Riješi [1+1/2sin(Omegat)]sin(8Omegat)

Procijeni

\frac{\sin(8t\Omega )\left(\sin(t\Omega )+2\right)}{2}

Faktor

2\sin(2t\Omega )\left(\sin(t\Omega )+2\right)\left(\cos(t\Omega )-\sin(t\Omega )\right)\left(\sin(t\Omega )+\cos(t\Omega )\right)\left(-8\left(\sin(t\Omega )\right)^{2}\left(\cos(t\Omega )\right)^{2}+1\right)

Kviz

Trigonometry

5 problemi slični sa:

[ 1 + \frac { 1 } { 2 } \sin ( \Omega t ) ] \sin ( 8 \Omega t )

Slični problemi iz web pretrage

Working out the power in a circuit when the current and voltage are in trigonometric waveform.

https://math.stackexchange.com/questions/2768001/working-out-the-power-in-a-circuit-when-the-current-and-voltage-are-in-trigonome

Hint: Use the product to sum formula \sin A \sin B = \frac {1}{2}[\cos (A - B) - \cos (A + B)] with A = \omega t - \pi/2 and B = \omega t. What do you notice?

If replacing i by -i doesn't change a complex valued function, then the function output is purely real - What's proof/intuition behind this? [closed]

https://math.stackexchange.com/q/2821308

Simply observe in general that z=x+iy \bar z=x-iy therefore z=\bar z \implies y=0 In your example, assuming \omega \neq 0 to be real and defining H(\omega)=F(i\omega) we have F(i\omega) = \dfrac{1}{i\omega} \left(e^{i\omega} - e^{-i\omega}\right)=\overline{F(i\omega)}=F(\overline{i\omega})=F(-i\omega) ...

Directional derivative doubt

https://math.stackexchange.com/q/2588955

The misunderstanding is in "since I know that for some (t,z) points both e and h are 0, that constant is 0". Let's see why by considering the simpler case where \omega=\beta=\eta=1. The ...

Why does \pi turn out to be negative

https://math.stackexchange.com/q/2738957

\lim_{\theta \to 0} -\frac{\theta\pi}{90} = \lim_{\theta \to 0} \sin(2\theta) The above is correct, but it does not imply the following line: \pi = -\lim_{\theta \to 0} \frac{90\sin(2\theta)}{\theta} ...

How to solve a differential equation involving \sin and \cos functions?

https://math.stackexchange.com/questions/684571/how-to-solve-a-differential-equation-involving-sin-and-cos-functions

\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} ...

How do I determine the sign of \sin\theta in this question.

https://math.stackexchange.com/q/673939/752

Your book is correct. We don't need to find either \alpha or \beta, i.e. to solve the given linear equation in \cos\theta and \sin\theta, which could be transformed into a quadratic equation ...

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\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]

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Teme

Teme

Slični problemi iz web pretrage

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