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 $x$. 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 $y^2=y^n, z^2=z^n$.
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

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

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.

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

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.
Weknow that cos ( A B) = cos A cos B + sin A sin B Hence A = (n + 1)x ,B = (n + 2)x Hence sin ( + 1) sin ( + 2) +cos ( + 1) cos ( + 2) = cos [ (n + 1)x (n + 2)x ] = cos [ nx + x nx 2x ] = cos [ nx nx x 2 x ] = cos (0 x ) = cos ( x) = cos x = R.H.S. Hence , L.H.S. = R.H.S. Hence proved rBFhNb.
  • b9bz2p70zc.pages.dev/346
  • b9bz2p70zc.pages.dev/442
  • b9bz2p70zc.pages.dev/391
  • b9bz2p70zc.pages.dev/427
  • b9bz2p70zc.pages.dev/373
  • b9bz2p70zc.pages.dev/260
  • b9bz2p70zc.pages.dev/414
  • b9bz2p70zc.pages.dev/282

    • sin n 1 x sin n 1 x