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 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 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. 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. 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) 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 ) tB8qou. 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 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 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? 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 $y^2=y^n, z^2=z^n$. 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