🐿️ Sin N 1 X Sin N 1 X

sinn+1)-sin(n-1)x= .. Rumlah Jumlah dan Selisih Sudut; Rumus jumlah dan selisih sinus/ kosinus/ tangen; Persamaan Trigonometri; TRIGONOMETRI; Matematika

I'm studying convergent sequences at the moment. And I came across this question in the section of Stolz Theorem. I realised that $\{x_n\}$ is monotonously decreasing and has a lower bound of $0$, so $\{x_n\}$ must be convergent, and the limit is $0$ let $L=\sinL$, then $L=0$. So to prove the original statement, I just need to prove lim nXn^2 → 3, and in order to prove that, I just need to prove $\lim \frac{1}{x_n^2} - \frac{1}{{x_{n-1}}^2} \to \frac{1}{3}$ by Stolz Theorem but I have no clue what to do from there. PS $x_{n+1}$ is $x$ sub $n+1$, and $x_n$ is outside the square root. Thanks guys

^{DetailedSolution. Given if I n = ∫π −π sinnx (1+πx)sinx dx,(1) i f I n = ∫ − π π s i n n x ( 1 + π x) s i n x d x, ( 1) I n = ∫π −π πxsinnx (1+πx)sinx dx.(2) I n = ∫ − π π π x s i n n x ( 1 + π x) s i n x d x. ( 2) On adding Eqs. (i) and (ii), we have.}

Question MediumOpen in AppSolutionVerified by TopprThe given equation is ...... i Let Therefore, from i, we get Since, both these values satisfy the given equation. Hence, the solutions of the given equation are .Video ExplanationWas this answer helpful? 00

Soif X n is the number of individuals alive in generation n, then X n+1 is the sum of X n -many independent, identically distributed random variables. Let's assume that X 0 = 1, p (0) > 0, and = k p (k) = E (X 1) 1. (a) If = 1 and 2 < , then there exist constants 0 < c 1 < c 2 < such that. c 1 /n < P ( X n 0 ) < c 2 /n. Chapter 3 Class 11 Trigonometric Functions Serial order wise Ex Check sibling questions Ex Ex 1 Important Ex 2 Important Ex 3 Important Ex 4 Ex 5 i Important Ex 5 ii Ex 6 Important Ex 7 Ex 8 Important Ex 9 Important Ex 10 You are here Ex 11 Important Ex 12 Ex 13 Important Ex 14 Ex 15 Ex 16 Important Ex 17 Ex 18 Important Ex 19 Ex 20 Ex 21 Important Ex 22 Important Ex 23 Important Ex 24 Ex 25 Ex 10 - Chapter 3 Class 11 Trigonometric Functions Last updated at May 29, 2023 by Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class Transcript E 10 Prove that sin + 1 sin + 2 +cos + 1 cos + 2 =cos Taking We know that cos A B = cos A cos B + sin A sin B Hence A = n + 1x ,B = n + 2x Hence sin + 1 sin + 2 +cos + 1 cos + 2 = cos [ n + 1x n + 2x ] = cos [ nx + x nx 2x ] = cos [ nx nx x 2 x ] = cos 0 x = cos x = cos x = Hence , = Hence proved Chapter 3 Class 11 Trigonometric Functions Serial order wise Ex Ex 1 Important Ex 2 Important Ex 3 Important Ex 4 Ex 5 i Important Ex 5 ii Ex 6 Important Ex 7 Ex 8 Important Ex 9 Important Ex 10 You are here Ex 11 Important Ex 12 Ex 13 Important Ex 14 Ex 15 Ex 16 Important Ex 17 Ex 18 Important Ex 19 Ex 20 Ex 21 Important Ex 22 Important Ex 23 Important Ex 24 Ex 25 Davneet Singh has done his from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

If[{Sin[(n + 1)x] + Sinx}/x] for lim x→0 = (1/2) then value of n is: (a) - 2.5 (b) - 0.5 (c) - 1.5 (d) - 1

S.}=sin (n+1) x sin (n+2) x+cos (n+1) x (n+2) x ) ( begin{aligned} &=text { Put }(n+1) x=A text { and }(n+2) x=mathbf{B} therefore text { L. } mathbf{H . S .} &=sin A sin B+cos A cos B &=cos A cos B+sin A sin B=cos (A-B) &=cos [(n+1) x-(n+2) x]=cos (n x+x-n x-2 x) &=cos (-x)=cos x=R .

Thecorrect option is An sin n - 1 x cos n + 1 xExplanation for the correct option :Given, d sin n x cos n x d xFor differentiating this apply theorem for product d u × v d x = u d v d x + v d u d x⇒ d sin n x cos n x d x = n sin n - 1 x cos x cos n x + sin n x - sin n x n ⇒ d sin n x cos n x d x = n sin n - 1 x cos x cos n x - sin x sin n x

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TheLimit of the Sequence n*sin(1/n) as n Approaches Infinity - YouTube. The Limit of the Sequence n*sin(1/n) as n Approaches InfinityPlease Subscribe here, thank you!!! sin(n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos ( ( n + 2 ) x − ( n − 1 ) x ) { ∵ cos ( A − B ) = sin A sin B + cos A cos B } ⇒ = cos ( ( n + 2 − n − 1 ) x )

Weknow that cos ( A B) = cos A cos B + sin A sin B Hence A = (n + 1)x ,B = (n + 2)x Hence sin ( + 1) sin ( + 2) +cos ( + 1) cos ( + 2) = cos [ (n + 1)x (n + 2)x ] = cos [ nx + x nx 2x ] = cos [ nx nx x 2 x ] = cos (0 x ) = cos ( x) = cos x = R.H.S. Hence , L.H.S. = R.H.S. Hence proved

sinA - sin B = 2 cos 1/2 (A+B) sin 1/2 (A-B) sin(x+1)x-sin(x-1)x = 2cos 1/2 ((x+1)x + (x-1)x) sin 1/2 ((x+1)x - (x-1)x) = 2cos 1/2(x²+x+x²-x) sin 1/2(x²+x-x²+x) = 2cos 1/2(2x²) sin 1/2(2x) = 2 cos x² sin x

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