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Question Number 137890 Answers: 0 Comments: 0

Question Number 137889 Answers: 0 Comments: 0

Question Number 137876 Answers: 1 Comments: 0

Question Number 137875 Answers: 1 Comments: 0

Proof by mathematical induction that f(n) = n^3 + 5n is a multiple of 6.

$Proof by mathematical induction that$ $f (n) = n^{3} + 5 n$ $is a multiple of 6 .$

Question Number 137874 Answers: 1 Comments: 0

.......nice ... .... calculus.... evaluate :: Ω=∫_0 ^( 1) (((ln(1−x))/x))^3 =?....

$. . . . . . . n i c e . . . . . . . c a l c u l u s . . . .$ $e v a l u a t e ::$ $Ω = \int_{0}^{1} {(\frac{l n (1 - x)}{x})}^{3} = ? . . . .$

Question Number 137871 Answers: 0 Comments: 0

Find the area bounded by the standard normal distribution p(−1.96≤Z≤2.5)

$Find the area bounded by the standard$ $normal distribution p (- 1.96 ⩽ Z ⩽ 2.5)$

Question Number 137869 Answers: 0 Comments: 0

prove that Arg(z) is harmonic function and find the conjecate this ?

$p r o v e t h a t A r g (z) i s h a r m o n i c f u n c t i o n$ $a n d f i n d t h e c o n j e c a t e t h i s ?$

Question Number 137864 Answers: 0 Comments: 4

if , acosα+bcosβ=Acos∅ proof that asinα+bsinβ=Asin∅

$if,$ $acos α + bcos β = Acos \emptyset$ $proof that$ $asin α + bsin β = Asin \emptyset$

Question Number 137856 Answers: 0 Comments: 0

Question Number 137855 Answers: 0 Comments: 0

Question Number 137861 Answers: 0 Comments: 1

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Question Number 137851 Answers: 1 Comments: 0

If 2^(2sin^2 θ−3sin θ+1) + 2^(2−2sin^2 θ+3sin θ) = 9 find the value of sin θ+cos θ .

$I f 2^{2 \sin^{2} θ - 3 \sin θ + 1} + 2^{2 - 2 \sin^{2} θ + 3 \sin θ} = 9$ $f i n d t h e v a l u e o f \sin θ + \cos θ .$

Question Number 137848 Answers: 1 Comments: 0

Solve the equation (x^2 −x+1)^2 +(y^2 +y+1)^2 =2021 x,y∈Z

$S o l v e t h e e q u a t i o n$ ${(x^{2} - x + 1)}^{2} + {(y^{2} + y + 1)}^{2} = 2021$ $x, y \in Z$

Question Number 137873 Answers: 0 Comments: 3

.......nice ............calculus....... 𝛗=Σ_(n=1) ^∞ ((sin(nx))/n) =(π/2)−(x/2) 𝛗=−Im(1−e^(ix) )=−Imln{(1−cos(x)−isin(x))} =−Im{ln((√((1−cos(x))^2 +sin^2 (x))) +itan^(−1) (((−sin(x))/(1−cos(x))))} =−Im{(√(2−2cos(x)_ )) −itan^(−1) (tan((π/2)−(x/2)))} =tan^(−1) (tan((π/2)−(x/2))) ........ 𝛗=(π/2)−(x/2) ...✓.....⟨∗⟩ ^(′′) ∫ ^(′′) both sides of ⟨∗⟩ −Σ_(n=1) ^∞ ((cos(nx))/n^2 )dx+C=((πx)/2)−(x^2 /4) x=0 ⇒ C=(π^2 /6) ...✓ Σ_(n=1) ^∞ ((cos(nx))/n^2 )=−((πx)/2)+(x^2 /4)+(π^2 /6) ....✓ x=π(1+(√(1/3)) ) .... Σ((cos(nπ(1+(√(1/3)) )))/n^2 )=(1/2)(−π^2 (1+(√(1/3)) )+((π^2 (1+(√(1/3)) )^2 )/2)+(π^2 /3)) =(π^2 /2)(−1−((√3)/3) +((1+2.((√3)/3) +(1/3))/2)+(1/3)) =(π^2 /2)(−1−(((√3) )/3)+(4/6)+(((√3) )/3)+(1/3))=0..✓✓ ......prepared by mr (Dwaipayan)..... m.n#

$. . . . . . . n i c e . . . . . . . . . . . . c a l c u l u s . . . . . . .$ $ϕ = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n} = \frac{π}{2} - \frac{x}{2}$ $ϕ = - I m (1 - e^{i x}) = - I m l n {(1 - c o s (x) - i s i n (x))}$ $= - I m {l n (\sqrt{{(1 - c o s (x))}^{2} + {s i n}^{2} (x)} + {i t a n}^{- 1} (\frac{- s i n (x)}{1 - c o s (x)})}$ $= - I m {\sqrt{2 - 2 c o s {(x)}_{}} - {i t a n}^{- 1} (t a n (\frac{π}{2} - \frac{x}{2}))}$ $= {t a n}^{- 1} (t a n (\frac{π}{2} - \frac{x}{2}))$ $. . . . . . . . ϕ = \frac{π}{2} - \frac{x}{2} . . . ✓ . . . . . ⟨ * ⟩$ $^{^{″}} \int^{^{″}} b o t h s i d e s o f ⟨ * ⟩$ $- \underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}} d x + C = \frac{π x}{2} - \frac{x^{2}}{4}$ $x = 0 \Rightarrow C = \frac{π^{2}}{6} . . . ✓$ $\underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}} = - \frac{π x}{2} + \frac{x^{2}}{4} + \frac{π^{2}}{6} . . . . ✓$ $x = π (1 + \sqrt{\frac{1}{3}}) . . . .$ $Σ \frac{c o s (n π (1 + \sqrt{\frac{1}{3}}))}{n^{2}} = \frac{1}{2} (- π^{2} (1 + \sqrt{\frac{1}{3}}) + \frac{π^{2} {(1 + \sqrt{\frac{1}{3}})}^{2}}{2} + \frac{π^{2}}{3})$ $= \frac{π^{2}}{2} (- 1 - \frac{\sqrt{3}}{3} + \frac{1 + 2 . \frac{\sqrt{3}}{3} + \frac{1}{3}}{2} + \frac{1}{3})$ $= \frac{π^{2}}{2} (- 1 - \frac{\sqrt{3}}{3} + \frac{4}{6} + \frac{\sqrt{3}}{3} + \frac{1}{3}) = 0 . . ✓ ✓$ $. . . . . . p r e p a r e d b y m r (D w a i p a y a n) . . . . .$ $You can't use 'macro parameter character #' in math mode$