Question:
(1) The range of $x$ is: $0^o \leqq x \leqq 180^o$. Solve $\cos^4 x > \sin^4 x$.
(2) The range of $x$ is: $0^o \leqq x < 360^o$. Solve $\cos 2x + \sin x < 0$.
Answer:
(1) Arrange the equation.
\begin{eqnarray}
\cos^4 x - \sin^4 x &>& 0 \\
(\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x) &>& 0 \\
(1)(\cos x + \sin x)(\cos x - \sin x) &>& 0
\end{eqnarray}
Therefore, we can state
\begin{eqnarray}
& & (\cos x + \sin x)>0 \ \mathrm{and} \ (\cos x - \sin x)>0 \\
\mathrm{or} & & \nonumber \\
& & (\cos x + \sin x)<0 \ \mathrm{and} \ (\cos x - \sin x)<0
\end{eqnarray}
In case of (4), within the $x$ range, when $\cos x > \sin x$, $0^o \leqq x < 45^o$. When $\cos x + \sin x > 0$, $0^o \leqq x < 135^o$ since the solution of $\cos x + \sin x = 0$ is $x=135^o$. Therefore, one range is
\[
0^o \leqq x < 45^o
\]
In case of (5), when $\cos x < \sin x$, $45^o < x \leqq 180^o$. When $\cos x + \sin x < 0$, $135^o < x \leqq 180^o$. Therefore, the other range is
\[
135^o < x \leqq 180^o
\]
Thus, the answer is $0^o \leqq x < 45^o$ and $135^o < x \leqq 180^o$.
(2) Use the double angle formula, $\cos 2x = 1-2\sin^2 x$ to rearrange the given expression.
\begin{eqnarray}
1-2\sin^2 x +\sin x &<& 0 \\
2\sin^2 x -\sin x-1 &>& 0 \\
(2\sin x + 1)(\sin x -1) &>& 0
\end{eqnarray}
$\sin x -1$ is always negative or zero, so the condition must be only
\[
2\sin x + 1 <0 \ \mathrm{and} \ \sin x -1<0
\]
Then, $2\sin x + 1 <0$ determines the range. Namely, $\sin x < -\frac{1}{2}$. For $\sin x = -\frac{1}{2}$, $x= -30^o$ or $210^o$. Therefore, the range must be
\[
210^o < x < 330^o
\]
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Showing posts with label trig function. Show all posts
Showing posts with label trig function. Show all posts
Thursday, February 25, 2016
Thursday, February 4, 2016
Parameter representation of derivatives
Question:
$x$ and $y$ are given as follows:
\begin{eqnarray*}
x &=& e^t \cos t \\
y &=& e^t \sin t
\end{eqnarray*}
Express $\frac{d^2y}{dx^2}$ in terms of $t$.
Answer:
Take the derivative with respect to $t$.
\begin{eqnarray*}
\frac{dx}{dt} &=& e^t (\cos t -\sin t) \\
\frac{dy}{dt} &=& e^t (\cos t +\sin t)
\end{eqnarray*}
Therefore,
\[ \frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}
= \frac{\cos t + \sin t}{\cos t - \sin t}\]
Now, $\frac{d^2y}{dx^2}$ is a little tricky.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dt}\frac{dy}{dx}\frac{dt}{dx} \]
Therefore,
\begin{eqnarray*}
\frac{d^2y}{dx^2} &=& \frac{d}{dt}\left(\frac{\cos t + \sin t}{\cos t - \sin t}\right)\frac{1}{e^t (\cos t -\sin t)} \\
&=& \frac{(-\sin t + \cos t)(\cos t - \sin t)-(\cos t + \sin t)(-\sin t - \cos t)}{e^t(\cos t - \sin t)^3} \\
&=& \frac{\cos^2 t - 2\cos t\sin t + \sin^2 t + \cos^2 t + 2\cos t\sin t + \sin^2 t}{e^t(\cos t - \sin t)^3} \\
&=& \frac{2(\cos^2 t + \sin^2 t)}{e^t(\cos t - \sin t)^3} \\
&=& \frac{2}{e^t(\cos t - \sin t)^3}
\end{eqnarray*}
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$x$ and $y$ are given as follows:
\begin{eqnarray*}
x &=& e^t \cos t \\
y &=& e^t \sin t
\end{eqnarray*}
Express $\frac{d^2y}{dx^2}$ in terms of $t$.
Answer:
Take the derivative with respect to $t$.
\begin{eqnarray*}
\frac{dx}{dt} &=& e^t (\cos t -\sin t) \\
\frac{dy}{dt} &=& e^t (\cos t +\sin t)
\end{eqnarray*}
Therefore,
\[ \frac{dy}{dx} = \frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}
= \frac{\cos t + \sin t}{\cos t - \sin t}\]
Now, $\frac{d^2y}{dx^2}$ is a little tricky.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dt}\frac{dy}{dx}\frac{dt}{dx} \]
Therefore,
\begin{eqnarray*}
\frac{d^2y}{dx^2} &=& \frac{d}{dt}\left(\frac{\cos t + \sin t}{\cos t - \sin t}\right)\frac{1}{e^t (\cos t -\sin t)} \\
&=& \frac{(-\sin t + \cos t)(\cos t - \sin t)-(\cos t + \sin t)(-\sin t - \cos t)}{e^t(\cos t - \sin t)^3} \\
&=& \frac{\cos^2 t - 2\cos t\sin t + \sin^2 t + \cos^2 t + 2\cos t\sin t + \sin^2 t}{e^t(\cos t - \sin t)^3} \\
&=& \frac{2(\cos^2 t + \sin^2 t)}{e^t(\cos t - \sin t)^3} \\
&=& \frac{2}{e^t(\cos t - \sin t)^3}
\end{eqnarray*}
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Tuesday, February 2, 2016
Trigonometric functions and complex variables
Question:
Define $z=x+iy$. Prove the following equation
\[ \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \tan z \]
Answer:
There are following relationships:
\begin{eqnarray*}
\sinh x &=& -i\sin ix \\
\cosh x &=& \cos ix \\
\sin A + \sin B &=& 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \\
\cos A + \cos B &=& 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} &=& \frac{\sin 2x + i(-i\sin i2y)}{\cos 2x + \cos i2y}\\
&=& \frac{\sin 2x + \sin i2y}{\cos 2x + \cos i2y} \\
&=& \frac{2\sin\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}{2\cos\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}} \\
&=& \frac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)} \\
&=& \frac{\sin(x+iy)}{\cos(x+iy)} \\
&=& \tan(x+iy) = \tan z
\end{eqnarray*}
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Define $z=x+iy$. Prove the following equation
\[ \frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} = \tan z \]
Answer:
There are following relationships:
\begin{eqnarray*}
\sinh x &=& -i\sin ix \\
\cosh x &=& \cos ix \\
\sin A + \sin B &=& 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \\
\cos A + \cos B &=& 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\end{eqnarray*}
Thus,
\begin{eqnarray*}
\frac{\sin 2x + i\sinh 2y}{\cos 2x + \cosh 2y} &=& \frac{\sin 2x + i(-i\sin i2y)}{\cos 2x + \cos i2y}\\
&=& \frac{\sin 2x + \sin i2y}{\cos 2x + \cos i2y} \\
&=& \frac{2\sin\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}}{2\cos\frac{2x+i2y}{2}\cos\frac{2x-i2y}{2}} \\
&=& \frac{2\sin(x+iy)\cos(x-iy)}{2\cos(x+iy)\cos(x-iy)} \\
&=& \frac{\sin(x+iy)}{\cos(x+iy)} \\
&=& \tan(x+iy) = \tan z
\end{eqnarray*}
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Wednesday, January 27, 2016
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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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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Monday, January 25, 2016
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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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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Sunday, January 24, 2016
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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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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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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Saturday, January 23, 2016
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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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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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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