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?
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!!! 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 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$\begingroup$ 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 .
Foreach of the following systems, determine whether the system is (a) linear, (b) (20 points) time-invariant, (c) causal, (dy memoryless, and (e) stable. (1) yln]xIn-11 sin [n] (2) y [n]-x [2n]+1 Time- Causal. sin (n+1)x sin (n+2)+cos (n+1)xcos (n+2)x = cosx - Brainly.in. 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 $y^2=y^n, z^2=z^n$. 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 Trigonometry Examples Popular Problems Trigonometry Simplify 1+sinx1-sinx Step 1Apply the distributive 2Multiply by .Step 3Rewrite using the commutative property of 4Multiply .Tap for more steps...Step to the power of .Step to the power of .Step the power rule to combine and . 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 $x$. 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. Closed1 hour ago. Improve this question I am not able to do this sum the question is Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x So, kindly help me in doing this sum. By l'Hopital's Rule, we can find lim_{x to infty}x sin1/x=1. Let us look at some details. lim_{x to infty}x sin1/x by rewriting a little bit, =lim_{x to infty}{sin1/x}/{1/x} by l'Ho[ital's Rule, =lim_{x to infty}{cos1/xcdot-1/x^2}/{-1/x^2} by cancelling out -1/x^2, =lim_{x to infty}cos1/x=cos0=1 Instead of l'Hopital's Rule, one can use the fundamental trigonometric limit lim_hrarr0sin h/h=1. The limit you are interested in can be written lim_xrarroosin 1/x/1/x. Now, as xrarroo, we know that 1/xrarr0 and we can think of the limit as lim_1/xrarr0sin 1/x/1/x. With h=1/x, this becomeslim_hrarr0sin h/h which is 1. Although it is NOT needed, here's the graph of the function graph{y = x sin1/x [ When you substitute in infinity, oo, you end up with the indeterminate form of oo*0. lim_x->oo xsin1/x=oo*sin1/oo=oo*sin0=oo*0 We still have options though. We now can fall back on L'Hopital's Rule which basically says to take the derivative of the numerator and denominator independently. Do not use the quotient rule. We need to rewrite this function so that is produces an indeterminate in the form oo/oo or 0/0. lim_x->oo sin1/x/x^-1=sin1/x/1/x=sin1/oo/1/oo=sin0/0=0/0 Applying L'Hopital lim_x->oosin1/x'/x^-1' =lim_x->oo-1*x^-2*cos1/x/-1*x^-2 Simplify the previous step =lim_x->oocos1/x=cos1/oo=cos0=1 E8jSP.