answer:Let ( z = -x - frac{pi}{3} ). Then ( frac{pi}{6} le z le frac{pi}{3} ), and ( frac{pi}{3} le 2z le frac{2pi}{3} ). Change the functions using ( z ): [tanleft(x + frac{5pi}{6}right) = tanleft(frac{pi}{2} - zright) = cot(z),] [tanleft(x + frac{pi}{3}right) = tanleft(frac{pi}{6} - zright) = cotleft(frac{pi}{2} - zright) = tan(z).] Thus, [ y = cot(z) - tan(z) + sin(z). ] This simplifies to [ y = frac{cos^2(z) - sin^2(z)}{sin(z)cos(z)} + sin(z) = frac{cos(2z)}{sin(z)cos(z)} + sin(z). ] Using the double angle identities, [ y = frac{cos(2z)}{frac{1}{2} sin(2z)} + sin(z) = frac{2cos(2z)}{sin(2z)} + sin(z). ] On (frac{pi}{3} le 2z le frac{2pi}{3}), (cos(2z)) is decreasing and (sin(2z)) increasing, making (frac{2cos(2z)}{sin(2z)}) decreasing. Similarly, (sin(z)) increases on (frac{pi}{6} le z le frac{pi}{3}). The maximum of (y) thus occurs at the lower bound (z = frac{pi}{6}). Calculating (y) at that point, [ yBig|_{z=frac{pi}{6}} = frac{2cosleft(frac{pi}{3}right)}{sinleft(frac{pi}{3}right)} + sinleft(frac{pi}{6}right) = frac{2 cdot frac{1}{2}}{frac{sqrt{3}}{2}} + frac{1}{2} = frac{2}{sqrt{3}} + frac{1}{2} = frac{4 + sqrt{3}}{2sqrt{3}} = boxed{frac{4 + sqrt{3}}{2sqrt{3}}}. ]

answer:To find the imaginary part of z=frac{1-2i}{i}, we first simplify z: begin{align*} z &= frac{1-2i}{i} &= frac{1-2i}{i} cdot frac{i}{i} &= frac{i(1-2i)}{i^2} &= frac{i-2i^2}{-1} &= frac{i-2(-1)}{-1} &= frac{i+2}{-1} &= -2 - i. end{align*} Therefore, the imaginary part of z is -1. So, the answer is boxed{-1}.

answer:# Problem 1: Calculate the expression: (1)-{1}^{2022}-[3×{(frac{2}{3})}^{2}-frac{8}{3}÷(-2)^{3}]. Step-by-Step Solution: 1. Start with the original expression: [ -1-[3×frac{4}{9}-frac{8}{3}÷(-8)] ] 2. Simplify inside the brackets: [ = -1-left[frac{4}{3}-frac{8}{3}×left(-frac{1}{8}right)right] ] 3. Continue simplifying: [ = -1-left(frac{4}{3}+frac{1}{3}right) ] 4. Combine the fractions: [ = -1-frac{5}{3} ] 5. Final calculation: [ = -frac{8}{3} ] Therefore, the final answer is boxed{-frac{8}{3}}. # Problem 2: Calculate the expression: frac{{2}^{3}}{3}×(-frac{1}{4}+frac{7}{12}-frac{5}{6})÷(-frac{1}{18}). Step-by-Step Solution: 1. Start with the original expression: [ =frac{8}{3}×(-frac{1}{4}+frac{7}{12}-frac{5}{6})×(-18) ] 2. Simplify the expression inside the parentheses: [ =(-frac{1}{4}+frac{7}{12}-frac{5}{6})×(-48) ] 3. Distribute the multiplication: [ =-frac{1}{4}×(-48)+frac{7}{12}×(-48)-frac{5}{6}×(-48) ] 4. Perform the calculations: [ =12-28+40 ] 5. Sum up the results: [ =24 ] Therefore, the final answer is boxed{24}.

answer:Since it is given that |vec{a}|=1 and vec{b}=(sqrt{3},1), we can use the Pythagorean theorem to find the magnitude of vec{b}. Hence, we have |vec{b}|=sqrt{sqrt{3}^2 + 1^2} = sqrt{3+1} = sqrt{4} = 2. Now, because vec{a} and vec{b} are orthogonal, their dot product is zero: vec{a}cdotvec{b}=0. Using this property, we can expand the square of the magnitude of the sum |2vec{a}+vec{b}|^2: begin{align*} |2vec{a}+vec{b}|^2 &= (2vec{a}+vec{b})cdot(2vec{a}+vec{b}) &= 4vec{a}cdotvec{a} + 4vec{a}cdotvec{b} + vec{b}cdotvec{b} &= 4|vec{a}|^2 + 4(vec{a}cdotvec{b}) + |vec{b}|^2 &= 4(1) + 4(0) + 2^2 &= 4 + 0 + 4 &= 8. end{align*} Taking the square root to find the magnitude, we get |2vec{a}+vec{b}| = sqrt{8} = 2sqrt{2}. Therefore, the correct answer is boxed{B: 2sqrt{2}}.