sin n 1 x sin n 1 x

Solucionatus problemas matemáticos con nuestro solucionador matemático gratuito, que incluye soluciones paso a paso. Nuestro solucionador matemático admite matemáticas básicas, pre-álgebra, álgebra, trigonometría, cálculo y mucho más. 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 ) 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} 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 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 Hủy Hợp Đồng Vay Tiền Online. 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 $\begingroup$ 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 . 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 $y^2=y^n, z^2=z^n$. 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 . 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 n 1 x sin n 1 x