💡Alright, let's see. I have this problem about three table runners overlapping to cover a table. The total area of the runners is 204 square inches. They cover 80% of the table, and there's some information about areas covered by exactly two layers and three layers. I need to find the area of the table.First, I'll try to understand what's given. Three runners have a combined area of 204 square inches. When they overlap to cover 80% of the table, some parts are covered by two layers, and some by three layers. Specifically, the area covered by exactly two layers is 24 square inches, and the area covered by three layers is 20 square inches.I think I need to relate the total area of the runners to the areas covered by different layers. Maybe I can use some kind of inclusion-exclusion principle here. Let me recall that principle. For three sets, the total area covered is the sum of the areas of each set minus the areas of their pairwise intersections plus the area of their triple intersection.But in this case, the runners are overlapping to cover the table, so the total area covered by the runners on the table is 80% of the table's area. Let me denote the table's area as A. So, 0.8A is the area covered by the runners.Now, the runners' total area is 204 square inches. But when they overlap, some areas are counted multiple times. Specifically, the area covered by exactly two layers is 24 square inches, and the area covered by three layers is 20 square inches.I think I can express the total area of the runners as the sum of the areas covered once, twice, and three times. Let me denote the area covered by exactly one layer as S. Then, the total area of the runners would be S (for the single layers) plus 2*24 (since the two-layer area is counted twice) plus 3*20 (since the three-layer area is counted three times).So, the equation would be:S + 2*24 + 3*20 = 204Let me compute that:S + 48 + 60 = 204S + 108 = 204S = 204 - 108S = 96So, the area covered by exactly one layer is 96 square inches.Now, the total area covered by the runners on the table is the sum of the single-layer, double-layer, and triple-layer areas:96 (single) + 24 (double) + 20 (triple) = 140 square inchesBut this total area covered is 80% of the table's area, so:0.8A = 140To find A, I'll divide 140 by 0.8:A = 140 / 0.8A = 175So, the area of the table is 175 square inches.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Total area of runners: 2042. Areas covered by two layers: 24, three layers: 203. Expressed total runners' area as S + 2*24 + 3*20 = 2044. Solved for S: 965. Total area covered on table: 96 + 24 + 20 = 1406. 140 is 80% of table area, so table area A = 140 / 0.8 = 175Everything seems to add up. I think I did it right.

💡Okay, so I need to figure out how many prime positive integers are divisors of 420. Hmm, let's start by recalling what prime numbers are. Prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, examples of prime numbers are 2, 3, 5, 7, 11, and so on.Now, the question is asking about the prime divisors of 420. That means I need to find all the prime numbers that can divide 420 without leaving a remainder. To do this, I think I should perform prime factorization on 420. Prime factorization is the process of breaking down a composite number into a product of its prime factors.Alright, let's start with the smallest prime number, which is 2. I'll check if 2 divides 420 evenly. Dividing 420 by 2 gives me 210. Okay, so 2 is a prime divisor. Now, I'll take 210 and see if it's divisible by 2 again. Dividing 210 by 2 gives me 105. So, 2 is a prime divisor again, but since we're looking for distinct prime factors, I'll note that 2 is a prime divisor once.Next, I'll move to the next prime number, which is 3. I'll check if 3 divides 105 evenly. Dividing 105 by 3 gives me 35. So, 3 is another prime divisor. Now, I'll take 35 and see if it's divisible by 3. Dividing 35 by 3 doesn't give me an integer, so I'll move on to the next prime number.The next prime number after 3 is 5. I'll check if 5 divides 35 evenly. Dividing 35 by 5 gives me 7. So, 5 is another prime divisor. Now, I'll take 7 and see if it's divisible by 5. Dividing 7 by 5 doesn't give me an integer, so I'll move on to the next prime number.The next prime number after 5 is 7. I'll check if 7 divides 7 evenly. Dividing 7 by 7 gives me 1. So, 7 is another prime divisor. Now, since I've reached 1, I know I've completed the prime factorization.So, putting it all together, the prime factors of 420 are 2, 3, 5, and 7. Each of these is a prime number, and they all divide 420 without leaving a remainder. Therefore, the number of prime positive integers that are divisors of 420 is 4.Wait, let me double-check to make sure I didn't miss any primes or count any extra. Starting from the smallest prime, 2, then 3, 5, and 7. Yeah, that seems correct. I don't think there are any other prime numbers between 2 and 7 that divide 420. For example, 11 is a prime number, but 11 doesn't divide 420 evenly because 420 divided by 11 is approximately 38.18, which isn't an integer. Similarly, 13 is a prime number, but 420 divided by 13 is about 32.31, which isn't an integer either. So, I think I've accounted for all the prime divisors of 420.Just to be thorough, let me list out all the factors of 420 and see which ones are prime. The factors of 420 are: 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, and 420. From this list, the prime numbers are 2, 3, 5, and 7. Yep, that matches what I found earlier with the prime factorization method.So, I'm confident that the number of prime positive integers that are divisors of 420 is indeed 4.

💡First, I need to determine the length of the bridge that a train can cross in 30 seconds. The train is 120 meters long and traveling at a speed of 45 km/hr.To make the units consistent, I'll convert the train's speed from kilometers per hour to meters per second. There are 1000 meters in a kilometer and 3600 seconds in an hour, so:45 km/hr = 45 × (1000 m / 3600 s) = 12.5 m/sNext, I'll calculate the total distance the train travels in 30 seconds at this speed:Distance = Speed × Time = 12.5 m/s × 30 s = 375 metersThis distance includes both the length of the train and the length of the bridge. To find the length of the bridge, I'll subtract the length of the train from the total distance:Length of the bridge = Total distance - Length of the train = 375 m - 120 m = 255 metersTherefore, the length of the bridge is 255 meters.

💡First, I need to determine how many bananas Elias bought initially.Elias ate 1 banana and has 11 bananas left.To find the total number of bananas he bought, I can add the number of bananas he ate to the number of bananas he has left.So, 11 bananas left plus 1 banana eaten equals 12 bananas bought.Therefore, Elias bought 12 bananas.

💡Alright, so I've got these three trigonometric identities to prove. Let me take them one by one and see if I can figure them out.Starting with part (a): cos frac{pi}{5} - cos frac{2pi}{5} = frac{1}{2}. Hmm, okay. I remember that there are some trigonometric identities involving cosines of angles that are fractions of pi, especially related to regular polygons. Maybe I can use some properties of a regular pentagon since 5 is in the denominator here.I recall that in a regular pentagon, the central angles are multiples of frac{2pi}{5}. Maybe if I consider the coordinates of the vertices, which can be represented using cosine and sine functions. Let me think about the real parts of these complex numbers on the unit circle.Wait, I also remember that there's an identity for the difference of cosines: cos A - cos B = -2 sin left( frac{A + B}{2} right) sin left( frac{A - B}{2} right). Maybe I can apply that here.Let me set A = frac{pi}{5} and B = frac{2pi}{5}. Plugging into the identity:cos frac{pi}{5} - cos frac{2pi}{5} = -2 sin left( frac{frac{pi}{5} + frac{2pi}{5}}{2} right) sin left( frac{frac{pi}{5} - frac{2pi}{5}}{2} right)Simplifying the arguments:frac{frac{pi}{5} + frac{2pi}{5}}{2} = frac{frac{3pi}{5}}{2} = frac{3pi}{10}frac{frac{pi}{5} - frac{2pi}{5}}{2} = frac{-frac{pi}{5}}{2} = -frac{pi}{10}So, the expression becomes:-2 sin left( frac{3pi}{10} right) sin left( -frac{pi}{10} right)I know that sin(-x) = -sin(x), so:-2 sin left( frac{3pi}{10} right) (-sin left( frac{pi}{10} right)) = 2 sin left( frac{3pi}{10} right) sin left( frac{pi}{10} right)Now, I need to find the value of this product. I remember that sin left( frac{pi}{10} right) and sin left( frac{3pi}{10} right) are related to the golden ratio. Specifically, sin left( frac{pi}{10} right) = frac{sqrt{5} - 1}{4} and sin left( frac{3pi}{10} right) = frac{sqrt{10 + 2sqrt{5}}}{4}.But multiplying these directly seems complicated. Maybe there's a simpler identity or relationship I can use. I recall that in a regular pentagon, the product of certain sine terms relates to the golden ratio. Alternatively, perhaps I can use the identity for the product of sines:sin A sin B = frac{1}{2} [cos(A - B) - cos(A + B)]Let me apply this:sin left( frac{3pi}{10} right) sin left( frac{pi}{10} right) = frac{1}{2} [cos left( frac{3pi}{10} - frac{pi}{10} right) - cos left( frac{3pi}{10} + frac{pi}{10} right)]Simplifying the arguments:frac{3pi}{10} - frac{pi}{10} = frac{2pi}{10} = frac{pi}{5}frac{3pi}{10} + frac{pi}{10} = frac{4pi}{10} = frac{2pi}{5}So, the expression becomes:frac{1}{2} [cos left( frac{pi}{5} right) - cos left( frac{2pi}{5} right)]Wait a minute, that's interesting. So, going back:2 sin left( frac{3pi}{10} right) sin left( frac{pi}{10} right) = 2 times frac{1}{2} [cos left( frac{pi}{5} right) - cos left( frac{2pi}{5} right)] = cos left( frac{pi}{5} right) - cos left( frac{2pi}{5} right)But that's exactly the left-hand side of our original equation! So, we've come full circle. Hmm, that doesn't seem to help directly. Maybe I need a different approach.Let me think about the exact values of cos frac{pi}{5} and cos frac{2pi}{5}. I recall that cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2 = frac{sqrt{5} + 1}{4} times 2 = frac{sqrt{5} + 1}{2} times frac{1}{2}? Wait, no, let me check.Actually, cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2 = frac{sqrt{5} + 1}{2} times frac{1}{2}? Hmm, I'm getting confused. Let me look it up or recall the exact value.I remember that cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2 = frac{sqrt{5} + 1}{2} times frac{1}{2}? Wait, no, actually, cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2 is incorrect.Let me derive it properly. Consider a regular pentagon and use the golden ratio. In a regular pentagon, the diagonal over the side is the golden ratio, phi = frac{1 + sqrt{5}}{2}. Using the law of cosines in a triangle formed by two sides and a diagonal, we can find the cosine of the internal angle.Alternatively, I can use the identity that in a regular pentagon, cos frac{pi}{5} = frac{phi}{2} = frac{1 + sqrt{5}}{4} times 2 = frac{1 + sqrt{5}}{4} times 2 = frac{1 + sqrt{5}}{2} times frac{1}{2}? Wait, I'm still confused.Let me use the exact value: cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2 = frac{sqrt{5} + 1}{2} times frac{1}{2}? No, actually, cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2 is not correct. Let me recall that cos 36^circ = cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2? Wait, no.I think the exact value is cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2? No, actually, it's cos frac{pi}{5} = frac{sqrt{5} + 1}{4} times 2? Hmm, I'm stuck here.Wait, let me use the identity for cos 36^circ. I know that cos 36^circ = frac{sqrt{5} + 1}{4} times 2? No, actually, cos 36^circ = frac{sqrt{5} + 1}{4} times 2 is incorrect. The correct value is cos 36^circ = frac{sqrt{5} + 1}{4} times 2? Wait, no.Let me derive it. Let theta = 36^circ, so 5theta = 180^circ, so sin 5theta = sin 180^circ = 0. Using the identity for sin 5theta:sin 5theta = 16 sin^5 theta - 20 sin^3 theta + 5 sin theta = 0Let x = sin theta, then:16x^5 - 20x^3 + 5x = 0x(16x^4 - 20x^2 + 5) = 0Since x neq 0, we have:16x^4 - 20x^2 + 5 = 0Let y = x^2, then:16y^2 - 20y + 5 = 0Solving for y:y = frac{20 pm sqrt{400 - 320}}{32} = frac{20 pm sqrt{80}}{32} = frac{20 pm 4sqrt{5}}{32} = frac{5 pm sqrt{5}}{8}Since sin 36^circ is positive and less than 1, x = sqrt{frac{5 - sqrt{5}}{8}}.But we need cos 36^circ. Using cos^2 theta + sin^2 theta = 1:cos^2 36^circ = 1 - frac{5 - sqrt{5}}{8} = frac{8 - 5 + sqrt{5}}{8} = frac{3 + sqrt{5}}{8}Thus, cos 36^circ = sqrt{frac{3 + sqrt{5}}{8}} = frac{sqrt{5} + 1}{4} times 2? Wait, let me compute:sqrt{frac{3 + sqrt{5}}{8}} = frac{sqrt{5} + 1}{4} times 2? Let me square frac{sqrt{5} + 1}{4} times 2:left( frac{sqrt{5} + 1}{4} times 2 right)^2 = left( frac{sqrt{5} + 1}{2} right)^2 = frac{5 + 2sqrt{5} + 1}{4} = frac{6 + 2sqrt{5}}{4} = frac{3 + sqrt{5}}{2}But frac{3 + sqrt{5}}{8} is not equal to frac{3 + sqrt{5}}{2}. So, my initial assumption is wrong. Therefore, cos 36^circ = sqrt{frac{3 + sqrt{5}}{8}}.Similarly, cos 72^circ = cos frac{2pi}{5} = sqrt{frac{5 - sqrt{5}}{8}}.Wait, let me verify:cos 72^circ = cos frac{2pi}{5}. Using the same method as above, but for theta = 72^circ, 5theta = 360^circ, so sin 5theta = sin 360^circ = 0. But that's similar to the previous case.Alternatively, I can use the identity cos 72^circ = 2 cos^2 36^circ - 1.Given that cos 36^circ = sqrt{frac{3 + sqrt{5}}{8}}, then:cos 72^circ = 2 left( frac{3 + sqrt{5}}{8} right) - 1 = frac{6 + 2sqrt{5}}{8} - 1 = frac{6 + 2sqrt{5} - 8}{8} = frac{-2 + 2sqrt{5}}{8} = frac{-1 + sqrt{5}}{4}Wait, that can't be right because cos 72^circ is positive. Let me check the calculation:cos 72^circ = 2 cos^2 36^circ - 1cos^2 36^circ = frac{3 + sqrt{5}}{8}So,cos 72^circ = 2 times frac{3 + sqrt{5}}{8} - 1 = frac{6 + 2sqrt{5}}{8} - 1 = frac{6 + 2sqrt{5} - 8}{8} = frac{-2 + 2sqrt{5}}{8} = frac{-1 + sqrt{5}}{4}Yes, that's correct. frac{-1 + sqrt{5}}{4} is approximately 0.3090, which is indeed cos 72^circ.So, now we have:cos frac{pi}{5} = cos 36^circ = sqrt{frac{3 + sqrt{5}}{8}}cos frac{2pi}{5} = cos 72^circ = frac{-1 + sqrt{5}}{4}Wait, but earlier I derived cos 72^circ = frac{-1 + sqrt{5}}{4}, which is approximately 0.3090, and sqrt{frac{5 - sqrt{5}}{8}} is also approximately 0.3090, so they are equal.Therefore, cos frac{pi}{5} = sqrt{frac{3 + sqrt{5}}{8}} and cos frac{2pi}{5} = frac{-1 + sqrt{5}}{4}.Now, let's compute cos frac{pi}{5} - cos frac{2pi}{5}:sqrt{frac{3 + sqrt{5}}{8}} - frac{-1 + sqrt{5}}{4}Let me rationalize sqrt{frac{3 + sqrt{5}}{8}}:sqrt{frac{3 + sqrt{5}}{8}} = frac{sqrt{3 + sqrt{5}}}{2sqrt{2}}But this seems complicated. Maybe instead, I can express both terms with a common denominator or find a way to simplify the expression.Alternatively, let me compute the numerical values:cos 36^circ approx 0.8090cos 72^circ approx 0.3090So, 0.8090 - 0.3090 = 0.5, which is frac{1}{2}. So, numerically, it checks out.But I need an algebraic proof. Let me try squaring both sides of the equation:left( cos frac{pi}{5} - cos frac{2pi}{5} right)^2 = left( frac{1}{2} right)^2 = frac{1}{4}Expanding the left side:cos^2 frac{pi}{5} - 2 cos frac{pi}{5} cos frac{2pi}{5} + cos^2 frac{2pi}{5} = frac{1}{4}We know cos^2 theta = frac{1 + cos 2theta}{2}, so:frac{1 + cos frac{2pi}{5}}{2} - 2 cos frac{pi}{5} cos frac{2pi}{5} + frac{1 + cos frac{4pi}{5}}{2} = frac{1}{4}Simplify:frac{1}{2} + frac{cos frac{2pi}{5}}{2} - 2 cos frac{pi}{5} cos frac{2pi}{5} + frac{1}{2} + frac{cos frac{4pi}{5}}{2} = frac{1}{4}Combine like terms:1 + frac{cos frac{2pi}{5} + cos frac{4pi}{5}}{2} - 2 cos frac{pi}{5} cos frac{2pi}{5} = frac{1}{4}Now, cos frac{4pi}{5} = cos (pi - frac{pi}{5}) = -cos frac{pi}{5}So, cos frac{2pi}{5} + cos frac{4pi}{5} = cos frac{2pi}{5} - cos frac{pi}{5}Thus, the equation becomes:1 + frac{cos frac{2pi}{5} - cos frac{pi}{5}}{2} - 2 cos frac{pi}{5} cos frac{2pi}{5} = frac{1}{4}Let me denote A = cos frac{pi}{5} and B = cos frac{2pi}{5} for simplicity.Then, the equation is:1 + frac{B - A}{2} - 2AB = frac{1}{4}Multiply both sides by 4 to eliminate denominators:4 + 2(B - A) - 8AB = 1Simplify:4 + 2B - 2A - 8AB = 1Bring all terms to one side:4 + 2B - 2A - 8AB - 1 = 0Simplify:3 + 2B - 2A - 8AB = 0Now, recall that from part (a), A - B = frac{1}{2}, so A = B + frac{1}{2}Substitute A = B + frac{1}{2} into the equation:3 + 2B - 2(B + frac{1}{2}) - 8(B + frac{1}{2})B = 0Simplify term by term:3 + 2B - 2B - 1 - 8B(B + frac{1}{2}) = 0Simplify further:3 - 1 - 8B^2 - 4B = 0Which is:2 - 8B^2 - 4B = 0Divide both sides by 2:1 - 4B^2 - 2B = 0Rearrange:-4B^2 - 2B + 1 = 0Multiply both sides by -1:4B^2 + 2B - 1 = 0Now, solve for B:B = frac{-2 pm sqrt{4 + 16}}{8} = frac{-2 pm sqrt{20}}{8} = frac{-2 pm 2sqrt{5}}{8} = frac{-1 pm sqrt{5}}{4}Since B = cos frac{2pi}{5} is positive, we take the positive root:B = frac{-1 + sqrt{5}}{4}Which matches our earlier result. Therefore, our assumption that A - B = frac{1}{2} leads to a consistent solution, confirming that cos frac{pi}{5} - cos frac{2pi}{5} = frac{1}{2}.Okay, that was a bit involved, but I think I managed to prove part (a).Moving on to part (b): frac{1}{sin frac{pi}{7}} = frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{3pi}{7}}This one looks trickier. I'm not immediately sure how to approach it. Maybe I can use some properties of sine functions or look for symmetries in the unit circle.I know that sin theta = sin (pi - theta), so sin frac{2pi}{7} = sin left( pi - frac{2pi}{7} right) = sin frac{5pi}{7} and similarly sin frac{3pi}{7} = sin frac{4pi}{7}. Not sure if that helps directly.Alternatively, perhaps I can use the identity for the sum of reciprocals of sines. I recall that in some cases, especially with angles that are fractions of pi, there are identities involving sums of reciprocals.Let me consider writing the equation as:frac{1}{sin frac{pi}{7}} - frac{1}{sin frac{2pi}{7}} - frac{1}{sin frac{3pi}{7}} = 0I need to show that this expression equals zero. Maybe I can find a common denominator or use some trigonometric identities to combine these terms.The common denominator would be sin frac{pi}{7} sin frac{2pi}{7} sin frac{3pi}{7}. Let's write each term with this denominator:frac{sin frac{2pi}{7} sin frac{3pi}{7} - sin frac{pi}{7} sin frac{3pi}{7} - sin frac{pi}{7} sin frac{2pi}{7}}{sin frac{pi}{7} sin frac{2pi}{7} sin frac{3pi}{7}} = 0Therefore, the numerator must be zero:sin frac{2pi}{7} sin frac{3pi}{7} - sin frac{pi}{7} sin frac{3pi}{7} - sin frac{pi}{7} sin frac{2pi}{7} = 0Let me denote A = frac{pi}{7}, so the angles become A, 2A, and 3A. Then, the equation becomes:sin 2A sin 3A - sin A sin 3A - sin A sin 2A = 0Factor out sin 3A from the first two terms:sin 3A (sin 2A - sin A) - sin A sin 2A = 0Now, use the identity sin 2A - sin A = 2 sin frac{3A}{2} cos frac{A}{2} - sin A. Wait, that might not be helpful.Alternatively, use the identity sin 2A - sin A = 2 cos frac{3A}{2} sin frac{A}{2}.Let me apply that:sin 3A times 2 cos frac{3A}{2} sin frac{A}{2} - sin A sin 2A = 0Simplify:2 sin 3A cos frac{3A}{2} sin frac{A}{2} - sin A sin 2A = 0Now, let's express sin 3A and sin 2A in terms of multiple angles.Recall that sin 3A = 3 sin A - 4 sin^3 A and sin 2A = 2 sin A cos A.But I'm not sure if that helps here. Maybe instead, use product-to-sum identities.For the first term: 2 sin 3A cos frac{3A}{2} sin frac{A}{2}Let me consider 2 sin 3A cos frac{3A}{2} = sin (3A + frac{3A}{2}) + sin (3A - frac{3A}{2}) = sin frac{9A}{2} + sin frac{3A}{2}But A = frac{pi}{7}, so frac{9A}{2} = frac{9pi}{14} and frac{3A}{2} = frac{3pi}{14}.So, the first term becomes:sin frac{9pi}{14} sin frac{A}{2} + sin frac{3pi}{14} sin frac{A}{2}Similarly, the second term is - sin A sin 2A = - sin A times 2 sin A cos A = -2 sin^2 A cos AThis seems to be getting more complicated. Maybe there's a better approach.Alternatively, consider using complex exponentials or roots of unity. Since 7 is a prime, the 7th roots of unity might have some symmetries that can help.The 7th roots of unity are e^{2pi ik/7} for k=0,1,2,3,4,5,6. The imaginary parts correspond to sine functions. Maybe I can relate the sum of reciprocals to some property of these roots.Alternatively, consider the identity that for n being an odd integer, sum_{k=1}^{n-1} frac{1}{sin frac{kpi}{n}} = cot frac{pi}{2n}But I'm not sure if that's applicable here. Let me check for n=7:sum_{k=1}^{6} frac{1}{sin frac{kpi}{7}} = cot frac{pi}{14}But our equation only involves k=1,2,3. Maybe there's a symmetry or pairing.Note that sin frac{kpi}{7} = sin frac{(7 - k)pi}{7}, so frac{1}{sin frac{kpi}{7}} = frac{1}{sin frac{(7 - k)pi}{7}}Therefore, the sum from k=1 to 6 can be paired as:frac{1}{sin frac{pi}{7}} + frac{1}{sin frac{6pi}{7}} = 2 frac{1}{sin frac{pi}{7}}Similarly, frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{5pi}{7}} = 2 frac{1}{sin frac{2pi}{7}}And frac{1}{sin frac{3pi}{7}} + frac{1}{sin frac{4pi}{7}} = 2 frac{1}{sin frac{3pi}{7}}So, the total sum is:2 left( frac{1}{sin frac{pi}{7}} + frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{3pi}{7}} right ) = cot frac{pi}{14}Therefore,frac{1}{sin frac{pi}{7}} + frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{3pi}{7}} = frac{1}{2} cot frac{pi}{14}But our equation is frac{1}{sin frac{pi}{7}} = frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{3pi}{7}}Which implies:frac{1}{sin frac{pi}{7}} - frac{1}{sin frac{2pi}{7}} - frac{1}{sin frac{3pi}{7}} = 0From the earlier sum, we have:frac{1}{sin frac{pi}{7}} + frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{3pi}{7}} = frac{1}{2} cot frac{pi}{14}But I need to relate this to the equation I have. Maybe I can express cot frac{pi}{14} in terms of the other terms.Alternatively, perhaps I can use the identity that cot frac{pi}{14} = tan frac{3pi}{7}, since frac{pi}{14} + frac{3pi}{7} = frac{pi}{14} + frac{6pi}{14} = frac{7pi}{14} = frac{pi}{2}, so cot frac{pi}{14} = tan frac{3pi}{7}But I'm not sure if that helps directly. Maybe I can use the fact that tan theta = frac{sin theta}{cos theta}, but I'm not seeing a clear path.Alternatively, perhaps I can use the identity for cot frac{pi}{14} in terms of sin functions. Let me recall that cot x = frac{cos x}{sin x}, so:cot frac{pi}{14} = frac{cos frac{pi}{14}}{sin frac{pi}{14}}But I don't see a direct connection to the terms in our equation.Maybe instead of trying to relate it to the sum, I can use another approach. Let me consider the equation:frac{1}{sin frac{pi}{7}} = frac{1}{sin frac{2pi}{7}} + frac{1}{sin frac{3pi}{7}}Multiply both sides by sin frac{pi}{7} sin frac{2pi}{7} sin frac{3pi}{7} to eliminate denominators:sin frac{2pi}{7} sin frac{3pi}{7} = sin frac{pi}{7} sin frac{3pi}{7} + sin frac{pi}{7} sin frac{2pi}{7}This is the same equation I had earlier. So, I need to show that:sin frac{2pi}{7} sin frac{3pi}{7} - sin frac{pi}{7} sin frac{3pi}{7} - sin frac{pi}{7} sin frac{2pi}{7} = 0Let me try to express these products as sums using the identity sin A sin B = frac{1}{2} [cos(A - B) - cos(A + B)]Applying this to each term:First term: sin frac{2pi}{7} sin frac{3pi}{7} = frac{1}{2} [cos left( frac{2pi}{7} - frac{3pi}{7} right) - cos left( frac{2pi}{7} + frac{3pi}{7} right)] = frac{1}{2} [cos left( -frac{pi}{7} right) - cos left( frac{5pi}{7} right)] = frac{1}{2} [cos frac{pi}{7} - cos frac{5pi}{7}]Second term: sin frac{pi}{7} sin frac{3pi}{7} = frac{1}{2} [cos left( frac{pi}{7} - frac{3pi}{7} right) - cos left( frac{pi}{7} + frac{3pi}{7} right)] = frac{1}{2} [cos left( -frac{2pi}{7} right) - cos left( frac{4pi}{7} right)] = frac{1}{2} [cos frac{2pi}{7} - cos frac{4pi}{7}]Third term: sin frac{pi}{7} sin frac{2pi}{7} = frac{1}{2} [cos left( frac{pi}{7} - frac{2pi}{7} right) - cos left( frac{pi}{7} + frac{2pi}{7} right)] = frac{1}{2} [cos left( -frac{pi}{7} right) - cos left( frac{3pi}{7} right)] = frac{1}{2} [cos frac{pi}{7} - cos frac{3pi}{7}]Now, substitute these back into the equation:frac{1}{2} [cos frac{pi}{7} - cos frac{5pi}{7}] - frac{1}{2} [cos frac{2pi}{7} - cos frac{4pi}{7}] - frac{1}{2} [cos frac{pi}{7} - cos frac{3pi}{7}] = 0Factor out frac{1}{2}:frac{1}{2} left[ (cos frac{pi}{7} - cos frac{5pi}{7}) - (cos frac{2pi}{7} - cos frac{4pi}{7}) - (cos frac{pi}{7} - cos frac{3pi}{7}) right] = 0Simplify inside the brackets:cos frac{pi}{7} - cos frac{5pi}{7} - cos frac{2pi}{7} + cos frac{4pi}{7} - cos frac{pi}{7} + cos frac{3pi}{7}Combine like terms:cos frac{pi}{7} - cos frac{pi}{7} = 0- cos frac{5pi}{7} + cos frac{4pi}{7} + cos frac{3pi}{7} - cos frac{2pi}{7}Now, note that cos frac{5pi}{7} = cos (pi - frac{2pi}{7}) = -cos frac{2pi}{7}Similarly, cos frac{4pi}{7} = cos (pi - frac{3pi}{7}) = -cos frac{3pi}{7}So, substitute these:- (-cos frac{2pi}{7}) + (-cos frac{3pi}{7}) + cos frac{3pi}{7} - cos frac{2pi}{7}Simplify:cos frac{2pi}{7} - cos frac{3pi}{7} + cos frac{3pi}{7} - cos frac{2pi}{7} = 0All terms cancel out, leaving 0. Therefore, the numerator is indeed zero, confirming that the original equation holds.Phew, that was a bit of a journey, but I think I managed to prove part (b) as well.Finally, part (c): sin 9^{circ} + sin 49^{circ} + sin 89^{circ} + ldots + sin 329^{circ} = 0This one seems to involve a sum of sines with angles increasing by 40 degrees each time, starting from 9 degrees up to 329 degrees. Let me list out all the terms to see the pattern.Starting at 9°, adding 40° each time:9°, 49°, 89°, 129°, 169°, 209°, 249°, 289°, 329°Wait, let's count how many terms there are. From 9° to 329°, with a common difference of 40°, the number of terms is:Number of terms = frac{329 - 9}{40} + 1 = frac{320}{40} + 1 = 8 + 1 = 9 terms.So, we have 9 terms in the sum.Now, I need to compute sum_{k=0}^{8} sin (9° + 40°k)I recall that the sum of sines with equally spaced angles can be expressed using the formula:sum_{k=0}^{n-1} sin (a + kd) = frac{sin frac{nd}{2}}{sin frac{d}{2}} sin left( a + frac{(n - 1)d}{2} right )Let me apply this formula here.Here, a = 9°, d = 40°, and n = 9.So, the sum becomes:frac{sin frac{9 times 40°}{2}}{sin frac{40°}{2}} sin left( 9° + frac{(9 - 1) times 40°}{2} right )Simplify:frac{sin 180°}{sin 20°} sin left( 9° + frac{8 times 40°}{2} right ) = frac{sin 180°}{sin 20°} sin left( 9° + 160° right ) = frac{0}{sin 20°} sin 169° = 0Since sin 180° = 0, the entire expression is zero.Therefore, the sum is zero.Alternatively, I can think about the symmetry of the sine function. Since sine has a period of 360°, and the angles are symmetrically distributed around the circle. Each term sin theta has a corresponding term sin (360° - theta) = -sin theta, so they cancel each other out.Looking at the terms:sin 9° and sin 351° (which is sin (360° - 9°) = -sin 9°)sin 49° and sin 311° (which is sin (360° - 49°) = -sin 49°)sin 89° and sin 271° (which is sin (360° - 89°) = -sin 89°)sin 129° and sin 231° (which is sin (180° + 51°) = -sin 51°, but wait, 231° is not directly related to 129°. Hmm, maybe my initial pairing is off.Wait, let's list all the terms:9°, 49°, 89°, 129°, 169°, 209°, 249°, 289°, 329°Now, let's pair them as follows:9° and 351° (which is 360° - 9°)49° and 311° (360° - 49°)89° and 271° (360° - 89°)129° and 231° (360° - 129° = 231°)169° and 191° (360° - 169° = 191°, but 191° is not in our list)Wait, 169° is in the list, but 191° is not. Similarly, 209° is in the list, but 151° is not. Hmm, maybe this approach isn't perfect.Alternatively, notice that 9° + 351° = 360°, 49° + 311° = 360°, 89° + 271° = 360°, 129° + 231° = 360°, and 169° + 191° = 360°, but since 191° isn't in our list, this might not help.Wait, but in our sum, we have 9 terms, which is odd, so the middle term is 169°, which is 180° - 11°, and sin 169° = sin (180° - 11°) = sin 11°. But I'm not sure if that helps.Alternatively, considering the sum as a vector addition on the unit circle. Each term sin theta represents the y-component of a unit vector at angle theta. If the vectors are symmetrically placed, their sum could be zero.Given that the angles are spaced at 40°, which is 360°/9, so they are equally spaced around the circle. Therefore, the vectors form a regular nonagon (9-sided polygon) on the unit circle. The sum of all vectors in a regular polygon centered at the origin is zero because of symmetry.Therefore, the sum of their y-components (which is our sum) is also zero.Hence, the sum sin 9^{circ} + sin 49^{circ} + sin 89^{circ} + ldots + sin 329^{circ} = 0.So, all three parts are proven.

💡To find the distance between the points ((2,5)) and ((-6,-1)), I can use the distance formula. The distance formula calculates the straight-line distance between two points in a plane.First, I'll identify the coordinates of the two points. The first point is ((2,5)), so (x_1 = 2) and (y_1 = 5). The second point is ((-6,-1)), so (x_2 = -6) and (y_2 = -1).Next, I'll apply the distance formula:[d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}]Substituting the values into the formula:[d = sqrt{(-6 - 2)^2 + (-1 - 5)^2}]Simplifying the differences inside the parentheses:[d = sqrt{(-8)^2 + (-6)^2}]Calculating the squares:[d = sqrt{64 + 36}]Adding the results:[d = sqrt{100}]Finally, taking the square root of 100 gives:[d = 10]So, the distance between the two points is 10 units.