🎇 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 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 Multiplyboth sides by n n. 1+sin(x) n n = kn 1 + sin ( x) n n = k n. Simplify the left side. Tap for more steps sin(x)+1 = kn sin ( x) + 1 = k n. Subtract 1 1 from both sides of the equation. sin(x) = kn−1 sin ( x) = k n - 1. Take the inverse sine of both sides of the equation to extract x x from inside the sine. 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 Solution Verified by Toppr. The given equation is. sin −1x+sin −1(1−x)=cos −1x. ⇒sin −1x+sin −1(1−x)= 2π−sin −1x. ⇒sin −1(1−x)= 2π−2sin −1x (i) Let sin −1x=y. ⇒x=siny. 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} . 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 $y^2=y^n, z^2=z^n$. 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 oo0. lim_x->oo xsin1/x=oosin1/oo=oosin0=oo0 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-1x^-2cos1/x/-1*x^-2 Simplify the previous step =lim_x->oocos1/x=cos1/oo=cos0=1 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