💡Alright, so I've got this problem here about a function and a triangle. Let me try to figure it out step by step.First, part (1) says: Given the function ( f(x) = 2sin x cdot cos^2 frac{phi}{2} + cos x sin phi - sin x ) which attains its minimum value at ( x = pi ). I need to find the value of ( phi ) where ( 0 < phi < pi ).Okay, so I need to analyze this function ( f(x) ). It looks like a combination of sine and cosine terms. Maybe I can simplify it first. Let me write it down again:( f(x) = 2sin x cdot cos^2 frac{phi}{2} + cos x sin phi - sin x )Hmm, I notice that there are two terms with ( sin x ). Let me factor those out:( f(x) = sin x cdot left( 2cos^2 frac{phi}{2} - 1 right) + cos x sin phi )Wait, ( 2cos^2 frac{phi}{2} - 1 ) is a trigonometric identity. Isn't that equal to ( cos phi )? Let me recall: yes, ( cos 2theta = 2cos^2 theta - 1 ). So, if I let ( theta = frac{phi}{2} ), then ( 2cos^2 frac{phi}{2} - 1 = cos phi ). Nice, that simplifies things.So, substituting that back in:( f(x) = sin x cdot cos phi + cos x cdot sin phi )Wait a second, that looks familiar. Isn't that the sine addition formula? ( sin(x + phi) = sin x cos phi + cos x sin phi ). So, that means:( f(x) = sin(x + phi) )Oh, that's much simpler! So, the function is just a sine function with a phase shift.Now, it's given that ( f(x) ) attains its minimum value at ( x = pi ). The sine function ( sin(theta) ) attains its minimum value of -1 when ( theta = frac{3pi}{2} + 2kpi ) for integer ( k ). So, for ( f(x) = sin(x + phi) ), the minimum occurs when ( x + phi = frac{3pi}{2} + 2kpi ).But since ( x = pi ) is where the minimum occurs, we can set up the equation:( pi + phi = frac{3pi}{2} + 2kpi )Solving for ( phi ):( phi = frac{3pi}{2} - pi + 2kpi = frac{pi}{2} + 2kpi )But ( phi ) is given to be between 0 and ( pi ), so ( k = 0 ) is the only possibility. Therefore, ( phi = frac{pi}{2} ).Okay, that seems straightforward. So, for part (1), ( phi = frac{pi}{2} ).Now, moving on to part (2). It says: In triangle ABC, sides ( a ), ( b ), and ( c ) are opposite to angles ( A ), ( B ), and ( C ), respectively. Given ( a = 1 ), ( b = ) [Wait, the original problem says "b=," which is incomplete. Maybe it's a typo? Let me check the original problem again.]Looking back, it says: "Given ( a = 1 ), ( b = ), and ( f(A) = frac{3}{2} ), find angle ( C )." Hmm, seems like the value of ( b ) is missing. Maybe it's supposed to be ( b = sqrt{2} ) or something? Since in the previous part, ( phi = frac{pi}{2} ), which might relate to a right triangle. Maybe ( b = sqrt{2} ). Let me assume that for now, or maybe it's another value. Wait, in the initial problem, the user wrote "b=," but in the second instance, they wrote "b=√2". Maybe it's a typo. Let me see.Wait, in the initial problem, the user wrote:"In triangle ABC, a, b, and c are the sides opposite to angles A, B, and C, respectively. Given a=1, b=, and f(A)=frac{3}{2}, find angle C."So, it's incomplete. But in the second instance, the user wrote "b=√2". Maybe that's the intended value. Let me proceed with ( b = sqrt{2} ).So, given ( a = 1 ), ( b = sqrt{2} ), and ( f(A) = frac{3}{2} ), find angle ( C ).First, since ( f(x) = sin(x + phi) ) and we found ( phi = frac{pi}{2} ), so ( f(x) = sin(x + frac{pi}{2}) ). Using the sine addition formula again, ( sin(x + frac{pi}{2}) = cos x ). So, ( f(x) = cos x ).Wait, let me verify that. ( sin(x + frac{pi}{2}) = sin x cos frac{pi}{2} + cos x sin frac{pi}{2} = sin x cdot 0 + cos x cdot 1 = cos x ). Yes, that's correct. So, ( f(x) = cos x ).Therefore, ( f(A) = cos A = frac{3}{2} ). Wait, hold on. The cosine of an angle can't be greater than 1. ( cos A = frac{3}{2} ) is impossible because the maximum value of cosine is 1. That doesn't make sense. Did I make a mistake?Wait, let me go back. The function ( f(x) ) was given as ( 2sin x cdot cos^2 frac{phi}{2} + cos x sin phi - sin x ). Then, after simplifying, I concluded that ( f(x) = sin(x + phi) ). But when ( phi = frac{pi}{2} ), then ( f(x) = sin(x + frac{pi}{2}) = cos x ). So, ( f(A) = cos A ).But the problem states that ( f(A) = frac{3}{2} ). But ( cos A ) can't be ( frac{3}{2} ). That's impossible. So, perhaps I made a mistake in simplifying ( f(x) ).Wait, let me double-check the simplification. Starting again:( f(x) = 2sin x cdot cos^2 frac{phi}{2} + cos x sin phi - sin x )Factor out ( sin x ):( f(x) = sin x cdot left( 2cos^2 frac{phi}{2} - 1 right) + cos x sin phi )As before, ( 2cos^2 frac{phi}{2} - 1 = cos phi ). So:( f(x) = sin x cdot cos phi + cos x cdot sin phi )Which is ( sin(x + phi) ). So, that seems correct.But then, when ( phi = frac{pi}{2} ), ( f(x) = sin(x + frac{pi}{2}) = cos x ). So, ( f(A) = cos A = frac{3}{2} ). But that's impossible because cosine can't exceed 1. So, perhaps I made a mistake in part (1)?Wait, let me check part (1) again. The function ( f(x) ) attains its minimum at ( x = pi ). So, ( f(pi) = sin(pi + phi) = -1 ). So, ( sin(pi + phi) = -1 ). But ( sin(pi + phi) = -sin phi ). So, ( -sin phi = -1 ), which implies ( sin phi = 1 ). Therefore, ( phi = frac{pi}{2} ). That seems correct.So, if ( phi = frac{pi}{2} ), then ( f(x) = cos x ). So, ( f(A) = cos A = frac{3}{2} ). But that's impossible. So, maybe the problem is misstated?Wait, looking back at the original problem, it says: "Given ( a=1 ), ( b= ), and ( f(A)=frac{3}{2} ), find angle ( C )." So, perhaps the value of ( b ) is not ( sqrt{2} ) but something else? Or maybe ( f(A) = frac{sqrt{3}}{2} ) instead of ( frac{3}{2} )? Because ( frac{sqrt{3}}{2} ) is a valid cosine value.Alternatively, maybe I misapplied the function. Let me check again.Wait, ( f(x) = sin(x + phi) ). So, ( f(A) = sin(A + phi) = frac{3}{2} ). But sine can't exceed 1 either. So, that's also impossible. Hmm, this is confusing.Wait, perhaps I made a mistake in the simplification of ( f(x) ). Let me go through it again.Starting with:( f(x) = 2sin x cdot cos^2 frac{phi}{2} + cos x sin phi - sin x )Let me compute ( 2cos^2 frac{phi}{2} ). Using the double-angle identity, ( cos phi = 2cos^2 frac{phi}{2} - 1 ). So, ( 2cos^2 frac{phi}{2} = cos phi + 1 ). Therefore, substituting back:( f(x) = sin x (cos phi + 1) + cos x sin phi - sin x )Simplify the ( sin x ) terms:( f(x) = sin x (cos phi + 1 - 1) + cos x sin phi )Which simplifies to:( f(x) = sin x cos phi + cos x sin phi )Which is indeed ( sin(x + phi) ). So, that part is correct.So, if ( f(x) = sin(x + phi) ), then ( f(A) = sin(A + phi) = frac{3}{2} ). But sine can't be more than 1. So, this is impossible. Therefore, there must be a mistake in the problem statement or my understanding.Wait, maybe ( f(A) = frac{sqrt{3}}{2} ) instead? Because ( frac{sqrt{3}}{2} ) is a valid value for sine. Let me assume that for a moment.If ( f(A) = sin(A + phi) = frac{sqrt{3}}{2} ), then ( A + phi = frac{pi}{3} ) or ( frac{2pi}{3} ). But since ( phi = frac{pi}{2} ), then ( A + frac{pi}{2} = frac{pi}{3} ) or ( frac{2pi}{3} ).Solving for ( A ):Case 1: ( A + frac{pi}{2} = frac{pi}{3} ) → ( A = frac{pi}{3} - frac{pi}{2} = -frac{pi}{6} ). Not possible, since angles can't be negative.Case 2: ( A + frac{pi}{2} = frac{2pi}{3} ) → ( A = frac{2pi}{3} - frac{pi}{2} = frac{pi}{6} ). Okay, that works.So, ( A = frac{pi}{6} ).Now, in triangle ABC, we have sides ( a = 1 ), ( b = sqrt{2} ), and angle ( A = frac{pi}{6} ). We need to find angle ( C ).Using the Law of Sines: ( frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ).First, compute ( frac{a}{sin A} ):( frac{1}{sin frac{pi}{6}} = frac{1}{frac{1}{2}} = 2 ).So, ( frac{b}{sin B} = 2 ) → ( sin B = frac{b}{2} = frac{sqrt{2}}{2} ).Therefore, ( B = frac{pi}{4} ) or ( B = frac{3pi}{4} ).But in a triangle, the sum of angles is ( pi ). So, if ( A = frac{pi}{6} ) and ( B = frac{pi}{4} ), then ( C = pi - frac{pi}{6} - frac{pi}{4} = pi - frac{2pi}{12} - frac{3pi}{12} = pi - frac{5pi}{12} = frac{7pi}{12} ).If ( B = frac{3pi}{4} ), then ( C = pi - frac{pi}{6} - frac{3pi}{4} = pi - frac{2pi}{12} - frac{9pi}{12} = pi - frac{11pi}{12} = frac{pi}{12} ).But we need to check if both solutions are valid. Since ( b = sqrt{2} ) is longer than ( a = 1 ), angle ( B ) must be larger than angle ( A ). ( frac{pi}{4} ) is larger than ( frac{pi}{6} ), and ( frac{3pi}{4} ) is also larger. So, both are possible.Therefore, angle ( C ) can be either ( frac{7pi}{12} ) or ( frac{pi}{12} ).But wait, in the original problem, it was stated that ( f(A) = frac{3}{2} ), which led to a contradiction. So, maybe the correct value is ( frac{sqrt{3}}{2} ), as I assumed earlier. Otherwise, the problem is inconsistent.Assuming that, then the answer for part (2) is angle ( C = frac{7pi}{12} ) or ( frac{pi}{12} ).But let me double-check if ( b = sqrt{2} ) is correct. If ( b ) was different, say ( b = 2 ), then ( sin B = frac{2}{2} = 1 ), so ( B = frac{pi}{2} ), and ( C = pi - frac{pi}{6} - frac{pi}{2} = frac{pi}{3} ). But since the problem didn't specify ( b ), I think my assumption of ( b = sqrt{2} ) is reasonable, especially since ( phi = frac{pi}{2} ) suggests a right angle somewhere.Alternatively, maybe ( b ) is not given, and I need to find it. But the problem says "Given ( a = 1 ), ( b = ), and ( f(A) = frac{3}{2} )", which is incomplete. So, perhaps there was a typo, and ( b = sqrt{2} ) is the intended value.In conclusion, for part (1), ( phi = frac{pi}{2} ). For part (2), assuming ( b = sqrt{2} ) and correcting ( f(A) ) to ( frac{sqrt{3}}{2} ), angle ( C ) is either ( frac{7pi}{12} ) or ( frac{pi}{12} ).But since the problem states ( f(A) = frac{3}{2} ), which is impossible, there might be an error in the problem statement. However, based on the given information and assuming a typo, the answer would be as above.

💡Okay, so I have this problem here: Let z be a complex number such that the absolute value of z, which is written as |z|, equals 8. I need to find the product of z and its conjugate, which is written as z multiplied by the conjugate of z, or z times overline{z}. Hmm, okay, let me think about how to approach this.First, I remember that complex numbers have a real part and an imaginary part. So, a complex number z can be written as z = a + bi, where a is the real part and b is the imaginary part, and i is the square root of -1. The conjugate of z, which is written as overline{z}, would then be a - bi. So, if z is a + bi, overline{z} is a - bi.Now, the problem says that the absolute value of z is 8. The absolute value of a complex number is also known as its modulus, and it represents the distance of the point from the origin in the complex plane. The formula for the modulus of z is |z| = sqrt(a^2 + b^2). So, in this case, sqrt(a^2 + b^2) = 8.I need to find z multiplied by its conjugate, which is z times overline{z}. Let me write that out using the expressions for z and overline{z}:z * overline{z} = (a + bi)(a - bi).Hmm, this looks like the product of a binomial and its conjugate. I remember that when you multiply a binomial by its conjugate, you get a difference of squares. So, (a + bi)(a - bi) should be equal to a^2 - (bi)^2.Let me compute that:(a + bi)(a - bi) = a^2 - (bi)^2.Now, (bi)^2 is equal to b^2 * i^2, and since i^2 is -1, this becomes b^2 * (-1) = -b^2. So, substituting back in:a^2 - (bi)^2 = a^2 - (-b^2) = a^2 + b^2.Wait a minute, that's interesting. So, z multiplied by its conjugate is equal to a^2 + b^2, which is exactly the square of the modulus of z. Because earlier, I wrote that |z| = sqrt(a^2 + b^2), so |z|^2 would be (sqrt(a^2 + b^2))^2, which is just a^2 + b^2.Therefore, z * overline{z} = |z|^2.Given that |z| = 8, then |z|^2 is 8 squared, which is 64. So, z multiplied by its conjugate is 64.Let me double-check my steps to make sure I didn't make any mistakes. I started by expressing z and its conjugate, multiplied them together, recognized the pattern of a difference of squares, expanded it, and realized that it simplifies to a^2 + b^2, which is the square of the modulus. Since the modulus is given as 8, squaring that gives 64. That seems correct.I think another way to think about it is that multiplying a complex number by its conjugate always gives a real number, specifically the square of its modulus. This is a useful property because it allows us to find the modulus squared without having to compute the square root, which can be handy in various applications.So, in summary, z times overline{z} equals |z| squared, which is 8 squared, so 64. I feel confident about this answer now.

💡Okay, so I have this problem here with vectors. It says that vectors a and b are parallel, and I need to find the value of 2x + y. Let me try to figure this out step by step.First, I remember that if two vectors are parallel, one must be a scalar multiple of the other. That means each component of vector a should be proportional to the corresponding component of vector b. So, if vector a is (2, 1, x) and vector b is (4, y, -1), there should be some constant k such that:2 = k * 4 1 = k * y x = k * (-1)Hmm, okay. So, let me write that down:1. 2 = 4k 2. 1 = yk 3. x = -kFrom the first equation, 2 = 4k, I can solve for k. Dividing both sides by 4 gives me k = 2/4, which simplifies to k = 1/2. So, k is 1/2.Now, using this value of k in the second equation, 1 = yk, which becomes 1 = y*(1/2). To solve for y, I can multiply both sides by 2, giving me y = 2. So, y is 2.Next, using k = 1/2 in the third equation, x = -k, which becomes x = -(1/2). So, x is -1/2.Now, the problem asks for 2x + y. Let me plug in the values I found:2x + y = 2*(-1/2) + 2Calculating 2*(-1/2) gives me -1. Then, adding 2 to that gives me -1 + 2 = 1.Wait, let me double-check to make sure I didn't make any mistakes. So, k is 1/2, y is 2, x is -1/2. Plugging back into the original vectors:Vector a would be (2, 1, -1/2) and vector b would be (4, 2, -1). Is vector a a scalar multiple of vector b?Let's see: If I multiply vector a by 2, I get (4, 2, -1), which is exactly vector b. So, yes, that checks out. Therefore, my calculations seem correct.So, 2x + y is indeed 1.

💡Okay, so I have this problem about an arithmetic sequence. Let me see... The sequence is denoted as {a_n}, and I'm told that the first term, a₁, is 5. Also, the sum of the sixth term and the eighth term is 58. I need to find the common difference, d.Alright, arithmetic sequences have a common difference between consecutive terms. That means each term is the previous term plus d. So, the nth term can be expressed as a₁ plus (n-1) times d. The formula for the nth term is a_n = a₁ + (n-1)d.Let me write that down for clarity:a_n = a₁ + (n - 1)dGiven that a₁ is 5, I can plug that into the formula:a_n = 5 + (n - 1)dNow, I need to find a₆ and a₈ because their sum is given as 58. Let me calculate each of these terms separately.Starting with a₆:a₆ = 5 + (6 - 1)da₆ = 5 + 5dOkay, that seems straightforward. Now, a₈:a₈ = 5 + (8 - 1)da₈ = 5 + 7dAlright, so a₆ is 5 + 5d and a₈ is 5 + 7d. The problem says that a₆ + a₈ = 58. Let me write that equation out:a₆ + a₈ = 58(5 + 5d) + (5 + 7d) = 58Now, let's simplify the left side of the equation. Combine like terms:5 + 5d + 5 + 7d = 58First, combine the constants:5 + 5 = 10Then, combine the terms with d:5d + 7d = 12dSo, putting it all together:10 + 12d = 58Now, I need to solve for d. Let's subtract 10 from both sides to isolate the term with d:10 + 12d - 10 = 58 - 1012d = 48Now, divide both sides by 12 to solve for d:12d / 12 = 48 / 12d = 4Wait, so the common difference d is 4? Let me double-check my calculations to make sure I didn't make a mistake.Starting from the beginning:a₁ = 5a₆ = 5 + 5da₈ = 5 + 7da₆ + a₈ = (5 + 5d) + (5 + 7d) = 10 + 12d = 5812d = 58 - 10 = 48d = 48 / 12 = 4Yes, that seems correct. So, the common difference d is indeed 4.Just to be thorough, let me write out the first few terms of the sequence with d = 4 to see if a₆ and a₈ add up to 58.a₁ = 5a₂ = a₁ + d = 5 + 4 = 9a₃ = a₂ + d = 9 + 4 = 13a₄ = a₃ + d = 13 + 4 = 17a₅ = a₄ + d = 17 + 4 = 21a₆ = a₅ + d = 21 + 4 = 25a₇ = a₆ + d = 25 + 4 = 29a₈ = a₇ + d = 29 + 4 = 33Now, let's add a₆ and a₈:a₆ + a₈ = 25 + 33 = 58Yes, that matches the given condition. So, my answer must be correct. The common difference d is 4.I think I've got it. The key was to express a₆ and a₈ in terms of a₁ and d, set up the equation from the given sum, and solve for d. Then, I verified by plugging the value back into the sequence to ensure that a₆ and a₈ indeed add up to 58. Everything checks out.

💡Okay, so I have this problem here: the terminal side of angle α passes through the point P(-x, -6), and cos α is given as -5/13. I need to find the value of 1/sin α + 1/tan α. Hmm, alright, let me try to figure this out step by step.First, I remember that in trigonometry, the cosine of an angle in standard position is equal to the x-coordinate divided by the radius (which is the distance from the origin to the point). So, cos α = x/r. In this case, cos α is given as -5/13, and the point is P(-x, -6). So, that means the x-coordinate is -x, and the y-coordinate is -6.Wait, hold on, the point is P(-x, -6), so the x-coordinate is actually negative x, and the y-coordinate is negative 6. So, in terms of the cosine formula, cos α = (x-coordinate)/r, which would be (-x)/r = -5/13. So, that gives me (-x)/r = -5/13. If I simplify that, the negatives cancel out, so x/r = 5/13. So, x = (5/13) * r.But I also know that the radius r is the distance from the origin to the point P(-x, -6). So, using the distance formula, r = sqrt[(-x)^2 + (-6)^2] = sqrt[x^2 + 36]. So, r = sqrt(x^2 + 36). Therefore, x = (5/13) * sqrt(x^2 + 36).Hmm, that seems a bit complicated, but maybe I can solve for x. Let me write that equation down:x = (5/13) * sqrt(x^2 + 36)Let me square both sides to eliminate the square root:x^2 = (25/169) * (x^2 + 36)Multiply both sides by 169 to get rid of the denominator:169x^2 = 25(x^2 + 36)Expand the right side:169x^2 = 25x^2 + 900Subtract 25x^2 from both sides:144x^2 = 900Divide both sides by 144:x^2 = 900 / 144Simplify that fraction:900 divided by 144 is... let's see, both are divisible by 36. 900 ÷ 36 = 25, and 144 ÷ 36 = 4. So, x^2 = 25/4.Taking the square root of both sides:x = ±5/2But wait, in the original point P(-x, -6), x is inside the negative. So, if x is positive, then -x would be negative, which is consistent with the coordinates given. If x were negative, then -x would be positive, which might not make sense because the point is in the third quadrant (both coordinates negative). So, I think x should be positive. Therefore, x = 5/2.So, the coordinates of point P are (-5/2, -6). Now, I can find the radius r:r = sqrt[(-5/2)^2 + (-6)^2] = sqrt[(25/4) + 36] = sqrt[(25/4) + (144/4)] = sqrt[169/4] = 13/2.Alright, so r is 13/2. Now, I need to find sin α and tan α.Starting with sin α, which is the y-coordinate divided by the radius. So, sin α = y/r = (-6)/(13/2) = (-6) * (2/13) = -12/13.Next, tan α is y/x, which is (-6)/(-5/2) = (-6) * (-2/5) = 12/5.Wait, hold on, tan α is sin α / cos α, right? Let me check that. Since sin α is -12/13 and cos α is -5/13, then tan α = (-12/13)/(-5/13) = (12/13)/(5/13) = 12/5. Yeah, that's correct.Now, the problem asks for 1/sin α + 1/tan α. Let's compute each term separately.First, 1/sin α is 1/(-12/13) = -13/12.Second, 1/tan α is 1/(12/5) = 5/12.So, adding them together: (-13/12) + (5/12) = (-13 + 5)/12 = (-8)/12.Simplify that fraction: divide numerator and denominator by 4, which gives -2/3.So, the value of 1/sin α + 1/tan α is -2/3.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Found x by using cos α = -5/13, which led me to x = 5/2.2. Calculated the radius r as 13/2.3. Found sin α as -12/13 and tan α as 12/5.4. Calculated 1/sin α as -13/12 and 1/tan α as 5/12.5. Added them together to get -2/3.Everything seems to check out. So, I think the answer is -2/3.**Final Answer**boxed{-dfrac{2}{3}}

💡To determine how many acute interior angles are in an obtuse triangle, let's start by recalling the definitions and properties of triangles.1. **Definition of an Obtuse Triangle**: An obtuse triangle is a triangle that has one angle greater than 90 degrees (obtuse angle) and two angles less than 90 degrees (acute angles).2. **Sum of Interior Angles in a Triangle**: The sum of the interior angles in any triangle is always 180 degrees.3. **Analyzing the Angles**: - Let the three angles of the triangle be ( A ), ( B ), and ( C ). - Without loss of generality, assume ( A ) is the obtuse angle, so ( A > 90^circ ). - The sum of the other two angles is ( B + C = 180^circ - A ). - Since ( A > 90^circ ), it follows that ( B + C < 90^circ ).4. **Determining Acute Angles**: - Both ( B ) and ( C ) must be less than 90 degrees because their sum is less than 90 degrees. - Therefore, both ( B ) and ( C ) are acute angles.5. **Conclusion**: An obtuse triangle has exactly two acute interior angles.