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

I have researched the question $\lim_{n \to \infty} n*\sin\frac{1}{n}$ quite profusely, and I know that it equals to 1, and I know why A You can use a change of variables and substitute, say, $m = \frac{1}{n}$ so that $m \to 0$ instead. B L'Hopital's rule The problem is, we haven't used either of these methods in class, so I am wondering if there is any other possible way to approach this question?

Calculus Examples Popular Problems Calculus Solve for x k=1+sinx/n Step 1Rewrite the equation as .Step 2Multiply both sides by .Step 3Simplify the left for more steps...Step .Tap for more steps...Step the common factor of .Tap for more steps...Step the common the and .Step 4Subtract from both sides of the 5Take the inverse sine of both sides of the equation to extract from inside the sine.

Inthe particular case of your question, we have the simple algebraic identity $$(n+1)x=nx+x.$$ When applying the sine function to this quantity, we need no further parentheses on the left-hand side; but parentheses must be introduced on the right-hand side because otherwise it would read as $$\sin nx+x,$$ which is the sum of $\sin nx$ and $x$.

$\begingroup$ Question Prove that $\sinnx \cosn+1x-\sinn-1x\cosnx = \sinx \cos2nx$ for $n \in \mathbb{R}$. My attempts I initially began messing around with the product to sum identities, but I couldn't find any way to actually use them. I also tried compound angles to expand the expression, but it became too difficult to work with. Any help or guidance would be greatly appreciated asked Jun 15, 2020 at 1531 $\endgroup$ 2 $\begingroup$The left-hand side is$$\begin{align}&\sin nx\cos nx\cos x-\sin nx\sin x-\sin nx\cos x-\cos nx\sin x\cos nx\\&=\cos^2nx-\sin^2nx\sin x\\&=\cos 2nx\sin x.\end{align}$$ answered Jun 15, 2020 at 1537 gold badges74 silver badges135 bronze badges $\endgroup$ $\begingroup$Use $\sina\cosb=\frac{1}{2}\sina-b+\sina+b$ $$ \sinnx \cosn+1x-\sinn-1x\cosnx $$ $$ =\frac{1}{2}\left\sin-x+\sin2n+1x-\sin-x-\sin2n-1x \right $$ $$ =\frac{1}{2}\left\sin2n+1x-\sin2n-1x \right $$Now use $\sina+b=\sina\cosb+\sinb\cosa$ $$ =\frac{1}{2}\left\sin2nx\cosx +\sinx\cos2nx-\sin2nx\cos-x-\sin-x\cos2nx \right $$Now use the parity of sine and cosine and you're done. $$ =\frac{1}{2}\left\sin2nx\cosx +\sinx\cos2nx-\sin2nx\cosx+\sinx\cos2nx \right $$ $$ =\sinx\cos2nx $$ answered Jun 15, 2020 at 1536 IntegrandIntegrand8,15415 gold badges41 silver badges69 bronze badges $\endgroup$ $\begingroup$ $$ \begin{align} \sinnx\cosn+1x &=\frac{\sinnx+n+1x+\sinnx-n+1x}2\tag1\\ &=\frac{\sin2n+1x-\sinx}2\tag2\\ \sinn-1x\cosnx &=\frac{\sin2n-1x-\sinx}2\tag3 \end{align} $$ Explanation $1$ identity $\sina\cosb=\frac{\sina+b+\sina-b}2$ $2$ simplify $3$ apply $2$ for $n-1$ Therefore, $$ \begin{align} \sinnx\cosn+1x-\sinn-1x\cosnx &=\frac{\sin2n+1x-\sin2n-1x}2\tag4\\ &=\sinx\cos2nx\tag5 \end{align} $$ Explanation $4$ subtract $3$ from $2$ $5$ identity $\sina-\sinb=2\sin\left\frac{a-b}2\right\cos\left\frac{a+b}2\right$ answered Jun 15, 2020 at 1822 robjohn♦robjohn337k35 gold badges446 silver badges832 bronze badges $\endgroup$ You must log in to answer this question. Not the answer you're looking for? Browse other questions tagged .

Itis the case to consider Laurent series, since both functions have a simple pole in zero. By definition: \frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^n \tag{1} hence: \frac{1}{1-e^{-x}}=\sum_{n\geq 0}\frac{B_n}{n!}(-1)^n x^{n-1} \tag{2}

sin Cosine calculator ► Sine calculation Calculation with sinangle degrad Expression Result Inverse sine calculator sin-1 Degrees First result Second result Radians First result Second result k = ...,-2,-1,0,1,2,... Arcsin calculator ► Sine table xdeg xrad sinx -90° -π/2 -1 -60° -π/3 -√3/2 -45° -π/4 -√2/2 -30° -π/6 -1/2 0° 0 0 30° π/6 1/2 45° π/4 √2/2 60° π/3 √3/2 90° π/2 1 See also Sine function Cosine calculator Tangent calculator Arcsin calculator Arccos calculator Arctan calculator Trigonometry calculator Degrees to radians conversion Radians to degrees conversion Degrees to degrees,minutes,seconds Degrees,minutes, seconds to degrees Write how to improve this page

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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.

If $n$ is even, then $$1= \cos^{n}x-\sin^{n}x \leq 1-0=1$$ with equality if and only if $\cos^{n}x=1, \sin^nx=0$. If $n$ is odd, $$1= \cos^{n}x-\sin^{n}x \,,$$ implies $\cosx \geq 0$ and $\sinx <0$. Let $\cosx=y, \sinx=-z$, with $y,z \geq 0$. $$y^n+z^n=1$$ $$y^2+z^2=1$$ Case 1 $n=1$ Then , since $0 \leq y,z \leq 1$ we have $$1 =y+z \geq y^2+z^2 =1$$ with equality if and only if $y=y^2, z=z^2$. Case 2 $n \geq 3$ Then , since $0 \leq y,z \leq 1$ we have $$1 =y^2+z^2 \geq y^n+z^n =1$$ with equality if and only if $y^2=y^n, z^2=z^n$. 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 . 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 Sin(n + 1)x - sin (n - 1)x = sin (nx + x) - sin(nx - x) = 2 cos ½ (nx + x + nx - x) sin ½ (nx + x - nx + x) = 2 cos ½ (2nx) sin ½ (2x) = 2 cos nx sin x C 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. but still positive), so each positive limit has a "basin of attraction" that includes an open inteval of values for $x_0$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $|x_{n+1}|=|x_n\sin x_n|\le|x_n|$, successive terms in any sequence always get closer to the origin).

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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

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sin n 1 x sin n 1 x