_{k=1}^{k_i} frac{c_{ik}}{(x - a_i)^k}]4. Combine the results to get:[R(x) = A(x) + sum_{i=1}^{n} sum_{k=1}^{k_i} frac{c_{ik}}{(x - a_i)^k}]Which is the desired form.I think this covers all the necessary steps. I should also mention that the constants ( c_{ik} ) can be found by equating coefficients or using other methods like the residue theorem, but the problem doesn't ask for that specifically, just to prove the existence of such a representation.Wait, does the problem assume that ( Q(x) ) is factored completely? Or do I need to address the factorization as part of the proof? I think since the problem mentions ( (x - a_i)^k ), it implies that ( Q(x) ) has been factored into linear terms, possibly with multiplicities. So, I don't need to worry about the factorization process itself, just use it as given.Also, since ( P ) and ( Q ) are coprime, the partial fraction decomposition doesn't have any polynomial terms beyond ( A(x) ), which is already separated out. So, the decomposition is purely into fractions with linear denominators.I think I've covered all the bases here. The key steps are polynomial division to separate the polynomial part and then partial fraction decomposition on the remaining proper rational function. Since ( P ) and ( Q ) are coprime, the decomposition is straightforward without any common factors complicating things.Just to recap:- Polynomial division gives ( R(x) = A(x) + frac{S(x)}{Q(x)} ).- Factor ( Q(x) ) into linear terms.- Decompose ( frac{S(x)}{Q(x)} ) into partial fractions with terms ( frac{c_{ik}}{(x - a_i)^k} ).- Combine everything to get the desired form.Yes, that seems solid. I don't see any gaps in this reasoning. Maybe I should write it out more formally to make sure it's clear.Another point to consider: what if ( Q(x) ) has a root at infinity? Wait, no, in polynomial terms, the degree is finite, so all roots are finite. So, we don't have to worry about terms like ( frac{1}{x^k} ) unless ( x = 0 ) is a root, but that's already covered by ( (x - a_i) ) terms where ( a_i = 0 ).Also, the problem doesn't specify whether the field is real or complex, but as I thought earlier, assuming complex numbers allows us to factor ( Q(x) ) completely into linear terms. If we were restricted to real numbers, we might have quadratic terms, but the problem's form suggests linear denominators, so complex factorization is likely intended.In conclusion, the proof involves two main steps: polynomial division and partial fraction decomposition, leveraging the coprimality of ( P ) and ( Q ) to ensure the decomposition is possible without common factors.

💡Okay, so I have this problem here where I need to find the exact value of the square root of the product of three terms: (2 - sin²(π/7)), (2 - sin²(2π/7)), and (2 - sin²(3π/7)). That is,[sqrt{left( 2 - sin^2 frac{pi}{7} right) left( 2 - sin^2 frac{2 pi}{7} right) left( 2 - sin^2 frac{3 pi}{7} right)}]Hmm, that looks a bit intimidating at first glance, but maybe I can break it down step by step.First, I remember that sin²θ can be expressed in terms of cos(2θ) using the double-angle identity:[sin^2 theta = frac{1 - cos 2theta}{2}]So, maybe I can rewrite each term inside the square root using this identity. Let me try that.For the first term, 2 - sin²(π/7):[2 - sin^2 frac{pi}{7} = 2 - frac{1 - cos frac{2pi}{7}}{2} = 2 - frac{1}{2} + frac{cos frac{2pi}{7}}{2} = frac{3}{2} + frac{cos frac{2pi}{7}}{2}]Similarly, for the second term, 2 - sin²(2π/7):[2 - sin^2 frac{2pi}{7} = 2 - frac{1 - cos frac{4pi}{7}}{2} = 2 - frac{1}{2} + frac{cos frac{4pi}{7}}{2} = frac{3}{2} + frac{cos frac{4pi}{7}}{2}]And for the third term, 2 - sin²(3π/7):[2 - sin^2 frac{3pi}{7} = 2 - frac{1 - cos frac{6pi}{7}}{2} = 2 - frac{1}{2} + frac{cos frac{6pi}{7}}{2} = frac{3}{2} + frac{cos frac{6pi}{7}}{2}]So now, the expression inside the square root becomes:[left( frac{3}{2} + frac{cos frac{2pi}{7}}{2} right) left( frac{3}{2} + frac{cos frac{4pi}{7}}{2} right) left( frac{3}{2} + frac{cos frac{6pi}{7}}{2} right)]Hmm, that's a product of three terms, each involving cosines of multiples of π/7. Maybe I can factor out the 1/2 from each term to simplify:[left( frac{1}{2} left( 3 + cos frac{2pi}{7} right) right) left( frac{1}{2} left( 3 + cos frac{4pi}{7} right) right) left( frac{1}{2} left( 3 + cos frac{6pi}{7} right) right)]Which simplifies to:[left( frac{1}{2} right)^3 left( 3 + cos frac{2pi}{7} right) left( 3 + cos frac{4pi}{7} right) left( 3 + cos frac{6pi}{7} right)]So that's (1/8) times the product of those three terms. Therefore, the entire expression becomes:[sqrt{ frac{1}{8} left( 3 + cos frac{2pi}{7} right) left( 3 + cos frac{4pi}{7} right) left( 3 + cos frac{6pi}{7} right) }]Which is equal to:[frac{1}{2sqrt{2}} sqrt{ left( 3 + cos frac{2pi}{7} right) left( 3 + cos frac{4pi}{7} right) left( 3 + cos frac{6pi}{7} right) }]Okay, so now I need to compute the product inside the square root. Let me denote this product as P:[P = left( 3 + cos frac{2pi}{7} right) left( 3 + cos frac{4pi}{7} right) left( 3 + cos frac{6pi}{7} right)]I need to find the value of P. Hmm, how can I compute this product? Maybe I can use some trigonometric identities or properties of roots of unity.I recall that the angles 2π/7, 4π/7, and 6π/7 are related to the seventh roots of unity. Specifically, they are the real parts of the primitive seventh roots of unity. Maybe I can use some properties from complex numbers or polynomials to find this product.Let me consider the seventh roots of unity. They are the solutions to the equation z^7 = 1. These roots are given by:[z_k = e^{2pi i k /7} quad text{for} quad k = 0, 1, 2, 3, 4, 5, 6]The primitive roots are those where k and 7 are coprime, so k = 1, 2, 3, 4, 5, 6. The real parts of these roots are cos(2πk/7). So, the cosines we have in our product are cos(2π/7), cos(4π/7), cos(6π/7), cos(8π/7), cos(10π/7), and cos(12π/7). But note that cos(8π/7) = cos(6π/7 - 2π) = cos(6π/7), similarly cos(10π/7) = cos(4π/7), and cos(12π/7) = cos(2π/7). So, the distinct cosines are cos(2π/7), cos(4π/7), and cos(6π/7).Therefore, the minimal polynomial for cos(2π/7) is a cubic equation, since there are three distinct cosines. Maybe I can find a polynomial whose roots are cos(2π/7), cos(4π/7), and cos(6π/7), and then use Vieta's formula to find the product of (3 + cos(2π/7)), (3 + cos(4π/7)), and (3 + cos(6π/7)).Let me try to find such a polynomial. Let’s denote x = cos θ, where θ = 2π/7, 4π/7, 6π/7.Using the identity for cos(7θ) = 0, since 7θ = 2π, 4π, 6π, etc., which are multiples of 2π, so cos(7θ) = 1. Wait, actually, cos(7θ) = cos(2πk) = 1 for integer k. So, maybe I can use the expansion of cos(7θ) in terms of cos θ.I remember that cos(7θ) can be expressed as a polynomial in cos θ. Let me recall the formula:cos(7θ) = 64 cos^7 θ - 112 cos^5 θ + 56 cos^3 θ - 7 cos θYes, that's correct. So, since cos(7θ) = 1, we have:64 cos^7 θ - 112 cos^5 θ + 56 cos^3 θ - 7 cos θ - 1 = 0But since θ = 2π/7, 4π/7, 6π/7, we can set x = cos θ, so x satisfies:64x^7 - 112x^5 + 56x^3 - 7x - 1 = 0But this is a seventh-degree equation, which is complicated. However, since we have three distinct roots x = cos(2π/7), cos(4π/7), cos(6π/7), maybe we can factor this polynomial into a product of a cubic and a quartic, where the cubic has these three roots.Let me try to factor this polynomial. Let me denote P(x) = 64x^7 - 112x^5 + 56x^3 - 7x - 1.I can try to factor out (x - 1) since x = 1 is a root (cos(0) = 1). Let me check:P(1) = 64 - 112 + 56 - 7 - 1 = 64 - 112 = -48; -48 +56=8; 8 -7=1; 1 -1=0. So yes, x=1 is a root.So, let's perform polynomial division to factor out (x - 1). Alternatively, I can use synthetic division.But maybe it's easier to note that since the roots are symmetric, perhaps the minimal polynomial for cos(2π/7) is a cubic. Let me check the degree.Since the Galois group of the cyclotomic polynomial for 7th roots of unity has degree φ(7)=6, but since we're dealing with cosines, which are real, the minimal polynomial should have degree 3.So, let me assume that the minimal polynomial is a cubic. Let me denote it as:8x^3 + 4x^2 - 4x -1 = 0Wait, I think I remember that the minimal polynomial for cos(2π/7) is 8x^3 + 4x^2 - 4x -1 = 0. Let me verify that.If x = cos(2π/7), then 8x^3 + 4x^2 - 4x -1 = 0.Let me compute 8x^3 + 4x^2 - 4x -1 when x = cos(2π/7). Hmm, I might need to use multiple-angle identities.Alternatively, I can use the identity for cos(7θ) = 0, but since 7θ = 2π, which is not zero, but 1. Wait, earlier I had:cos(7θ) = 64 cos^7 θ - 112 cos^5 θ + 56 cos^3 θ - 7 cos θBut for θ = 2π/7, 7θ = 2π, so cos(7θ) = cos(2π) = 1.Therefore, 64x^7 - 112x^5 + 56x^3 - 7x = 1, where x = cos(2π/7). So,64x^7 - 112x^5 + 56x^3 - 7x -1 = 0Similarly for x = cos(4π/7) and x = cos(6π/7). So, these three cosines are roots of the equation:64x^7 - 112x^5 + 56x^3 - 7x -1 = 0But since this is a seventh-degree polynomial, and we know x=1 is a root, we can factor it as (x -1)(something). Let me perform polynomial division.Divide P(x) = 64x^7 - 112x^5 + 56x^3 - 7x -1 by (x -1).Using synthetic division:Coefficients of P(x): 64, 0, -112, 0, 56, 0, -7, -1Wait, actually, P(x) is 64x^7 + 0x^6 -112x^5 + 0x^4 +56x^3 +0x^2 -7x -1Divide by (x -1). Using synthetic division with root 1:Bring down 64Multiply by 1: 64Add to next coefficient: 0 +64=64Multiply by1:64Add to next coefficient: -112 +64= -48Multiply by1: -48Add to next coefficient:0 + (-48)= -48Multiply by1: -48Add to next coefficient:56 + (-48)=8Multiply by1:8Add to next coefficient:0 +8=8Multiply by1:8Add to next coefficient:-7 +8=1Multiply by1:1Add to last coefficient:-1 +1=0So, the quotient polynomial is:64x^6 +64x^5 -48x^4 -48x^3 +8x^2 +8x +1So, P(x) = (x -1)(64x^6 +64x^5 -48x^4 -48x^3 +8x^2 +8x +1)Now, let's see if this sixth-degree polynomial can be factored further. Maybe it can be factored into two cubics or a quadratic and a quartic.Alternatively, since we're interested in the minimal polynomial for cos(2π/7), which is a real root, perhaps we can factor out another quadratic or cubic.Alternatively, perhaps the minimal polynomial is a cubic, so maybe the sixth-degree polynomial can be factored into two cubics, each with three roots.But this might be complicated. Alternatively, perhaps I can use the identity that relates the product (3 + cos(2π/7))(3 + cos(4π/7))(3 + cos(6π/7)).Let me denote y = 3 + x, where x is cos(2π/7), cos(4π/7), or cos(6π/7). Then, the product P is equal to (y1)(y2)(y3), where y1 = 3 + x1, y2 = 3 + x2, y3 = 3 + x3, and x1, x2, x3 are the roots of the minimal polynomial.If I can express P in terms of the roots of the minimal polynomial, I can use Vieta's formula.Given that x1, x2, x3 are roots of the minimal polynomial, say, ax^3 + bx^2 + cx + d =0, then:x1 + x2 + x3 = -b/ax1x2 + x1x3 + x2x3 = c/ax1x2x3 = -d/aThen, P = (3 + x1)(3 + x2)(3 + x3) = 27 + 9(x1 + x2 + x3) + 3(x1x2 + x1x3 + x2x3) + x1x2x3So, if I can find the sums and products of the roots, I can compute P.Earlier, I thought the minimal polynomial is 8x^3 +4x^2 -4x -1=0. Let me check if that's correct.If x = cos(2π/7), then 8x^3 +4x^2 -4x -1=0.Let me compute 8x^3 +4x^2 -4x -1 when x = cos(2π/7). Hmm, I might need to use multiple-angle identities.Alternatively, I can use the identity for cos(7θ) = 1, which gives an equation in terms of cos θ.Earlier, we had:cos(7θ) = 64 cos^7 θ - 112 cos^5 θ + 56 cos^3 θ - 7 cos θ = 1So, 64x^7 - 112x^5 +56x^3 -7x -1=0But since x = cos(2π/7), this is satisfied.But perhaps we can factor this equation in terms of the minimal polynomial.Wait, if the minimal polynomial is 8x^3 +4x^2 -4x -1=0, then let me check if 8x^3 +4x^2 -4x -1 divides into the seventh-degree polynomial.Let me perform polynomial division of 64x^7 -112x^5 +56x^3 -7x -1 by 8x^3 +4x^2 -4x -1.Alternatively, maybe it's easier to note that if 8x^3 +4x^2 -4x -1=0, then 64x^6 + 32x^5 -32x^4 -8x^3=0Wait, perhaps not. Alternatively, let me try to express higher powers of x in terms of lower powers using the minimal polynomial.Given 8x^3 = -4x^2 +4x +1So, x^3 = (-4x^2 +4x +1)/8Similarly, x^4 = x * x^3 = x*(-4x^2 +4x +1)/8 = (-4x^3 +4x^2 +x)/8But x^3 can be replaced again:x^4 = (-4*(-4x^2 +4x +1)/8 +4x^2 +x)/8Simplify:= (16x^2 -16x -4)/8 +4x^2 +x)/8= (2x^2 -2x -0.5 +4x^2 +x)/8= (6x^2 -x -0.5)/8Similarly, x^5 = x * x^4 = x*(6x^2 -x -0.5)/8 = (6x^3 -x^2 -0.5x)/8Replace x^3:= (6*(-4x^2 +4x +1)/8 -x^2 -0.5x)/8= (-24x^2 +24x +6)/8 -x^2 -0.5x)/8= (-24x^2 +24x +6 -8x^2 -4x)/8= (-32x^2 +20x +6)/8Similarly, x^6 = x * x^5 = x*(-32x^2 +20x +6)/8 = (-32x^3 +20x^2 +6x)/8Replace x^3:= (-32*(-4x^2 +4x +1)/8 +20x^2 +6x)/8= (128x^2 -128x -32)/8 +20x^2 +6x)/8= (16x^2 -16x -4 +20x^2 +6x)/8= (36x^2 -10x -4)/8Similarly, x^7 = x * x^6 = x*(36x^2 -10x -4)/8 = (36x^3 -10x^2 -4x)/8Replace x^3:= (36*(-4x^2 +4x +1)/8 -10x^2 -4x)/8= (-144x^2 +144x +36)/8 -10x^2 -4x)/8= (-144x^2 +144x +36 -80x^2 -32x)/8= (-224x^2 +112x +36)/8Now, let's plug x^7 into the original equation:64x^7 -112x^5 +56x^3 -7x -1 =0Substitute x^7, x^5, x^3:64*(-224x^2 +112x +36)/8 -112*(-32x^2 +20x +6)/8 +56*(-4x^2 +4x +1)/8 -7x -1 =0Simplify each term:64*(-224x^2 +112x +36)/8 = 8*(-224x^2 +112x +36) = -1792x^2 +896x +288-112*(-32x^2 +20x +6)/8 = 14*(-32x^2 +20x +6) = -448x^2 +280x +8456*(-4x^2 +4x +1)/8 = 7*(-4x^2 +4x +1) = -28x^2 +28x +7Now, combine all terms:-1792x^2 +896x +288 -448x^2 +280x +84 -28x^2 +28x +7 -7x -1 =0Combine like terms:x^2 terms: -1792 -448 -28 = -2268x terms: 896 +280 +28 -7 = 1200 -7=1193constants:288 +84 +7 -1=378So, overall:-2268x^2 +1193x +378=0Wait, but this should equal zero for x = cos(2π/7), but it's a quadratic equation, which suggests that our minimal polynomial assumption might be incorrect, or perhaps I made a mistake in the calculations.Alternatively, maybe the minimal polynomial is indeed 8x^3 +4x^2 -4x -1=0, and the higher powers can be expressed in terms of lower powers, but perhaps I made an error in the substitution.Alternatively, perhaps I should accept that the minimal polynomial is 8x^3 +4x^2 -4x -1=0, and use Vieta's formula on that.Assuming that, let me denote the minimal polynomial as:8x^3 +4x^2 -4x -1=0So, for roots x1, x2, x3:x1 + x2 + x3 = -4/8 = -1/2x1x2 + x1x3 + x2x3 = -4/8 = -1/2x1x2x3 = 1/8Wait, no, in the standard form ax³ + bx² + cx + d =0, the sums are:x1 + x2 + x3 = -b/a = -4/8 = -1/2x1x2 + x1x3 + x2x3 = c/a = (-4)/8 = -1/2x1x2x3 = -d/a = -(-1)/8 = 1/8So, yes, that's correct.Now, going back to P = (3 + x1)(3 + x2)(3 + x3)Expanding this:P = 27 + 9(x1 + x2 + x3) + 3(x1x2 + x1x3 + x2x3) + x1x2x3Substituting the values from Vieta's formula:= 27 + 9*(-1/2) + 3*(-1/2) + (1/8)Compute each term:27 is just 27.9*(-1/2) = -9/23*(-1/2) = -3/21/8 is 0.125So, adding them up:27 - 9/2 - 3/2 + 1/8Convert all to eighths to add:27 = 216/8-9/2 = -36/8-3/2 = -12/81/8 = 1/8So,216/8 -36/8 -12/8 +1/8 = (216 -36 -12 +1)/8 = (216 -48 +1)/8 = (169)/8So, P = 169/8Therefore, going back to the original expression:[sqrt{ frac{1}{8} times frac{169}{8} } = sqrt{ frac{169}{64} } = frac{13}{8}]So, the exact value is 13/8.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I converted each term 2 - sin²θ to 3/2 + cos(2θ)/2, which seems correct.2. Factored out 1/2 from each term, leading to (1/8) times the product of (3 + cos(2π/7)), etc. That seems correct.3. Then, I needed to compute the product P = (3 + x1)(3 + x2)(3 + x3), where x1, x2, x3 are cos(2π/7), cos(4π/7), cos(6π/7). I used the minimal polynomial 8x³ +4x² -4x -1=0, which I think is correct.4. Using Vieta's formula, I found the sums and products of the roots, then expanded P to get 27 + 9S + 3Q + R, where S = sum of roots, Q = sum of products two at a time, R = product of roots.5. Plugging in the values: S = -1/2, Q = -1/2, R =1/8.6. Calculated P = 27 - 9/2 - 3/2 +1/8 = 27 -6 - 3/2 +1/8. Wait, hold on, 9*(-1/2) is -9/2, which is -4.5, and 3*(-1/2) is -1.5, so total is 27 -4.5 -1.5 +0.125 = 27 -6 +0.125=21.125, which is 169/8, since 169 divided by 8 is 21.125.Yes, that seems correct.7. Then, the entire expression is sqrt(1/8 * 169/8) = sqrt(169/64) =13/8.So, yes, the exact value is 13/8.I think that makes sense. I didn't see any mistakes in my calculations, so I'm confident that the answer is 13/8.**Final Answer**boxed{dfrac{13}{8}}