A parallel plate capacitor, dielectric constants

Question:
There is a parallel plate capacitor of area $L^2$ and separation distance of $d$. A half of the space, $d/2$, is filled with a material of dielectric constant, $\kappa_1$. The other half is filled with the other material of constant, $\kappa_2$. What is the capacitance of this capacitor if the free space capacitance is $C_0$?

Answer:
The capacitance of a parallel plate capacitor is
$$C = \frac{\epsilon_0\kappa A}{d}$$
where $A=L^2$. When a capacitor does not have any dielectrics (free space between plates), it will be
$$C_0 = \frac{\epsilon_0 A}{d}$$
We can consider the particular capacitor is like series connection of two different capacitors. Namely,
$$\frac{1}{C_{\mathrm{Total}}} = \frac{1}{C_1}+\frac{1}{C_2} \\ \rightarrow C_{\mathrm{Total}} = \frac{C_1C_2}{C_1+C_2}$$
From the condition, we have each capacitance of $C_1$ and $C_2$.
$$C_{1,2} =  \frac{\epsilon_0\kappa_{1,2} A}{\frac{1}{2}d} = \frac{2\epsilon_0\kappa_{1,2} A}{d}$$
Then, we can rewrite it in terms of $C_0$.
$$C_{1,2} = 2\kappa_{1,2}C_0$$

Plug in the above.
$$C_{\mathrm{Total}} = \frac{2\kappa_1C_0\times 2\kappa_2 C_0}{2\kappa_1C_0+2\kappa_2 C_0}    \\       = \frac{4\kappa_1\kappa_2 C^2_0}{2C_0(\kappa_1+\kappa_2)}   \\       =  \frac{2\kappa_1\kappa_2 C_0}{\kappa_1+\kappa_2}$$

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De Moivre's theorem and application

Question:
By using De Moivre's theorem, simplify the following expressions:
(1) $\left(\frac{1+\sqrt{3}i}{2}\right)^{10}$
(2) $\left(\frac{\sqrt{3}+i}{1+i}\right)^{6}$

Answer:
De Moivre's theorem shows
$$(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$$
(1) First, we obtain the magnitude.
$$\left| \frac{1+\sqrt{3}i}{2} \right| = \sqrt{\left(\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2} = 1$$
The argument $\theta$ ranges from 0 to 360 degrees. Comparing with $\cos\theta + i\sin\theta$, we can have
$$\cos\theta = \frac{1}{2}, \sin\theta = \frac{\sqrt{3}}{2}$$
Therefore, $\theta = 60^o$. Use the theorem.
$$(\cos 60^o + i\sin 60^o)^{10} = \cos(10\times60^o)+i\sin(10\times60^o)   \\                      = \cos 600^o + i\sin 600^o   \\                      = \cos 240^o + i\sin 240^o    \\                      = -\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

(2)
Let us separate the numerator and denominator.
$$\frac{(\sqrt{3}+i)^6}{(1+i)^6}$$
For the numerator, the magnitude and argument are
$$|\sqrt{3}+i| = 2   \\     \cos\theta_1 = \frac{\sqrt{3}}{2}, \sin\theta_1 = \frac{1}{2} \rightarrow \theta_1 = 30^o$$
Thus, $\sqrt{3}+i = 2(\cos 30^o + i\sin 30^o)$.
For the denominator, we have
$$|1+i| = \sqrt{2}   \\     \cos\theta_2 = \frac{1}{\sqrt{2}}, \sin\theta_2 = \frac{1}{\sqrt{2}} \rightarrow \theta_2 = 45^o$$
Thus, $1+i = \sqrt{2}(\cos 45^o + i\sin 45^o)$.
We have
$$\frac{\sqrt{3}+i}{1+i}=\frac{2(\cos 30^o + i\sin 30^o)}{\sqrt{2}(\cos 45^o + i\sin 45^o)} \\     = \sqrt{2}[\cos(-15^o)+i\sin(-15^o)]$$
Use the theorem.
$$(\sqrt{2}[\cos(-15^o)+i\sin(-15^o)])^6 = 2^3[\cos(-6\times 15^o) + i\sin(-6\times 15^o)]  \\                = 8[\cos(-90^o) + i\sin(-90^o)]  \\                =-8i$$

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Dropping a coin and finding out the depth

Question:
You released a coin in order to find the depth of a well. The time between dropping the coin and hearing it hit the bottom is 2.059 s. The speed of sound is $v_s=$343 m/s. Calculate the depth, $d$, of the well.

Answer:
The total time $T$ is divided into the time hitting the bottom, $t_1$ and time the sound traveling, $t_2$. Namely,
$$T=t_1+t_2$$
We have
$$d = \frac{1}{2}gt_1^2 \\   t_2 = \frac{d}{v_s}$$
Since $t_1 = T - t_2$, the depth can be expressed as
$$d = \frac{1}{2}g(T-\frac{d}{v_s})^2$$
Expand and arrange it in terms of $d$.
$$d^2 - d(\frac{2v^2}{g}+2v_sT)+v_s^2T^2=0$$
Plug $v_s=$343 m/s and $g=$9.8 m/s$^2$ into the above.
$$d^2 -25,422.5d + 498,770.7 = 0$$
The solution is
$$d=\frac{-(25,422.5)\pm\sqrt{(25,422.5)^2-4\times 498,770.7}}{2} \\    = 19.6 \mathrm{m}$$
The other solution is neglected.

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Equilibrium of forces

Question:
When this system is in equilibrium, find the tension in cord b, $T_b$. The hanging mass is 2.00 kg.

Answer:
As shown in the second figure, the tensions are vectors on $x$-$y$ coordinate. Since these are at equilibrium, the summation of the vectors must be zero.

In order to calculate, we can use the component method.
  $$T_{ax}=T_a \cos 270^o=0, T_{ay}=T_a \sin 270^o = -mg  \\ T_{bx}=T_b \cos 0^o=T_b, T_{by}=T_b \sin 0^o=0 \\ T_{cx}=T_c \cos 150^o, T_{cy}=T_c \sin 150^o$$
Thus,
$$T_{ax}+T_{bx}+T_{cx}=T_b+T_c\cos 150^o = 0 \\    T_{ay}+T_{by}+T_{cy}=-mg+T_c\sin 150^o = 0$$
From the second equation, we have
$$T_c = \frac{mg}{\sin 150^o}=39.2 \mathrm{N}$$
Then, plug it in the first equation.
$$T_b = T_c\sin 150^o = 33.9 \mathrm{N}$$












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Find the extrema: an optimization problem with Lagrange multipliers

Question:
By using Lagrange multipliers, find the extrema of $x^2+2y^2+3z^2$ when $3x+2y+z=-1$.

Answer:
Using a Lagrange multiplier, we define the following function:
$$F(x,y,z) =  x^2+2y^2+3z^2 - \lambda (3x+2y+z+1)$$
Take each partial derivative.
$$F_x = 2x-3\lambda = 0  \\ F_y = 4y-2\lambda = 0    \\ F_z = 6z-\lambda = 0   \\ F_{\lambda} = 3x+2y+z+1 = 0$$
They have four unknowns and four equations. Solve for each variable.
$$x=-9/34  \\   y=-3/34  \\   z=1/34  \\   \lambda = -6/34$$
If you plug $x$, $y$, and $z$, we obtain $F = 3/34$ as the extremum.

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Finding the extrema; an optimization problem

Question:
Find the extrema of $x^2+2y^2+3z^2$ when $3x+2y+z=-1$.

Answer:
From the condition, we can solve for $z$.
$$z = -(3x+2y+1)$$
Thus, we can define a function.
$$F(x,y) = x^2+2y^2+3(-(3x+2y+1))$$
It is supposed to find the extrema of this function. Take the partial derivatives.
$$\frac{\partial F(x,y)}{\partial x} = 56x+36y+18,  \\   \frac{\partial F(x,y)}{\partial y} = 36x+28y+12,   \\    \frac{\partial^2 F(x,y)}{\partial x^2} = 56,  \\     \frac{\partial^2 F(x,y)}{\partial y^2} = 28,  \\       \frac{\partial^2 F(x,y)}{\partial x \partial y} = 36$$
The value of the extremum is found when $\frac{\partial F(x,y)}{\partial x} = 0$ and $\frac{\partial F(x,y)}{\partial y} = 0$. Namely,
$$x = -\frac{9}{34}  \\    y = -\frac{3}{34}$$
In order to find if this is maximum or minimum, calculate $D = F_{xx}F_{yy}-F^2_{xy}$. Then, use the following condition:
If $D>0$ and $F_{xx}>0$, then it is a minimum.
If $D>0$ and $F_{xx}<0$, then it is a maximum.
If $D<0$, then it is not a extremum.
If $D=0$, then it cannot be determined in this method.
For this problem, $D = 272>0$ and $F_{xx}=56>0$, so $F(-9/34,-3/34) = 3/34$ is the minimum.

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Lagrangian function of a sphere on a cylindrical surface

Question:
There is an open cylinder of radius $R$., which is stationary. A hollow sphere of radius $\rho$ and mass $m$ rolls on the surface without slipping. Find the Lagrangian function.

Answer:
The Lagrangian function is defined as
$$L = KE - PE$$
where $KE$ and $PE$ represent kinetic energy and potential energy, respectively.
The mechanical potential energy is the gravitational force $\times$ height, which is taken from the center. The height varies with the angle $\theta$. Thus,
$$PE = -mg(R-\rho)\cos\theta$$
The negative sign indicates the height is directed to the center.
Now the kinetic energy has two parts, translational and rotational kinetic energies.
$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
The translational velocity, $v$, is along with the trace of circle whose radius is $R-\rho$. Therefore,
$$v = (R-\rho)\omega$$
The angular velocity can be expressed by
$$\omega = \theta' = \frac{d\theta}{dt}$$
The moment of inertia of hollow sphere is
$$I = \frac{2}{3}m\rho^2$$
Plug everything into above.
$$L = \frac{1}{2}m(R-\rho)^2 \theta '^2 + \frac{1}{2}\frac{2}{3}m\rho^2 \theta '^2 + mg(R-\rho)\cos\theta$$
We can rewrite it as
$$L = \frac{m\theta'^2}{6}(3R^2 -6R\rho + 5\rho^2)+ mg(R-\rho)\cos\theta$$

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Residue theorem and integral of a complex-valued function

Question:
Find the value of
$$\int_C \frac{dz}{z^2+2iz-4}$$
The contour, $C$, is a circle that has center $z=1$ and radius $\sqrt{2}$. It is directed positively.

Answer:
The integrand can be reduced in two factors.
$$I = \int_C \frac{dz}{z^2+2iz-4} = \int_C \frac{dz}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))}$$
The value when the denominator becomes zero is the singular point in the contour. $z=-(\sqrt{3}+i)$ and $z=\sqrt{3}-i$ can be the points, but only $z=\sqrt{3}-i$ is the pole inside contour.
Calculate the residue.
$$\mathrm{Res}(f(z):\sqrt{3}-i) = \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)f(z) \\                            = \lim_{z\rightarrow \sqrt{3}-i}(z-(\sqrt{3}-i)\frac{1}{(z+(\sqrt{3}+i))(z-(\sqrt{3}-i))} \\                            = \lim_{z\rightarrow \sqrt{3}-i}\frac{1}{z+(\sqrt{3}+i)}    \\                            = \frac{1}{2\sqrt{3}}$$
Then, using the residue theorem, we have the value of integral:
$$I = 2\pi i \times \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}\pi i}{3}$$

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Mechanical model of the proton (angular momentum etc.)

Question:
The proton's mass and radius are 1.67 $\times$ 10$^{-27}$ kg and 1.00 $\times$ 10$^{-15}$ m. This system is governed by quantum and classical regimes. Namely, the proton is assumed to have a rotational uniform solid spherical body and a half spin of quantum angular momentum. Find the equatorial velocity of the proton.

Answer:
In terms of classical sense, the moment of inertia of proton is
$$I = \frac{2}{5}mr^2$$
Therefore, the angular momentum becomes
$$L = I\omega = \frac{2}{5}mr^2\omega$$
For the quantum perspective, the spin angular momentum is given as
$$S = \frac{1}{2}\hbar$$
Therefore,
$$\frac{2}{5}mr^2\omega  = \frac{1}{2}\hbar$$
Solve for the angular velocity.
$$\omega = \frac{1}{2}\hbar \frac{5}{2mr^2}$$
The equatorial velocity is given by
$$v = r\omega =  \frac{5\hbar}{4mr} \\       = \frac{5}{4}\frac{1.055\times 10^{-34}\mathrm{Js}}{1.673\times 10^{-27}\mathrm{kg}\times 1.00\times 10^{-15}\mathrm{m}}$$
Then we have
$$v = 7.88 \times 10^7 \mathrm{m/s}$$

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Sound intensity

Question:
There is a speaker emitting sound with a uniform power of 250 W. At what distance will the intensity be just below the threshold of pain, which is 1.00 W/m$^2$?

Answer:
The sound intensity is given by
$$I = \frac{P}{A}$$
The sound emitted uniformly in space, so it is spherically propagated. The area of sphere is $4\pi r^2$. Thus,
$$I=\frac{P}{4\pi r^2}$$
Solve for the distance.
$$r^2 = \frac{P}{4\pi I} \\     r = \sqrt{\frac{P}{4\pi I}}$$
Plug in the numbers.
$$r = \sqrt{\frac{250 (\mathrm{W})}{4\pi \times 1.00 (\mathrm{W/m}^2)}}$$
Therefore, the distance is
$$r = 4.46 \mathrm{m}$$

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Simultaneous ordinary differential equation 1

Question:
Solve for the following simultaneous differential equations:
$$\left\{ \begin{array}{ll}            \frac{dx}{dt}+ay = e^{bt}  & (1)\\            \frac{dy}{dt}-ax = 0  & (2)           \end{array}    \right.$$
where $a$ and $b$ are real numbers.

Answer:
Take differentiation of equation (1).
$$x'' + ay' = be^{bt}$$
Use (2).
$$x'' + a^2x = be^{bt}$$
The general solution of this is given when right hand side is equal to zero.
$$x'' + a^2x = 0$$
Thus,
$$x(t) = C_1\cos at + C_2\sin at$$
To solve for the particular solution, due to the expression of right hand side, we can assume that it is $x=Ae^{bt}$. Then, substitute it into the above equation:
$$(Ae^{bt})'' + a^2(Ae^{bt})' = Ab^2e^{bt} + Aa^2e^{bt} = (a^2 + b^2)Ae^{bt}$$
This must be equal to $be^{bt}$, so
$$A = \frac{b}{a^2 + b^2}$$
The particular solution becomes $\frac{b}{a^2 + b^2}e^{bt}$. Therefore,
$$x(t) = C_1\cos at + C_2\sin at + \frac{b}{a^2 + b^2}e^{bt}$$
Plug this solution into (1) to obtain the solution of $y(t)$.
$$(C_1\cos at + C_2\sin at + \frac{b}{a^2 + b^2}e^{bt})'+ay=e^{bt}  \\    -C_1a\sin at + C_2a\cos at + \frac{b^2}{a^2 + b^2}e^{bt} + ay = e^{bt} \\   ay = C_1a\sin at - C_2a\cos at - \frac{b^2}{a^2 + b^2}e^{bt} + e^{bt}$$
The total solution of $y$ becomes
$$y(t) = C_1\sin at - C_2\cos at + \frac{a}{a^2 + b^2}e^{bt}$$

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Normalization of a wave function

Question:
The wave function for the one dimensional harmonic oscillator with the potential energy, $\frac{1}{2}kx^2$, is given as
$$\Phi_0 = C\exp(-ax^2)$$
Find the constant $C$ when the wave function is normalized. Note that you can use the normalized
Gaussian distribution: $\frac{1}{\sqrt{2\pi}\sigma}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2\sigma^2})dx = 1$.

Answer:
For the normalized wave function, it has to satisfy:
$$\int^{\infty}_{-\infty}\Phi_0\Phi^*_0 dx = C^2\int^{\infty}_{-\infty}\exp(-2ax^2) dx = 1$$
where $\Phi^*_0$ is the complex conjugate of $\Phi_0$. Compare this with the Gaussian normal distribution.
$$2a = \frac{1}{2\sigma^2}  \\    \sigma = \frac{1}{\sqrt{4a}}$$
Plug it in the formula of Gaussian distribution.
$$\frac{1}{\sqrt{2\pi}(1/\sqrt{4a})}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2(1/4a)})dx = 1$$
We can see
$$C^2 = \frac{1}{\sqrt{2\pi}(1/\sqrt{4a})} = \frac{1}{\sqrt{2\pi}}\sqrt{4a} \\            = \sqrt{\frac{2a}{\pi}}$$
Therefore,
$$C = (2a/\pi)^{1/4}$$

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Integral using Beta and Gamma functions

Question:
Find the value of
$$\int^{\infty}_{-\infty}\frac{dx}{x^6+1}$$

Answer:
We can reduce the above integral into
$$\int^{\infty}_{-\infty}\frac{dx}{x^6+1} = 2\int^{\infty}_{0}\frac{dx}{x^6+1}$$
Replace $x$ with $t^{\frac{1}{6}}$. Then, we have $dx=\frac{1}{6}t^{-\frac{5}{6}}dt$. Thus,
$$\int^{\infty}_{0}\frac{dx}{x^6+1} = \frac{1}{6}\int^{\infty}_{0}\frac{t^{-\frac{5}{6}}}{t+1}dt$$
Remember that the Beta function is defined as follows:
$$B(p,q) = \int^{\infty}_{0}\frac{x^{p-1}}{(1+x)^{p+q}}dx \\ B(p,q)=B(q,p)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$
Therefore, the above expression becomes
$$\frac{1}{6}\int^{\infty}_{0}\frac{t^{-\frac{5}{6}}}{t+1}dt = \frac{1}{6}B(\frac{1}{6},\frac{5}{6}) \\    = \frac{1}{6}\frac{\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6}+\frac{5}{6})} \\    =  \frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})$$
since $\Gamma(1)=1$.
Recall the following formula:
$$\Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin\pi s}$$
The parameter satisfies 0<s<1.
Therefore,
$$\frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6}) = \frac{1}{6}\frac{\pi}{\sin(\pi/6)} = \frac{\pi}{3}$$
Hence,
$$\int^{\infty}_{-\infty}\frac{dx}{x^6+1} = \frac{2\pi}{3}$$

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Newton's equation of motion

Question:
The objects A and B have masses, 20.0 kg and 10.0 kg, respectively. They are placed on a frictionless surface. The force (29.4 N) is acted as shown and these objects move together.
(1) Find the acceleration of the objects.
(2) What is the force exerting B from A?

Answer:
From the second figure, the forces exerting B from A and vice versa are equal due to the action and re-action pair.

(1) You can set up the Newton's equation for object A:
$$\sum F_A = F - N = m_A a$$
The equation for object B is:
$$\sum F_B = N = m_B a$$
Notice that the acceleration of each object is the same since they move together. Solve for the acceleration from them. Plug $N=m_B a$ into $F - N = m_A a$.
$$F - m_B a = m_A a \\     (m_A + m_B)a = F  \\    a = \frac{F}{m_A + m_B}$$
Therefore,
$$a = \frac{29.4 \mathrm{N}}{20.0 \mathrm{kg} + 10.0 \mathrm{kg}} = 0.98 \mathrm{m/s^2}$$

(2) From the solution in (1), we can find the force exerting B from A, which is $N$ in the figure. Thus,
$$N = m_B a = 10.0\mathrm{kg}\times 0.98\mathrm{m/s^2} = 9.8\mathrm{N}$$

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Differentiation with trigonometric and logarithmic functions

Question:
Here is a function of $x$.
$$y=\alpha \cos x + 2 - \cos x \log\frac{1+\cos x}{1-\cos x}$$
$\alpha$ is a constant, and the range of $x$ is $0<x<\pi$.
Calculate $\frac{d^2 y}{dx^2}+\frac{\cos x}{\sin x}\frac{dy}{dx}$.

Answer:
Define each term.
$$y_1 =  \alpha \cos x \\     y_2 = 2 \\     y_3 = \log\frac{1+\cos x}{1-\cos x}$$
Calculate each derivative.
$$y_1' = -\alpha \sin x  \\   y_2' = 0 \\ $$
The above derivatives are easy, but the third one is a little complicated:
$$y_3' = (\log|1+\cos x| - \log|1-\cos x|)' \\          = \frac{(1+\cos x)'}{1+\cos x} - \frac{(1-\cos x)'}{1-\cos x}   \\          = \frac{-\sin x}{1+\cos x} - \frac{\sin x}{1-\cos x}   \\         = \frac{-\sin x(1-\cos x) - \sin x(1+\cos x)}{1 - \cos^2x}  \\         = \frac{-2\sin x}{\sin^2 x}   \\         = \frac{-2}{\sin x}$$
Since $y=y_1+y_2-\cos x y_3$, the derivative becomes $y'=y_1'+y_2'-[(\cos x)' y_3+\cos x y_3']$. Namely,
$$\frac{dy}{dx} = -\alpha\sin x + \sin x \log\frac{1+\cos x}{1-\cos x} + \frac{2\cos x}{\sin x}$$
The second derivative is calculated as follows:
$$\frac{d^2y}{dx^2} = -\alpha\cos x + \cos x \log\frac{1+\cos x}{1-\cos x} + \sin x\frac{-2}{\sin x} + \frac{2((\cos x)'\sin x - \cos x(\sin x)')}{\sin^2x}  \\            = -\alpha\cos x + \cos x \log\frac{1+\cos x}{1-\cos x} - 2 - \frac{2}{\sin^2x}$$
You can notice that the first three terms of above is equal to $-y$. Therefore,
$$\frac{d^2y}{dx^2} = -y -\frac{2}{\sin^2 x}$$
The next term becomes
$$\frac{\cos x}{\sin x}\frac{dy}{dx} = -\alpha\cos x + \cos x\log\frac{1+\cos x}{1-\cos x} + \frac{2\cos^2 x}{\sin^2 x}$$
This can be rewrite as
$$\frac{\cos x}{\sin x}\frac{dy}{dx} = -y + 2 + \frac{2\cos^2 x}{\sin^2 x}$$
Thus, we have
$$\frac{d^2 y}{dx^2}+\frac{\cos x}{\sin x}\frac{dy}{dx} = -2y$$

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A problem with trigonometric addition theorem

Question:
Show that $\frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\sin\alpha - \sin\beta}=\sin\alpha + \sin\beta$.

Answer:
Use $\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha$.
The given expression will become
$$\frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\sin\alpha - \sin\beta} = \frac{(\sin\alpha\cos\beta + \sin\beta\cos\alpha)(\sin\alpha\cos\beta - \sin\beta\cos\alpha)}{\sin\alpha - \sin\beta}      = \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\sin\alpha - \sin\beta}$$

Use $\sin^2\theta+\cos^2\theta=1$ to eliminate $\cos^2\alpha$ and $\cos^2\beta$.
$$\frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\sin\alpha - \sin\beta}        = \frac{\sin^2\alpha(1-\sin^2\beta) - \sin^2\beta(1-\sin^2\alpha)}{\sin\alpha - \sin\beta}       =  \frac{\sin^2\alpha-\sin^2\alpha\sin^2\beta - \sin^2\beta + \sin^2\beta\sin^2\alpha}{\sin\alpha - \sin\beta}      = \frac{\sin^2\alpha - \sin^2\beta}{\sin\alpha - \sin\beta}      = \frac{(\sin\alpha + \sin\beta)(\sin\alpha - \sin\beta)}{\sin\alpha - \sin\beta}      = \sin\alpha + \sin\beta$$
QED

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Trigonometric functions with a parametric representation

Question:
Here is a set of simultaneous equations, $\sin x = \cos y = \sin (y-x)$. The ranges are $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$. Find $x$ and $y$.

Answer:
It is easier to equate the equations with a parameter. Thus,
$$\sin x = \cos y = \sin (y-x) = t$$
From the conditions, we can notice that $\sin x = \cos y = t > 0$.  If we use $\sin^2\theta + \cos^2\theta = 1$, we can have the following expressions:
$$\cos x = \sqrt{1-t^2};  \sin y = \sqrt{1-t^2}$$
Now, we can rewrite $\sin (y-x)$ as follows:
$$\sin (y-x) = \sin y\cos x - \sin x\cos y  \\                     = \sqrt{1-t^2}\cdot\sqrt{1-t^2}-t\cdot t  \\                    = (1-t^2) - t^2  \\                    = 1 - 2t^2$$
Since $\sin(y-x) = t$, we have
$$1 - 2t^2 = t$$
Therefore,
$$2t^2 + t - 1 = 0  \\     (2t - 1)(t - 1) = 0$$
The solution is that $t=\frac{1}{2}$ because $t>0$.
This gives $\sin x = \cos y = \frac{1}{2}$. Hence we obtain $x = \frac{\pi}{6}$ and $y = \frac{\pi}{3}$.

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Trigonometric functions 2

Question:
When $\tan\theta = -2$, find $\cos 2\theta$ and $\sin 2\theta$.

Answer:
This might be a little tricky, but if you come up with a relationship between $\tan^2 \theta$ and $\cos^2 \theta$ first. Then, $\cos^2 \theta$ can be converted into $\cos 2\theta$.

Remember
$$1+\tan^2\theta = \frac{1}{\cos^2\theta}$$
This can be derived from $\sin^2\theta + \cos^2\theta = 1$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$.

Since $\tan\theta = -2$, we have
$$\cos^2\theta = \frac{1}{1+\tan^2\theta} = \frac{1}{1+(-2)^2} = \frac{1}{5}$$

Now, recall the double-angle formula as for cosine:
$$\cos 2\theta = 2\cos^2\theta -1$$
Plug above result into this.
$$\cos 2\theta = 2\times\frac{1}{5} -1 = -\frac{3}{5}$$
Also remember
$$\sin 2\theta = 2\sin\theta\cos\theta$$
Modify the right hand side.
$$\sin 2\theta = 2\sin\theta\cos\theta \times \frac{\cos\theta}{\cos\theta} = 2\tan\theta\cos^2\theta$$
Thus, we have
$$\sin 2\theta = 2\times(-2)\times\frac{1}{5}=-\frac{4}{5}$$

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Let's start with tangent!

Question:
Calculate $(\tan 15^o + \sqrt{3})^2$ without a calculator.

Answer:
It is not easy to obtain $\tan 15^o$ without a calculator. In this case, you may want to come up with angles which can easily convert into specific numbers, such as 30, 45, 60, 90, etc. In fact, 15 = 45 - 30, so $\tan 15^o = \tan(45^o - 30^o)$. Thus, we can use the addition theorem.
$$\tan(45^o - 30^o) = \frac{\tan 45^o - \tan 30^o}{1 + \tan 45^o \tan 30^o}$$
If we use typical triangles, we know $\tan 45^o = 1$ and $\tan 30^o = 1/\sqrt{3}$. Therefore,
$$\tan 15^o = \frac{1 - 1/\sqrt{3}}{1 + 1 \times 1/\sqrt{3}}$$
Multiply $\sqrt{3}$ by both numerator and denominator.
$$= \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$$
$$= \frac{\sqrt{3}-1}{\sqrt{3}+1}$$
$$= \frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$$
$$= \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$
Hence, we have
$$(\tan 15^o + \sqrt{3})^2 = 2^2 = 4$$

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