Citation Form | MLA... \( +- \pm \times \div \neq=<> \) \( \wedge \vee \cup \cap \in \mid \infty \leq \geq \) \begin{tabular}{|c|c|c|c|c|c|} \hline & \( \sim \) & & & & \\ \hline \end{tabular} \begin{tabular}{|c|c|c|c|c|c|c|c|} \hline\( \alpha \) & \( \beta \) & \( \theta \) & \( \lambda \) & \( \mu \) & \( \pi \) & \( \sigma \) & \( \omega \) \\ \hline \( \log \) & \( \ln \) & \( e \) & \( \overline{\mathbf{x}} \) & \( \overline{\mathbf{d}} \) & \( \hat{\mathbf{p}} \) & \( A_{0} \) & \( v_{0} \) \\ \hline \end{tabular} \[ \sin (\square) \cos (\square) \tan (\square) \csc (\square) \sec (\square) \cot (\square) \] \[ \cos (2 x)= \] \[ \sin ^{\#}(\square) \cos ^{\#}(\square) \tan ^{\#}(\square) \csc ^{\mathrm{W}}(\square) \sec ^{\#}(\square) \cot ^{\#}(\square) \] Submit Answer \( 56^{\circ} \mathrm{F} \) Cloudy Search Keypad Keyboard Shortcuts
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However, the question itself is somewhat unclear and seems to mix several unrelated elements. Assuming the core of the question is to explain how to use these symbols in a mathematical context, let's proceed with a general guide on how to approach mathematical Show more…
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The exponential function $f(x)=e^{x}$ can be defined for exponents that are complex numbers, $a+b i,$ where $i=\sqrt{-1}$. Complex-valued functions may be differentiated just like real-valued functions, and previous derivative rules carry over in this setting. For example, $\left(e^{a x}\right)^{\prime}=a e^{a x}$ holds if $a$ and $x$ are complex numbers. Furthermore, in the complex-number domain there is an important formula (known as Euler's formula) relating exponential and trigonometric functions: $e^{i x}=\cos x+i \sin x$. (These topics are all addressed formally when we develop the theory of power series in Chapter $10 .)$ (a) Using Euler's formula, prove $$ \sin x=\frac{e^{i x}-e^{-i x}}{2 i} \text { and } \cos x=\frac{e^{i x}+e^{-i x}}{2} $$ (b) Using the expressions from (a) and the derivative rule for $e^{x}$, prove the derivative rules for $\sin x$ and $\cos x$.
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From the previous problem, $\sigma^{\prime}=\varepsilon_{0} \frac{\varepsilon-1}{\varepsilon} E_{0} \cos \theta$ (a) Then $\oint \quad \vec{E} \cdot \overrightarrow{d S}=\frac{1}{\varepsilon_{0}} Q=\pi R^{2} E_{0} \cos \theta \frac{\varepsilon-1}{\varepsilon}$ (b) $\oint \vec{D} \cdot \overrightarrow{d l}=\left(D_{1 t}-D_{2 t}\right) l=\left(\varepsilon_{0} E_{0} \sin \theta-\varepsilon \varepsilon_{0} E_{0} \sin \theta\right)=-(\varepsilon-1) \varepsilon_{0} E_{0} l \sin \theta$
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$\xi_{1}=\Lambda \sin (\omega t)$ $\xi_{2}=\Lambda \sin \left(\omega t+60^{\circ}\right)$ (2) $\xi_{3}^{2}=\Lambda \sin \left(\omega t+120^{\circ}\right)$ (3) From $(1)+(2)$, $\xi^{\prime}=\Lambda^{\prime} \sin (\omega t+\phi)$ $\tan \phi=\frac{A \sin 60^{\circ}}{A+A \cos 60}=\frac{\sqrt{3} / 2}{3 / 2}=\frac{1}{\sqrt{3}}$ $=A^{\prime} \sin \left(\omega t+\frac{\pi}{6}\right)$ $A^{\prime}=\sqrt{A^{2}+A^{2}+2 A^{2} \cos 60^{\circ}}$ $\therefore \quad \phi=30^{\circ} \quad \therefore \quad \Lambda^{\prime}=\sqrt{3} \Lambda$again from $(3)+(4)$ $\xi^{\prime \prime}=A^{n} \sin \left[\left(\omega t+30^{\circ}\right)+\delta \mid\right.$ $A^{\prime \prime}=\sqrt{\Lambda^{2}+\Lambda^{\prime 2}+2 \Lambda \Lambda^{\prime} \cos \phi}$ $=\sqrt{A^{2}+3 A^{2}+2 \sqrt{3} A^{2} \times \cos 90^{\circ}}$ requircd amplitude $\Lambda^{\prime \prime}=2 \Lambda$
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