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.
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
a) Prove the reduction formulaintegrate cos^n(x)dx = 1/n * cos^(n-1) * x sin x + (n-1)/n * integral cos^(n-2)x dx
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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.
but still positive), so each positive limit has a "basin of attraction" that includes an open inteval of values for $x_0$ slightly greater than the limit (and likewise, for each negative limit, an open interval of values slightly less than the limit -- since $|x_{n+1}|=|x_n\sin x_n|\le|x_n|$, successive terms in any sequence always get closer to the origin).
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
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
sinn+1)-sin(n-1)x= .. Rumlah Jumlah dan Selisih Sudut; Rumus jumlah dan selisih sinus/ kosinus/ tangen; Persamaan Trigonometri; TRIGONOMETRI; Matematika
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sin n 1 x sin n 1 x
