💡Okay, so I have these triple angle formulas for sine and cosine. They are:- sin(3θ) = 3 sinθ - 4 sin³θ- cos(3θ) = 4 cos³θ - 3 cosθAnd from these, they've derived these expressions:- cos(3θ)/cosθ = 4 cos²θ - 3- sin(3θ)/sinθ = 3 - 4 sin²θHmm, interesting. So, these expressions are basically simplifying the ratio of the triple angle function to the single angle function. I can see how that might be useful for certain types of problems.Now, the question is asking what kind of trigonometric function problem can be constructed using these formulas. Let me think about how these expressions can be applied.First, maybe solving equations involving triple angles? For example, if I have an equation like cos(3θ) = something, I could express it in terms of cosθ using the formula. Similarly for sine.Alternatively, perhaps simplifying expressions that involve multiple angles. For instance, if I have a product of terms like (4 cos²θ - 3)(4 cos²3θ - 3)... and so on, maybe I can relate that to some multiple angle identity.Wait, in the previous problem, they showed that the product of these terms equals cos(3^n θ)/cosθ. So, maybe a problem where you have to find such a product or relate it to a specific angle.Another thought: Maybe proving identities using these triple angle formulas. For example, proving that the product of certain terms equals a specific trigonometric function.Or perhaps solving for θ when given a ratio involving triple angles. For example, if I have cos(3θ)/cosθ = some value, I can set up an equation and solve for θ.Let me think of a specific problem. Maybe something like: Given that θ is 9 degrees, find the value of [ (4 cos²9° - 3)(4 cos²27° - 3)(4 cos²81° - 3)(4 cos²243° - 3) ].Wait, that seems a bit involved, but using the formula, this product would equal cos(3^4 * 9°)/cos9°. Since 3^4 is 81, 81 * 9° is 729°, which is equivalent to 9° because 729° - 2*360° = 9°. So, cos(729°)/cos9° = cos9°/cos9° = 1. So, the product equals 1.That's a neat result. So, the problem could be to compute this product and show that it equals 1.Alternatively, maybe a problem where you have to express a multiple angle function in terms of single angle functions, using these identities.Another idea: Maybe solving for θ when given a ratio involving triple angles. For example, if sin(3θ)/sinθ = 2, find θ.Using the identity, sin(3θ)/sinθ = 3 - 4 sin²θ = 2. So, 3 - 4 sin²θ = 2 => 4 sin²θ = 1 => sin²θ = 1/4 => sinθ = ±1/2. So, θ = 30°, 150°, etc. That could be a problem.Or, using the cosine version: If cos(3θ)/cosθ = 1, find θ. Then, 4 cos²θ - 3 = 1 => 4 cos²θ = 4 => cos²θ = 1 => cosθ = ±1 => θ = 0°, 180°, etc.So, these kinds of problems where you set up an equation using the ratio identities and solve for θ.Another angle (no pun intended): Maybe proving that a certain product of terms equals a specific trigonometric function, using these identities. For example, proving that the product of (4 cos²3^k θ - 3) from k=0 to n-1 equals cos(3^n θ)/cosθ.That seems like a good problem, especially since it involves induction or recursive application of the triple angle formula.Alternatively, maybe using these identities to express higher multiple angles in terms of single angles. For example, expressing cos(9θ) in terms of cosθ by applying the triple angle formula multiple times.Wait, actually, that's similar to the product problem. Because cos(9θ) = cos(3*3θ) = 4 cos³(3θ) - 3 cos(3θ). But cos(3θ) itself can be expressed as 4 cos³θ - 3 cosθ. So, substituting, we get cos(9θ) in terms of cosθ.But that might get complicated quickly, as each application introduces higher powers.Alternatively, maybe expressing sin(3θ) in terms of sinθ, and then using that to find higher multiples.But perhaps the most straightforward problem is the one where you have to compute a product of terms like (4 cos²θ - 3)(4 cos²3θ - 3)... and show that it equals 1 or some specific value.Another thought: Maybe using these identities to solve for θ in a more complex equation. For example, combining both sine and cosine ratios.Wait, for example: If I have both sin(3θ)/sinθ and cos(3θ)/cosθ given, maybe I can relate them somehow. Since sin(3θ)/sinθ = 3 - 4 sin²θ and cos(3θ)/cosθ = 4 cos²θ - 3.But since sin²θ + cos²θ = 1, maybe I can set up a system of equations or find a relationship between these two ratios.Alternatively, maybe using these identities to find exact values of trigonometric functions for specific angles, like 9°, 27°, etc., which are related through tripling.Wait, 9°, 27°, 81°, 243°, etc., are all related by tripling. So, maybe a problem where you have to find the exact value of cos9°, using these identities and the fact that cos243° is related to cos9°.But that might be a bit involved, as it would require solving a polynomial equation, since the triple angle formula leads to a cubic equation.Alternatively, maybe a problem where you have to express tanθ in terms of these ratios. Since tanθ = sinθ/cosθ, and we have expressions for sin3θ/sinθ and cos3θ/cosθ, maybe we can relate tan3θ to tanθ.Wait, tan3θ = sin3θ/cos3θ. Using the identities, sin3θ = 3 sinθ - 4 sin³θ and cos3θ = 4 cos³θ - 3 cosθ. So, tan3θ = (3 sinθ - 4 sin³θ)/(4 cos³θ - 3 cosθ). Maybe simplifying that expression.Alternatively, expressing tan3θ in terms of tanθ. Let me recall that tan3θ can be expressed as (3 tanθ - tan³θ)/(1 - 3 tan²θ). Maybe deriving that formula using the given identities.So, that could be another problem: Derive the formula for tan3θ using the given sine and cosine triple angle formulas.Alternatively, maybe using these identities to solve for θ in a real-world problem, like in physics or engineering, where triple angles come into play.But perhaps the most straightforward problem is the product one, where you have to compute the product of these terms and show it equals 1 or some specific value, as in the example I thought of earlier.So, to sum up, possible problems include:1. Solving equations involving ratios of triple angle functions to single angle functions.2. Proving identities involving products of these ratios.3. Expressing higher multiple angles in terms of single angles.4. Deriving formulas for tan3θ using the given identities.5. Finding exact values of trigonometric functions for specific angles using these identities.I think the product problem is a good candidate because it demonstrates the recursive application of the triple angle formula and leads to an elegant result.Alright, so the problem could be:"Given θ = 9°, compute the value of [ (4 cos²9° - 3)(4 cos²27° - 3)(4 cos²81° - 3)(4 cos²243° - 3) ] and show that it equals 1."This problem uses the given identities and shows how the product of these terms simplifies to 1, leveraging the periodicity of the cosine function.Alternatively, to make it more general, the problem could be:"Prove that for any θ, the product from k=0 to n-1 of (4 cos²3^k θ - 3) equals cos(3^n θ)/cosθ."This is a more general problem that demonstrates the recursive nature of the triple angle formula and how these products relate to higher multiple angles.Either way, both problems effectively utilize the given triple angle identities and lead to interesting results.

💡Okay, so I have this problem about folding a triangular piece of paper. Let me try to visualize it first. The triangle is named ABC, and the base AB is 20 cm long. When the paper is folded over the base, there's a crease DE that's parallel to AB. After folding, a smaller triangle projects below the base, and its area is 81% of the area of triangle ABC. I need to find the length of DE.Hmm, folding the paper over the base... So, if I imagine folding the triangle along DE, which is parallel to AB, the part of the triangle above DE gets folded down over the base AB. The area that projects below AB is 81% of the original triangle's area. That means the area of the folded part is 81% of ABC's area.Wait, actually, when you fold the paper, the part that projects below the base is the area of the folded triangle. So, if the original triangle ABC has an area, say, A, then the area of the folded part is 0.81A. That makes sense.Since DE is parallel to AB, triangles ABC and the folded triangle (let's call it CDE, maybe?) are similar. Similar triangles have areas that are proportional to the square of the ratio of their corresponding sides. So, if the area ratio is 0.81, then the ratio of their sides should be the square root of 0.81.Let me write that down. Let’s denote the ratio of similarity as k. Then, the area ratio is k² = 0.81. So, k = sqrt(0.81) = 0.9. That means each side of the smaller triangle is 0.9 times the corresponding side of the original triangle.But wait, is the folded triangle similar to ABC? Let me think. When you fold the paper along DE, the crease DE becomes the base of the folded triangle, and since DE is parallel to AB, the triangles ABC and the folded triangle should indeed be similar.So, if DE is the base of the smaller triangle, and AB is the base of the original triangle, then DE = k * AB. Since AB is 20 cm, DE should be 0.9 * 20 cm.Calculating that, 0.9 * 20 = 18 cm. So, DE should be 18 cm long.But let me double-check. If the area of the folded triangle is 81% of ABC, then the ratio of areas is 0.81, so the ratio of sides is sqrt(0.81) = 0.9. Therefore, DE is 0.9 times AB, which is 18 cm. That seems right.Wait, but sometimes when folding, the ratio can be tricky. Is the area of the folded part 81% of the original, or is the area of the remaining part 81%? Let me read the problem again."The area of the triangle that projects below the base is 81% of the area of the triangle ABC." So, the folded part, which is below the base, is 81% of ABC's area. That means the area of the folded triangle is 0.81 times the area of ABC. So, yes, the ratio is 0.81, and the sides are scaled by 0.9.Therefore, DE is 0.9 * 20 = 18 cm. I think that's correct.Just to visualize again: when you fold the triangle over DE, the part that's folded down has an area of 81% of ABC. So, the crease DE is parallel to AB, and the triangle CDE is similar to ABC with a scaling factor of 0.9. Therefore, DE is 18 cm.I don't see any mistakes in this reasoning. The key was recognizing the similarity of the triangles and understanding that the area ratio relates to the square of the side ratio. So, taking the square root of 0.81 gives the scaling factor for the sides, which is 0.9, and multiplying that by the original base length gives DE.Yeah, I think that's solid. So, DE is 18 cm long.**Final Answer**The length of DE is boxed{18} cm.

💡Okay, so I have this problem where integers a, b, c satisfy two equations: a + b - c = 1 and a² + b² - c² = -1. I need to find the sum of all possible values of a² + b² + c². Hmm, let me try to figure this out step by step.First, the given equations are:1. a + b - c = 12. a² + b² - c² = -1I think I can use the first equation to express c in terms of a and b. Let me rearrange equation 1:c = a + b - 1Okay, so now I can substitute this expression for c into equation 2. Let me do that:a² + b² - (a + b - 1)² = -1Now, I need to expand the squared term. Let's compute (a + b - 1)²:(a + b - 1)² = (a + b)² - 2(a + b) + 1= a² + 2ab + b² - 2a - 2b + 1So, substituting back into equation 2:a² + b² - [a² + 2ab + b² - 2a - 2b + 1] = -1Let me simplify this step by step. Distribute the negative sign:a² + b² - a² - 2ab - b² + 2a + 2b - 1 = -1Now, let's combine like terms:(a² - a²) + (b² - b²) + (-2ab) + (2a) + (2b) - 1 = -1So, this simplifies to:-2ab + 2a + 2b - 1 = -1Hmm, let me add 1 to both sides to simplify further:-2ab + 2a + 2b = 0I can factor out a 2 from the terms:2(-ab + a + b) = 0Divide both sides by 2:-ab + a + b = 0Let me rewrite this equation:ab = a + bHmm, interesting. So, ab = a + b. Maybe I can rearrange this to factor it. Let me subtract a and b from both sides:ab - a - b = 0Now, I can add 1 to both sides to make it factorable:ab - a - b + 1 = 1Now, this factors into:(a - 1)(b - 1) = 1Ah, nice! So, (a - 1)(b - 1) = 1. Since a and b are integers, the possible integer pairs that multiply to 1 are (1,1) and (-1,-1). Let me consider these cases.Case 1: (a - 1) = 1 and (b - 1) = 1So, a = 2 and b = 2.Case 2: (a - 1) = -1 and (b - 1) = -1So, a = 0 and b = 0.Are there any other cases? Well, since 1 is prime, the only integer pairs are (1,1) and (-1,-1). So, these are the only possibilities.Now, let's find c for each case using c = a + b - 1.Case 1: a = 2, b = 2c = 2 + 2 - 1 = 3Case 2: a = 0, b = 0c = 0 + 0 - 1 = -1Okay, so now we have two sets of solutions:1. (a, b, c) = (2, 2, 3)2. (a, b, c) = (0, 0, -1)Now, let's compute a² + b² + c² for each case.Case 1:a² + b² + c² = 2² + 2² + 3² = 4 + 4 + 9 = 17Case 2:a² + b² + c² = 0² + 0² + (-1)² = 0 + 0 + 1 = 1So, the possible values of a² + b² + c² are 17 and 1. The problem asks for the sum of all possible values, so 17 + 1 = 18.Wait, let me double-check my work to make sure I didn't make any mistakes.Starting from the beginning, I expressed c in terms of a and b correctly. Then, substituting into the second equation, I expanded the square correctly and simplified step by step. Factoring led me to (a - 1)(b - 1) = 1, which seems right. The integer solutions for this are indeed (2,2) and (0,0). Calculating c for each case also seems correct. Then, computing a² + b² + c² for each case gives 17 and 1, which add up to 18. I think this is correct.I don't see any errors in my reasoning, so I feel confident that the sum of all possible values of a² + b² + c² is 18.**Final Answer**boxed{18}

💡Okay, so I have this problem where I need to find the sum of the digits of the integer N, which is given by the sum of a series of numbers like 9, 99, 999, 9999, and so on, up to a number with 400 digits, all of which are 9s. That seems pretty big! Let me try to break this down step by step.First, I notice that each term in the series is a number made up entirely of 9s. The first term is 9 (which is one digit), the next is 99 (two digits), then 999 (three digits), and so on, up to a term with 400 digits. So, in total, there are 400 terms in this series, right? Each term has an increasing number of 9s, starting from one up to 400.Now, I need to find the sum of all these terms, which is N, and then find the sum of the digits of N. That sounds a bit complicated, but maybe I can find a pattern or a formula to simplify the process.Let me think about each term individually. Each term is a number consisting of all 9s. For example, 9 is 9, 99 is 99, 999 is 999, etc. I remember that such numbers can be expressed as 10^n - 1, where n is the number of digits. Let me verify that:- For n=1: 10^1 - 1 = 10 - 1 = 9. That's correct.- For n=2: 10^2 - 1 = 100 - 1 = 99. That's correct too.- For n=3: 10^3 - 1 = 1000 - 1 = 999. Yep, that works.So, each term in the series can be written as 10^n - 1, where n is the number of digits in that term. Therefore, the entire series N can be expressed as the sum from n=1 to n=400 of (10^n - 1). Let me write that out:N = (10^1 - 1) + (10^2 - 1) + (10^3 - 1) + ... + (10^400 - 1)Hmm, this looks like a sum of two separate series: one is the sum of 10^n from n=1 to 400, and the other is the sum of (-1) from n=1 to 400. So, I can split this into two parts:N = (10^1 + 10^2 + 10^3 + ... + 10^400) - (1 + 1 + 1 + ... + 1)The first part is a geometric series, and the second part is just adding 1 a total of 400 times, which is 400. So, N can be rewritten as:N = (Sum from n=1 to 400 of 10^n) - 400Now, I need to compute the sum of the geometric series. The formula for the sum of a geometric series is S = a*(r^n - 1)/(r - 1), where a is the first term, r is the common ratio, and n is the number of terms. In this case, a = 10, r = 10, and n = 400.Plugging these values into the formula:Sum = 10*(10^400 - 1)/(10 - 1) = 10*(10^400 - 1)/9So, the sum of the geometric series is (10*(10^400 - 1))/9.Therefore, N becomes:N = (10*(10^400 - 1))/9 - 400Hmm, that looks a bit messy, but maybe I can simplify it further. Let me compute this step by step.First, let's compute (10*(10^400 - 1))/9:(10*(10^400 - 1))/9 = (10^401 - 10)/9So, N = (10^401 - 10)/9 - 400To combine these terms, I can write 400 as 3600/9, so that both terms have the same denominator:N = (10^401 - 10 - 3600)/9 = (10^401 - 3610)/9Hmm, okay. So, N is equal to (10^401 - 3610)/9. Now, I need to figure out what this number looks like when written out in decimal form because I need to find the sum of its digits.Let me think about 10^401. That's a 1 followed by 401 zeros. So, 10^401 is a very large number. When I subtract 3610 from it, I'm effectively subtracting 3610 from the last few digits of this number. Let me try to visualize this.10^401 is:1 followed by 401 zeros.Subtracting 3610 would affect the last few digits. Let me write it out:10^401 = 1000...000 (with 401 zeros)Subtracting 3610:1000...000- 3610= 999...9996390Wait, let me check that. If I subtract 3610 from 10000, I get 6390. So, in the case of 10^401, subtracting 3610 would result in a number that has a lot of 9s followed by 6390.Specifically, 10^401 - 3610 would be a number that has 397 nines followed by 6390. Let me confirm:Starting from the right, the last four digits would be 0000 - 3610. But since we can't subtract 3610 from 0000, we need to borrow from the preceding digits. So, the last four digits become 10000 - 3610 = 6390. The preceding digits, which were all zeros, become 9s because we've borrowed 1 from each of them. Since there are 401 digits in total, subtracting 4 digits (3610) would leave 397 digits before the last four. So, those 397 digits would all be 9s.Therefore, 10^401 - 3610 is a number that has 397 nines followed by 6390.Now, N is equal to (10^401 - 3610)/9. So, I need to divide this number by 9. Let me think about how dividing a number consisting of all 9s followed by some digits by 9 affects its digits.I know that 999...999 divided by 9 is 111...111. For example, 999/9 = 111, 9999/9 = 1111, etc. So, if I have a number that is 397 nines followed by 6390, dividing by 9 would result in a number that is 397 ones followed by the result of 6390 divided by 9.Let me compute 6390 divided by 9:6390 ÷ 9 = 710Because 9*700 = 6300, and 9*10 = 90, so 6300 + 90 = 6390. Therefore, 6390 ÷ 9 = 710.So, putting it all together, (10^401 - 3610)/9 is a number that has 397 ones followed by 710.Therefore, N is equal to 111...111710, where there are 397 ones followed by 710.Now, I need to find the sum of the digits of N. Let's break this down:- The first 397 digits are all 1s.- The last three digits are 7, 1, and 0.So, the sum of the digits is the sum of 397 ones plus the sum of 7, 1, and 0.Calculating that:Sum = (397 * 1) + 7 + 1 + 0 = 397 + 8 = 405Therefore, the sum of the digits of N is 405.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Expressed each term as 10^n - 1: Correct.2. Summed the series to get (10*(10^400 - 1))/9 - 400: Correct.3. Simplified to (10^401 - 10)/9 - 400: Correct.4. Converted 400 to 3600/9 to combine terms: Correct.5. Resulted in (10^401 - 3610)/9: Correct.6. Subtracted 3610 from 10^401 to get 397 nines followed by 6390: Correct.7. Divided by 9 to get 397 ones followed by 710: Correct.8. Summed the digits: 397*1 + 7 + 1 + 0 = 405: Correct.Everything seems to check out. I don't see any errors in my reasoning. So, I'm confident that the sum of the digits of N is 405.

💡Okay, so I need to find the area of a parallelogram formed by two vectors, a and b. The vectors are given in terms of other vectors p and q. Let me write down what I know:Vector a is 2p minus 3q, so a = 2p - 3q.Vector b is 3p plus q, so b = 3p + q.The magnitudes of p and q are given: |p| = 4 and |q| = 1.The angle between p and q is π/6, which is 30 degrees.Alright, so I remember that the area of a parallelogram formed by two vectors is equal to the magnitude of their cross product. So, the area S is |a × b|.First, I need to compute the cross product of a and b. Since a and b are expressed in terms of p and q, I can expand the cross product using the distributive property.Let me write that out:a × b = (2p - 3q) × (3p + q)Now, I'll distribute the cross product:= 2p × 3p + 2p × q - 3q × 3p - 3q × qSimplify each term:= 6(p × p) + 2(p × q) - 9(q × p) - 3(q × q)I remember that the cross product of any vector with itself is zero, so p × p = 0 and q × q = 0. That simplifies things:= 0 + 2(p × q) - 9(q × p) - 0Now, I also recall that the cross product is anti-commutative, meaning q × p = - (p × q). So, let's substitute that in:= 2(p × q) - 9(- (p × q))= 2(p × q) + 9(p × q)Combine like terms:= (2 + 9)(p × q)= 11(p × q)So, the cross product a × b is 11 times (p × q). Now, to find the magnitude of this cross product, I need |a × b| = |11(p × q)| = 11|p × q|.The magnitude of the cross product |p × q| is equal to |p||q|sin(theta), where theta is the angle between p and q. So, let's compute that:|p × q| = |p||q|sin(theta) = 4 * 1 * sin(π/6)I know that sin(π/6) is 0.5, so:|p × q| = 4 * 1 * 0.5 = 2Therefore, |a × b| = 11 * 2 = 22So, the area of the parallelogram is 22.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded the cross product correctly: Yes, I distributed each term.2. Remembered that p × p and q × q are zero: Correct.3. Applied the anti-commutative property correctly: Yes, q × p became -p × q.4. Combined the coefficients correctly: 2 + 9 is 11, so that's right.5. Calculated |p × q| correctly: 4 * 1 * sin(π/6) = 2, which seems right.6. Multiplied by 11: 11 * 2 = 22, correct.I think everything checks out. So, the area should be 22.

💡Okay, so I have this problem where I'm given two functions, f(x) and g(x). They're defined on all real numbers, R. The problem says that f(x) is an odd function and g(x) is an even function. I also know that when you add them together, f(x) + g(x), you get 3^x. My task is to find the value of f(1). Alright, let's break this down. First, I need to recall what it means for a function to be odd or even. An odd function satisfies the property f(-x) = -f(x) for all x in its domain. On the other hand, an even function satisfies g(-x) = g(x) for all x in its domain. So, knowing that, I can use these properties to set up some equations.Given that f(x) + g(x) = 3^x, I can write that down as equation (1). Now, if I plug in -x into both sides of this equation, I should get another equation. Let's see, f(-x) + g(-x) would equal 3^{-x}. But since f is odd and g is even, I can substitute f(-x) with -f(x) and g(-x) with g(x). So, equation (2) becomes -f(x) + g(x) = 3^{-x}.Now, I have two equations:1. f(x) + g(x) = 3^x2. -f(x) + g(x) = 3^{-x}Hmm, so I have a system of two equations with two variables: f(x) and g(x). Maybe I can solve this system to find expressions for f(x) and g(x). Let me try subtracting equation (2) from equation (1). If I subtract equation (2) from equation (1), I get:[f(x) + g(x)] - [-f(x) + g(x)] = 3^x - 3^{-x}Simplifying the left side:f(x) + g(x) + f(x) - g(x) = 2f(x)So, 2f(x) = 3^x - 3^{-x}Therefore, f(x) = (3^x - 3^{-x}) / 2Alright, so that gives me an expression for f(x). Now, I need to find f(1). Let's plug x = 1 into this expression.f(1) = (3^1 - 3^{-1}) / 2Calculating each term:3^1 is 3, and 3^{-1} is 1/3. So,f(1) = (3 - 1/3) / 2Let me compute the numerator first:3 - 1/3 = (9/3 - 1/3) = 8/3So, f(1) = (8/3) / 2 = (8/3) * (1/2) = 4/3Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make any mistakes.1. I started with f(x) + g(x) = 3^x.2. Then, using the properties of odd and even functions, I substituted f(-x) and g(-x) to get -f(x) + g(x) = 3^{-x}.3. I set up the system of equations and subtracted the second equation from the first to eliminate g(x).4. That gave me 2f(x) = 3^x - 3^{-x}, so f(x) = (3^x - 3^{-x}) / 2.5. Plugging in x = 1, I got f(1) = (3 - 1/3) / 2 = (8/3) / 2 = 4/3.Everything seems to check out. I don't see any errors in my calculations. So, I think f(1) is indeed 4/3.Just to be thorough, let me verify if f(x) as defined is indeed an odd function. If f(x) = (3^x - 3^{-x}) / 2, then f(-x) should equal -f(x).Let's compute f(-x):f(-x) = (3^{-x} - 3^{x}) / 2 = -(3^x - 3^{-x}) / 2 = -f(x)Yes, that works out. So, f(x) is indeed odd, which aligns with the given condition. Similarly, let's check g(x). From equation (1), g(x) = 3^x - f(x). Plugging in f(x):g(x) = 3^x - (3^x - 3^{-x}) / 2 = (2*3^x - 3^x + 3^{-x}) / 2 = (3^x + 3^{-x}) / 2Now, let's check if g(x) is even:g(-x) = (3^{-x} + 3^{x}) / 2 = (3^x + 3^{-x}) / 2 = g(x)So, g(x) is indeed even, which also matches the given condition. Therefore, my solution is consistent with the properties of odd and even functions.I think I've covered all the bases here. I used the definitions of odd and even functions to set up a system of equations, solved for f(x), and then evaluated it at x = 1. I also double-checked that the resulting functions satisfy the required properties. So, I feel confident that the value of f(1) is 4/3.**Final Answer**The value of (f(1)) is boxed{dfrac{4}{3}}.