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2 \sqrt{ 3 } \sin ( \frac{ \pi }{ 3 } )

Cael gwerth \sin(\frac{\pi }{3}) o’r tabl gwerthoedd trigonometrig.

2\sqrt{3}\times \frac{\sqrt{3}}{2}

Canslo 2 a 2.

\sqrt{3}\sqrt{3}

Lluosi \sqrt{3} a \sqrt{3} i gael 3.

Ffactor

Cwis

Trigonometry

2 \sqrt{ 3 } \sin ( \frac{ \pi }{ 3 } )

Problemau tebyg o chwiliad gwe

Determining the convergence of \sqrt{n} \sin(\pi/\sqrt{n})

https://math.stackexchange.com/q/381799

If your limit seems to be \infty \times 0, then you can't just conclude it's \infty or 0 or some definite value. The result depends on how strongly each factor approaches its limit. As others ...

Calculating the Inverse Fourier Transform of \frac{1}{\sqrt{2\pi}k}\sin k

https://math.stackexchange.com/q/1603550

Perhaps what the hint means is f(x)=\frac{-1}{2\pi\mathrm{i}}\frac{\partial}{\partial x}\int_{-\infty}^\infty dk\frac{\sin k}{k^2}e^{-\mathrm{i}kx}=\frac{1}{2\pi}\frac{\partial}{\partial x}\int_{-\infty}^\infty dk\frac{\sin k}{k^2}\sin(kx)\ , ...

How does \sin (\frac{3\pi}{4}) = \frac{1}{\sqrt{2}}?

https://math.stackexchange.com/q/1766655

That's simply because \;\sin\dfrac{3\pi}4=\sin\dfrac\pi4=\dfrac1{\sqrt 2}

prove that \int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8} [duplicate]

https://math.stackexchange.com/q/2604921

Just extending what you've got so far. Let's note z=e^{i\frac{5\pi}{8}} and recall that \cos{z}=\frac{e^{iz}+e^{-iz}}{2}, then: 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right]= 2 \pi i \left(\frac{-1}{8}\right) \left[z+z^3+z^5+z^7\right]=\\ 2 \pi i \left(\frac{-1}{8}\right) z \left[1+z^2+z^4+z^6\right] = 2 \pi i \left(\frac{-1}{8}\right) z \left[1+z^2+z^4(1+z^2)\right]=\\ 2 \pi i \left(\frac{-1}{8}\right) z (1+z^2)(1+z^4)= 2 \pi i \left(\frac{-1}{8}\right) z^4 (z^{-1}+z)(z^{-2}+z^2)=\\ 2 \pi i \left(\frac{-1}{8}\right) e^{i\frac{5\pi}{2}} 2\cos\left(\frac{5\pi}{8}\right)2 \cos\left(\frac{5\pi}{4}\right)=\pi i(-1)i \cos\left(\frac{\pi}{2}+\frac{\pi}{8}\right)\left(-\frac{1}{\sqrt{2}}\right)=\\ \frac{\pi}{\sqrt{2}}\sin\left(\frac{\pi}{8}\right)

Turning points on 2\sin x - x

https://math.stackexchange.com/questions/372575/turning-points-on-2-sin-x-x

Your answer is right. The answers given are the points on the Cartesian plane (x,f(x)) whereas you have found only the abscissa.Here your f(x)=2\sin x -x

Finding \int x \sqrt{1- x^2 \over 1 + x^2}dx.

https://math.stackexchange.com/questions/2238770/finding-int-x-sqrt1-x2-over-1-x2dx

Note that \frac{d}{dx}\arcsin(x^2)=\frac{2x}{\sqrt{1-x^4}} and \frac{d}{dx}2\arcsin\left(\frac{\sqrt{1+x^2}}{\sqrt 2}\right)=\frac{2x}{\sqrt{1-x^4}} So, the error in the OP is in the term \arcsin\left(\frac{\sqrt{1+x^2}}{2}\right) ...

Rhannu

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2\sqrt{3}\times \frac{\sqrt{3}}{2}

Cael gwerth \sin(\frac{\pi }{3}) o’r tabl gwerthoedd trigonometrig.

\sqrt{3}\sqrt{3}

Canslo 2 a 2.

Lluosi \sqrt{3} a \sqrt{3} i gael 3.

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Rhannu

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