Rozwiąż c^2=1+sinθ+sin^2θ/2! | Microsoft Math Solver

Rozwiąż względem c (complex solution)

c=-\frac{\sqrt{4\sin(\theta )-\cos(2\theta )+5}}{2}

c=\frac{\sqrt{4\sin(\theta )-\cos(2\theta )+5}}{2}

Rozwiąż względem c

c=\frac{\sqrt{2\left(\left(\sin(\theta )\right)^{2}+2\sin(\theta )+2\right)}}{2}

c=-\frac{\sqrt{2\left(\left(\sin(\theta )\right)^{2}+2\sin(\theta )+2\right)}}{2}

Rozwiąż względem θ

\theta =\pi +2\pi n_{22}+arcSin(1+\left(-1\right)\left(-1+2c^{2}\right)^{\frac{1}{2}})\text{, }n_{22}\in \mathrm{Z}

\theta =\left(-1\right)arcSin(1+\left(-1\right)\left(-1+2c^{2}\right)^{\frac{1}{2}})+2\pi n_{21}\text{, }n_{21}\in \mathrm{Z}

\theta =\pi +2\pi n_{48}+arcSin(1+\left(-1+2c^{2}\right)^{\frac{1}{2}})\text{, }n_{48}\in \mathrm{Z}

\theta =\left(-1\right)arcSin(1+\left(-1+2c^{2}\right)^{\frac{1}{2}})+2\pi n_{47}\text{, }n_{47}\in \mathrm{Z}

Wykres

Quiz

Trigonometry

c ^ { 2 } = 1 + \sin \theta + \frac { \sin ^ { 2 } \theta } { 2 ! }

Podobne zadania z wyszukiwania w sieci web

Definite integral of \int_0^\pi \frac{d\theta}{a+\sin^2 \theta} \quad (a>0)

https://math.stackexchange.com/questions/1733250/definite-integral-of-int-0-pi-fracd-thetaa-sin2-theta-quad-a0

Through the substitution \theta=\arctan t: \begin{eqnarray*}\int_{0}^{\pi}\frac{d\theta}{a+\sin^2\theta} &=& 2\int_{0}^{\pi/2}\frac{d\theta}{a+\sin^2\theta}\\&=&2\int_{0}^{+\infty}\frac{dt}{(1+t^2)\left(a+\frac{t^2}{1+t^2}\right)}\\&=&\frac{2}{a}\int_{0}^{+\infty}\frac{dt}{1+\left(1+\frac{1}{a}\right)t^2}\\&=&\frac{2}{a\sqrt{1+\frac{1}{a}}}\int_{0}^{+\infty}\frac{dt}{1+t^2}\\&=&\color{red}{\frac{\pi}{\sqrt{a^2+a}}}.\end{eqnarray*}

By considering the real part of the integral

https://math.stackexchange.com/questions/1665740/by-considering-the-real-part-of-the-integral

\frac 1 {4+\sin^2\theta} = \frac 1 {(2+i\sin\theta)(2-i\sin\theta)} = \frac A {2+i\sin\theta} + \frac B {2-i\sin\theta} If you've worked with partial fractions before, then do the same thing ...

Contour complex integration with specifications

https://math.stackexchange.com/q/208610

The function e^x has the Maclaurin series 1 + x + \frac{x^2}{2!} + \mathcal{O}(x^3) if we make the substitution x\mapsto \frac{1}{z^2} then we get the Laurent series e^\frac{1}{z^2} = 1 + \frac{1}{z^2} + \frac{1}{2!\,z^4} + \mathcal{O}\left(\frac{1}{z^6}\right) ...

Manipulating Complex Exponentials

https://math.stackexchange.com/questions/1633998/manipulating-complex-exponentials

Using Euler's identity e^{ix}=\cos x+i\sin x e^{j\pi/2}=\cos\dfrac\pi2+j\sin\dfrac\pi2=j \implies e^{-j\pi/2}=j^{-1}=-j as j^2=-1

What is the largest radius that the circular track can have such that the car remains in contact with the track? [closed]

https://physics.stackexchange.com/q/429024

The first method is incorrect; it equates the potential energy at the top of the loop with the kinetic energy as the car enters the loop. It thus assumes zero kinetic energy (and thus zero velocity) ...

Normal Integration methods failing. \int_0^1 x^{n-1}\sin\left(\frac{\pi x}2\right){\rm d}x

https://math.stackexchange.com/questions/1275409/normal-integration-methods-failing-int-01-xn-1-sin-left-frac-pi-x2-rig

For (1),use this well known result \lim_{n\to\infty}n\int_{0}^{1}x^{n-1}f(x)dx=f(1).f\in C[0,1] see 1 for (2),we have general \lim_{n\to\infty}\dfrac{\int_{0}^{1}g(x)[f(x)]^{n+1}dx}{\int_{0}^{1}g(x)[f(x)]^ndx}=\max_{x\in[0,1]}\{f(x)\};f,g\in C[0,1],{\rm{and}}\; f,g>0 ...

Udostępnij

Skopiowano do schowka

Przykłady

Równanie kwadratowe

{ x } ^ { 2 } - 4 x - 5 = 0

Trygonometria

4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Równania równoważne

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Różniczkowanie

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Całkowanie

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Granice

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

Tematy

Tematy

Podobne zadania z wyszukiwania w sieci web

Udostępnij

Przykłady