• ### Shortcuts for x^2+y^2?

Best answer: If your willing to go into the realm of complex numbers, then you can have a shortcut. Imagine you have x² + 1, this would factor to (x + i)(x - i). The same could be applied to x² + y²: x² + y² = (x + iy)(x - iy) or (y + ix)(y - ix)
Best answer: If your willing to go into the realm of complex numbers, then you can have a shortcut. Imagine you have x² + 1, this would factor to (x + i)(x - i). The same could be applied to x² + y²: x² + y² = (x + iy)(x - iy) or (y + ix)(y - ix)
4 answers · Mathematics · 2 years ago
• ### Al wants to visit TN, NY, MA, TX, OH, CA, PA, OK, OR, & ND. If he decides to visit some, all, or none, how many travel options does he have?

Best answer: You either visit each location or you don't. So that means that for each state, there are 2 options. By the counting principle, the total number of options is the product of the number of possibilities for all options. Since there are 10 options, there are 2^10 or 1024 travel options that involve visiting all, some, or none of the... show more
Best answer: You either visit each location or you don't. So that means that for each state, there are 2 options. By the counting principle, the total number of options is the product of the number of possibilities for all options. Since there are 10 options, there are 2^10 or 1024 travel options that involve visiting all, some, or none of the options.
1 answer · Mathematics · 11 months ago
• ### Pre-Calc question help. I honestly have no idea what I am doing. Thanks for the help in advance?

Best answer: The triangle on the left is a cross section of the cone and cylinder. The triangle on the right is a "zoomed-in" portion of the right side of the triangle on the left. All you need to do is set up the proportion in the picture. Solving for h, you get 8 - 4/3r. The volume of a cylinder is πr²h. Now use h = 10 - 5r/3 and... show more
Best answer: The triangle on the left is a cross section of the cone and cylinder. The triangle on the right is a "zoomed-in" portion of the right side of the triangle on the left. All you need to do is set up the proportion in the picture. Solving for h, you get 8 - 4/3r. The volume of a cylinder is πr²h. Now use h = 10 - 5r/3 and solve for V: V = πr²(10 - 5r/3) V = 10πr² - 5πr³/3 Now use r = 6 - 3h/5 to solve for V in terms of h: V = π(6 - 3h/5)²h V = π(36 - 36h/5 + 9h²/25)h V = 36hπ - 36πh²/5 + 9πh³/25
2 answers · Mathematics · 11 months ago
• ### If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8 m from the dock?

Best answer: This is the picture that you should have to aid in solving this problem. From it, you get: x² + y² = L² Differentiate with respect to time: d/dt[x² + y²] - d/d[L²] 2x*dx/dt + 2y*dy/dt = 2L*dL/dt Now you want to find how fast the boat is approaching the dock which is dx/dt. Solve for dx/dt: x*dx/dt = L*dL/dt - y*dy/dt dx/dt =... show more
Best answer: This is the picture that you should have to aid in solving this problem. From it, you get: x² + y² = L² Differentiate with respect to time: d/dt[x² + y²] - d/d[L²] 2x*dx/dt + 2y*dy/dt = 2L*dL/dt Now you want to find how fast the boat is approaching the dock which is dx/dt. Solve for dx/dt: x*dx/dt = L*dL/dt - y*dy/dt dx/dt = (L*dL/dt - y*dy/dt)/x x = 8 (Given) y = 1 (Given) dy/dt = 0 (the height of the pulley is constant) dL/dt = 1 (Given) L = √65 (Calculated from Pythagorean theorem) dx/dt = (√65*1 - 1*0)/8 dx/dt = √65/8 m/s ≈ 1.007782219 m/s
1 answer · Mathematics · 11 months ago

Best answer: You have: cos(x)tan(x) + √3cos(x) = 0 Factor out cos(x) cos(x)(tan(x) + √3) = 0 This means that cos(x) = 0 or tan(x) + √3 = 0 cos(x) = 0 when x = 0 or x = π These occur with a period of 2π, but since it is at multiples of π, then cos(x) = 0, when x = 0 + πk tan(x) + √3 = 0 tan(x) = -√3 tan(x) = -√3 when x = 2π/3 or x =... show more
Best answer: You have: cos(x)tan(x) + √3cos(x) = 0 Factor out cos(x) cos(x)(tan(x) + √3) = 0 This means that cos(x) = 0 or tan(x) + √3 = 0 cos(x) = 0 when x = 0 or x = π These occur with a period of 2π, but since it is at multiples of π, then cos(x) = 0, when x = 0 + πk tan(x) + √3 = 0 tan(x) = -√3 tan(x) = -√3 when x = 2π/3 or x = 5π/3 tan(x) has a period of π, so then tan(x) = --√3 (tan(x) + √3 = 0) when x = 2π/3 + πk or x = 5π/3 + πk You could actually write this as tan(x) = --√3 (tan(x) + √3 = 0) when x = 2π/3 + πk This would actually encompass both 2π/3 + πk and 5π/3 + πk It's probably looking for: x = 0 + 2πk x = 2π/3 + πk x = π + 2πk
1 answer · Mathematics · 12 months ago
• ### Prove (explain how) the sum of harmonics y = a cos(2πvt) + b sin(2πvt) can be expressed as a single harmonic oscillator, that is?

Best answer: First thing, linear combinations of sinusoids can only be combined if the frequency of each sinusoid is the same. For instance: 5 sin⁡(2x)+4cos⁡(2x-5) sin⁡(4x-5)-9 cos⁡(4x+2) can all be transformed into the form A sin⁡(Bx-C). However: 3 sin⁡(3x)+8cos⁡(x+3) 6sin⁡(3x-5)-9 cos⁡(8x+1) cannot be transformed into the form A sin⁡(Bx-C). ... show more
Best answer: First thing, linear combinations of sinusoids can only be combined if the frequency of each sinusoid is the same. For instance: 5 sin⁡(2x)+4cos⁡(2x-5) sin⁡(4x-5)-9 cos⁡(4x+2) can all be transformed into the form A sin⁡(Bx-C). However: 3 sin⁡(3x)+8cos⁡(x+3) 6sin⁡(3x-5)-9 cos⁡(8x+1) cannot be transformed into the form A sin⁡(Bx-C). To combine linear combinations of sine and cosine functions you need to know the difference formula for sine: sin⁡(x-y)=sin⁡(x) cos⁡(y)-cos⁡(x) sin⁡(y) Begin with the linear combination: a sin⁡(Bx)+b cos⁡(Bx) You want to write this as a single sinusoid: a sin⁡(Bx)+b cos⁡(Bx)=A sin⁡(Bx-C) Begin by re-writing the right hand side using the difference formula: a sin⁡(Bx)+b cos⁡(Bx)=A(sin⁡(Bx) cos⁡(C)-cos⁡(Bx) sin⁡(C) ) a sin⁡(Bx)+b cos⁡(Bx)=A sin⁡(Bx) cos⁡(C)-A cos⁡(Bx) sin⁡(C) a sin⁡(Bx)+b cos⁡(Bx)=A 〖 cos⁡(C)sin〗⁡(Bx)-A sin⁡(C) cos⁡(Bx) Since both sides are equal, the coefficients of sin⁡(Bx) and cos⁡(Bx) are also equal: b=-A sin⁡(C) a=A cos⁡(C) Divide the top equation by the bottom equation to get: b/a=-(A sin⁡(C))/(A cos⁡(C) ) This reduces to: -b/a=tan⁡(C) Solve for C: C=tan^(-1)⁡(-b/a) Return to the equations equating the coefficients: b=-A sin⁡(C) a=A cos⁡(C) Square both sides of both equations: b^2=(-A sin⁡(C) )^2 a^2=(A cos⁡(C) )^2 This turns into: b^2=A^2 sin^2⁡(C) a^2=A^2 cos^2⁡(C) Add both equations to get: a^2+b^2=A^2 sin^2⁡(C)+A^2 cos^2⁡(C) a^2+b^2=A^2 (sin^2⁡(C)+cos^2⁡(C) ) Remember that sin^2⁡(C)+cos^2⁡(C)=1: A^2=a^2+b^2 A=√(a^2+b^2 ) To recap, given: a sin⁡(Bx)+b cos⁡(Bx) This can be re-written as: A sin⁡(Bx-C) A=√(a^2+b^2 ) C=tan^(-1)⁡(-b/a) Now for when the phase shift is different: Suppose you're given: A sin⁡(ωt+α)+B cos⁡(ωt+β) To write this as a single sinusoid you want: A sin⁡(ωt+α)+B cos⁡(ωt+β)=C sin⁡(ωt+θ) To solve this, you need to know the sum formulas for both sine and cosine: sin⁡(x+y)=sin⁡(x) cos⁡(y)+cos⁡(x) sin⁡(y) cos⁡〖(x+y)=cos⁡(x) cos⁡(y)-sin⁡(x) sin⁡(y) 〗 Also to help keep track of which item belongs to which trig function on the left hand side we'll redefine some of them: A=A_s B=A_c α=θ_s β=θ_c Re-write using the modifications: A_s sin⁡(ωt+θ_s )+A_c cos⁡(ωt+θ_c )=A sin⁡(ωt+θ) Re-write both sides using sum and difference formulas: A_s cos⁡(θ_s ) sin⁡(ωt)+A_s sin⁡(θ_s ) cos⁡(ωt)+A_c cos⁡(θ_c ) cos⁡(ωt)-A_c sin⁡(θ_c ) sin⁡(ωt)=A cos⁡(θ) sin⁡(ωt)+A sin⁡(θ) cos⁡(ωt) Re-write the left hand side by grouping sin⁡(ωt) and cos⁡(ωt): (A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) ) sin⁡(ωt)+(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) ) cos⁡(ωt)=A cos⁡(θ) sin⁡(ωt)+A sin⁡(θ) cos⁡(ωt) Equate the coefficients of sin⁡(ωt) and cos⁡(ωt): A sin⁡(θ)=A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) A cos⁡(θ)=A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) Divide the top equation by the bottom top to get: (A sin⁡(θ))/(A cos⁡(θ) )=(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) ) Simplify and solve for θ: tan⁡(θ)=(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) ) θ=tan^(-1)⁡((A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) )) Now return to the equations that equated the coefficients: A sin⁡(θ)=A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) A cos⁡(θ)=A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) Square both sides: (A sin⁡(θ) )^2=(A_s sin⁡(θ_s )+A_c cos⁡(θ_c ) )^2 (A cos⁡(θ) )^2=(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) )^2 A^2 sin^2⁡(θ)=〖A_s〗^2 sin^2⁡(θ_s )+2A_s A_c sin⁡〖(θ_s ) cos⁡(θ_c ) 〗+〖A_c〗^2 cos^2⁡(θ_c ) A^2 cos^2⁡(θ)=〖A_c〗^2 sin^2⁡(θ_c )-2A_s A_c sin⁡〖(θ_c ) cos⁡(θ_s ) 〗+〖A_s〗^2 cos^2⁡(θ_s ) Add both equations and reduce (remember that sin^2⁡(x)+cos^2⁡(x)=1): A^2 sin^2⁡(θ)+A^2 cos^2⁡(θ)=〖A_s〗^2 sin^2⁡(θ_s )+2A_s A_c sin⁡〖(θ_s ) cos⁡(θ_c ) 〗+〖A_c〗^2 cos^2⁡(θ_c )+〖A_c〗^2 sin^2⁡(θ_c )-2A_s A_c sin⁡〖(θ_c ) cos⁡(θ_s ) 〗+〖A_s〗^2 cos^2⁡(θ_s ) A^2=〖A_c〗^2+〖A_s〗^2+2A_s A_c (sin⁡〖(θ_s ) cos⁡(θ_c ) 〗-cos⁡(θ_s ) sin⁡(θ_c ) ) Remember the difference formula for sine: sin⁡(x-y)=sin⁡(x) cos⁡(y)-cos⁡(x) sin⁡(y) Reduce the right hand side and solve for A: A^2=〖A_c〗^2+〖A_s〗^2+2A_s A_c (sin⁡〖(θ_s ) cos⁡(θ_c ) 〗-cos⁡(θ_s ) sin⁡(θ_c ) ) A^2=〖A_c〗^2+〖A_s〗^2+2A_s A_c sin⁡(θ_s-θ_c ) A=√(〖A_c〗^2+〖A_s〗^2+2A_s A_c sin⁡(θ_s-θ_c ) ) To recap, given: A_s sin⁡(ωt+θ_s )+A_c cos⁡(ωt+θ_c ) This can be re-written as: A sin⁡(ωt+θ) A=√(〖A_c〗^2+〖A_s〗^2+2A_s A_c sin⁡(θ_s-θ_c ) ) θ=tan^(-1)⁡((A_s sin⁡(θ_s )+A_c cos⁡(θ_c ))/(A_s cos⁡(θ_s )-A_c sin⁡(θ_c ) )) Hope all of this helps
1 answer · Mathematics · 11 months ago
• ### Rationalize the denominator?

Best answer: In this instance, 4/(1-√3) you multiply by the conjugate of the denominator which in this case is 1+√3 to get: 4/(1-√3) x (1+√3)/(1+√3) = 4(1+√3)/(1 - 3) = 4(1+√3)/-2 = -2(1+√3)
Best answer: In this instance, 4/(1-√3) you multiply by the conjugate of the denominator which in this case is 1+√3 to get: 4/(1-√3) x (1+√3)/(1+√3) = 4(1+√3)/(1 - 3) = 4(1+√3)/-2 = -2(1+√3)
2 answers · Mathematics · 12 months ago
• ### 9 – 5 ÷ (8 – 3) × 3 + 6?

Best answer: PEMDAS is a bit of "marketing". You don't strictly do multiplication before any division. Think of it more as steps. On the first step is parenthesis. The next step is exponents. On the next step is multiplication and division. On the last step is addition and subtraction. Since multiplication and division are on the... show more
Best answer: PEMDAS is a bit of "marketing". You don't strictly do multiplication before any division. Think of it more as steps. On the first step is parenthesis. The next step is exponents. On the next step is multiplication and division. On the last step is addition and subtraction. Since multiplication and division are on the same step you do them simultaneously. Same with addition and subtraction. You do them simultaneously from left to right as you read. So that is why you divide -5 by 5 before multiplying 5 by 3.
6 answers · Mathematics · 12 months ago
• ### Can anyone help me with this.....How many six-digit palindromic numbers are there which are greater than 250000?pls explain it ty?

Best answer: Consider the number abcdef, where each of the digits may be different or the same. So in order for abcdef to be a palindrome you have def = cba. Basically the first 3 numbers would necessarily determine the last 3 numbers. But since you want the number to be greater than 250000, the first 3 numbers must be at least 250. So this now... show more
Best answer: Consider the number abcdef, where each of the digits may be different or the same. So in order for abcdef to be a palindrome you have def = cba. Basically the first 3 numbers would necessarily determine the last 3 numbers. But since you want the number to be greater than 250000, the first 3 numbers must be at least 250. So this now turns into the problem of determining how many 3 digit numbers are greater than or equal to 250. So you have all 3 digit numbers from 250 to 999. If you take the difference of 999 and 250 you get 749 but that only includes one of the end points. So you need to add 1 to that to get both end points. So there are 750 6 digit palindromes greater than 250000.
3 answers · Mathematics · 12 months ago
• ### Can you solve this math problem?

Best answer: This is the picture that you should have: A couple of things to note: 475 = x + y tan(18°) = x/L tan(2°) = y/L Solve for y: y = 475 - x Substitute: tan(18°) = x/L tan(2°) = (475 - x)/L Since you want to solve for L, solve for x in one of the equations: x = Ltan(18°) Substitute this into tan(2°) = (475 - x)/L: tan(2°) = (475 -... show more
Best answer: This is the picture that you should have: A couple of things to note: 475 = x + y tan(18°) = x/L tan(2°) = y/L Solve for y: y = 475 - x Substitute: tan(18°) = x/L tan(2°) = (475 - x)/L Since you want to solve for L, solve for x in one of the equations: x = Ltan(18°) Substitute this into tan(2°) = (475 - x)/L: tan(2°) = (475 - Ltan(18°))/L Now solve for L: tan(2°) = (475 - Ltan(18°))/L Ltan(2°) = 475 - Ltan(18°) 475 = Ltan(2°) + Ltan(18°) 475 = L(tan(2°) + tan(18°)) L = 475/(tan(2°) + tan(18°)) L ≈ 1320.029416 feet
3 answers · Mathematics · 1 year ago
• ### Hard Vector Curvature Problem, Help!!?

Best answer: So try this: r(t) = -4sin(t)i - 4sin(t)j + cos(t)k κ(t) = |T(t)|/|r'(t)| (curvature) r'(t) -4cos(t)i - 4cos(t)j - sin(t)k |r'(t)| = √((-4cost)² + (-4cost)² + (-sint)²) |r'(t)| = √(16cos²t + 16cos²t + sin²t) |r'(t)| = √(32cos²t + sin²t) Now you can pull out a cos²t: |r'(t)| = √(cos²t(32 +... show more
Best answer: So try this: r(t) = -4sin(t)i - 4sin(t)j + cos(t)k κ(t) = |T(t)|/|r'(t)| (curvature) r'(t) -4cos(t)i - 4cos(t)j - sin(t)k |r'(t)| = √((-4cost)² + (-4cost)² + (-sint)²) |r'(t)| = √(16cos²t + 16cos²t + sin²t) |r'(t)| = √(32cos²t + sin²t) Now you can pull out a cos²t: |r'(t)| = √(cos²t(32 + sin²t/cos²t)) |r'(t)| = cos(t)√(32 + tan²t) Now find T(t) (T(t) = r'(t)/|r'(t)|) r'(t) = -4cos(t)i - 4cos(t)j - sin(t)k |r'(t)| = cos(t)√(32 + tan²t) T(t) = (-4cos(t)i - 4cos(t)j - sin(t)k)/(cos(t)√(32 + tan²t)) T(t) = -4/√(32 + tan²t)i - 4/√(32 + tan²t)j - tant/√(32 + tan²t)k |T(t)| = √((-4/√(32 + tan²t))² + (-4/√(32 + tan²t))² + (tant/√(32 + tan²t))²) |T(t)| = √(16/(32 + tan²t)) + 16/(32 + tan²t)) + tan²t/(32 + tan²t))) |T(t)| = √((16 + 16 + tan²t)/(32 + tan²t)) |T(t)| = √((32 + tan²t)/(32 + tan²t)) |T(t)| = √1 = 1 Now back to κ(t): κ(t) = |T(t)|/|r'(t)| κ(t) = 1/√(32cos²t + sin²t) Now at this point you could try a few different representations: κ(t) = sec(t)/√(32 + tan²t) κ(t) = sec(t)/√(31 + sec²t) κ(t) = 1/√(31cos²t + 1) I could go on and on with these. Try κ(t) = 1/√(32cos²t + sin²t) and see if that works.
1 answer · Mathematics · 1 year ago

Best answer: This problem involves a few steps. First you have to calculate the payment on the 30 year mortgage, the remaining balance at 10 years on the 30 year mortgage and the payment on the 15 year mortgage. The formula for a payment on a mortgage is: p = rL(1 + r)^t/((1 + r)^t - 1) Where: r = monthly rate L = Loan amount t = length of the... show more
Best answer: This problem involves a few steps. First you have to calculate the payment on the 30 year mortgage, the remaining balance at 10 years on the 30 year mortgage and the payment on the 15 year mortgage. The formula for a payment on a mortgage is: p = rL(1 + r)^t/((1 + r)^t - 1) Where: r = monthly rate L = Loan amount t = length of the loan For the 30 year mortgage: r = 0.007 ---> (.084/12) L = \$127,262.85 ( \$149,721.00*.85 ---> Amount after 15% down) t = 360 (30 years * 12 months/year) The payment for the 30 year mortgage is \$969.54. Now you need to calculate the balance of this loan at 10 years. The balance, B, is given by the following eqaution: B = L(1 + r)^t - p/r*((1 + r)^t - 1) Where: r = 0.007 ---> (.084/12) L = \$127,262.85 ( \$149,721.00*.85 ---> Amount after 15% down) t = 120 (10 years * 12 months/year) p = \$969.54 (as calculated previously) After 10 years the balance of the loan is \$112,539.90 and there are 240 payments left. So this person will have to make 240 payments of \$969.54. This means this person will pay \$232,688.72 (240 * \$969.54) over the next 20 years, but only \$112,539.90 is principal (the remaining amount owed). So that means that the difference of \$232,688.72 and \$112,539.90 is the amount of interest this person will pay over the next 20 years. The amount of interest they will pay is \$120,148.81 (\$232,688.72 - \$112,539.90). Now calculate the payment on a 15 year mortgage using the payment formula above and the following amounts: L = \$112,539.90 (the remaining balance on the mortgage after 10 years) r = 0.0045 (.054/12) t = 180 (15 years * 12 months/year) You should get a payment of \$913.58 for the 15 year loan. Again, the person will make 180 payments of \$913.58 for a total of \$164,445.09 (180 * \$913.58). But only \$112,539.90 of that amount is principal, the rest is interest. The total amount of interest paid on the 15 year mortgage is \$51,905.19 (\$164,445.09 - \$112,539.90). Now to calculate the interest savings, subtract the 2 interest amounts. \$120,148.81 - \$51,905.19 = \$68,243.62 This means that the person would save \$68,243.62 by switching to the 15 year mortgage.
3 answers · Mathematics · 2 years ago
• ### What does ||c'(t)|| mean?

Best answer: c(t) is probably a vector valued function. c'(t) is the derivative of that function. The bars indicate the magnitude (the length of the vector) of that function.
Best answer: c(t) is probably a vector valued function. c'(t) is the derivative of that function. The bars indicate the magnitude (the length of the vector) of that function.
2 answers · Mathematics · 2 years ago
• ### A closed box with square base is to be built to house an ant colony...?

Best answer: The picture should help
Best answer: The picture should help
1 answer · Mathematics · 2 years ago

2 answers · Mathematics · 2 years ago
• ### What is an even rational function and what is an odd rational function? How do you know which one it is based on the equation and/or graph?

Best answer: The easiest way to tell if a function is even, odd, or neither is to look at it's equation. Even functions only have terms with even exponents. Odd functions have terms that only have odd exponents. Equations that contain terms with both odd and even exponents are neither odd nor even. For example: 3x² + 7 → even (remember that... show more
Best answer: The easiest way to tell if a function is even, odd, or neither is to look at it's equation. Even functions only have terms with even exponents. Odd functions have terms that only have odd exponents. Equations that contain terms with both odd and even exponents are neither odd nor even. For example: 3x² + 7 → even (remember that the 7 is really 7x⁰) exponents are 0 and 2, both even numbers, function is even 5x³ - 9x → odd since exponents are 1 and 3 5x³ + 3x² - 9x - 7 → neither odd nor even Now for rational functions, there'e one more layer to consider. For rational functions: Even/Even → even function Odd/Odd → even function Even/Odd → odd function Odd/Even → odd function
4 answers · Mathematics · 2 years ago
• ### Y=arc csc(2sqrt3/3) give the exact value of y in radians. Rmember that each inverse trig function has a certain interval.?

Best answer: Given y = csc⁻¹(x), that is the same as y = sin⁻¹(1/x) You have y = csc⁻¹((2√3)/3). This becomes: y = sin⁻¹(3/(2√3)) 3/(2√3) = (3√3)/6 = √3/2 y = sin⁻¹(√3/2) The range of csc⁻¹(x) is [−π/2,0) ∪ (0,π/2] (sin⁻¹(x) is [−π/2,π/2], nearly identical). So you need to find when sin is √3/2 and is in the range [−π/2,π/2]. That occurs at π/3. show more
Best answer: Given y = csc⁻¹(x), that is the same as y = sin⁻¹(1/x) You have y = csc⁻¹((2√3)/3). This becomes: y = sin⁻¹(3/(2√3)) 3/(2√3) = (3√3)/6 = √3/2 y = sin⁻¹(√3/2) The range of csc⁻¹(x) is [−π/2,0) ∪ (0,π/2] (sin⁻¹(x) is [−π/2,π/2], nearly identical). So you need to find when sin is √3/2 and is in the range [−π/2,π/2]. That occurs at π/3.
4 answers · Mathematics · 2 years ago

Best answer: Given a + bi, to find θ, use tan⁻¹(b/a) to find the reference angle. Then based upon the angle you get, you have to put it into the correct quadrant. Think of a as x and b as y. So in the first example, θ = tan⁻¹(2/2), where both x and y are positive (QI). In the second example you have θ = tan⁻¹(4/-3) where x is negative and y is... show more
Best answer: Given a + bi, to find θ, use tan⁻¹(b/a) to find the reference angle. Then based upon the angle you get, you have to put it into the correct quadrant. Think of a as x and b as y. So in the first example, θ = tan⁻¹(2/2), where both x and y are positive (QI). In the second example you have θ = tan⁻¹(4/-3) where x is negative and y is positive (QII). Then to find r use √(a² + b²). For a) θ = tan⁻¹(2/2) = π/4, and since it is in QI, that's the end of it. For r you get r = √(2² + 2²) = √8 = 2√2 θ = π/4 r = 2√2 For c) θ = tan⁻¹(4/-3). There is no exact value, you get θ = -0.9272952, which means the reference angle is 0.9272952. You want it in QII, so that means that θ = tan⁻¹(4/-3) + π. That is as exact as you can get, or you can write it as θ = 2.2142974. As for t, you get r = √((-3)² + 4²) = √(25) = 5 θ = tan⁻¹(4/-3) + π ≈ 2.2142974 r = 5
2 answers · Mathematics · 2 years ago