n\frac{1}{\sin(x)},\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1} View solution steps. Steps for Solving Linear Equation. \frac { 1 } { n } = \sin ( x ) Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by n. 1=n\sin(x) sinn+1)-sin(n-1)x= .. Rumlah Jumlah dan Selisih Sudut; Rumus jumlah dan selisih sinus/ kosinus/ tangen; Persamaan Trigonometri; TRIGONOMETRI; Matematika produk jasa profesi dan profesionalisme dimulai dengan melakukan. 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 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 >>Class 11>>Maths>>Trigonometric Functions>>Trigonometric Functions of Sum and Difference of Two angles>>Prove sinn + 1x sinn + 2 x + cos n Open in AppUpdated on 2022-09-05SolutionVerified by TopprTo prove- Proof Hence any question of Trigonometric Functions with-Was this answer helpful? 00More From ChapterLearn with Videos Practice more questions $\begingroup$ 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 . 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

sin n 1 x sin n 1 x