7
By DeMoivre's identity,
$\cos(3x)+i\sin(3x)=(\cos(x)+i\sin(x))^3=$
$\cos^3(x)+3i\cos^2(x)\sin(x)-3\cos(x)\sin^2(x)-i\sin^3(x)$.
Grouping the imaginary terms together, we see that $\sin(3x)=3\cos^2(x)\sin(x)-\sin^3(x)$.
We can rewrite $3\cos^2(x)\sin(x)$ as $3\sin(x)(1-\sin^2(x))$, which gives us $\sin(3x)=3\sin(x)-4\sin^3(x)$.
Finally, we can write $\...
7
The identity is
$$\sin^3x=\frac{3\sin x-\sin(3x)}{4}$$
Proof:
\begin{align*}
4\sin^3x-3\sin x &= \sin x(4\sin^2x-3)
\\
&=\sin x(1-4\cos^2x) \quad \text{Pythagorean identity}
\\
&=\sin x(2\cos^2x-\cos(2x)-4\cos^2x) \quad \text{Double angle cosine}
\\
&=-\sin x(2\cos^2x+\cos(2x))
\\
&=-[(2\sin x\cos x)\cos x+\sin x\cos(2x)]
...
7
Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
From $\pi < \frac{22}{7}$ we deduce $\frac{3 \pi}{10} < \frac{66}{70} < 1$, and so
$$2 \sin 1 > 2 \sin \frac{3 \pi}{10} = 2 \cdot \frac{1}{4}(1 + \sqrt{5}) > ...
7
There is rule to satisfy either to find the identities or to find any particular value of a trigonometric function with that prime in the denominator and a multiple of $ \pi$ in the numerator.
Considering the case of $5$, it is a Fermat prime and hence it can be constructed using straight edges and compass. Because, $F_n = 2^{2^n}+1$ and $F_1=5$. And since ...
6
Note that\begin{align}(\forall z\in\mathbb C):\bigl\lvert\cos(z)\bigr\rvert&=\left\lvert1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\cdots\right\rvert\\&\leqslant1+\frac{\lvert z\rvert^2}{2!}+\frac{\lvert z\rvert^4}{4!}+\frac{\lvert z\rvert^6}{6!}+\cdots\\&\leqslant1+\lvert z\rvert+\frac{\lvert z\rvert^2}{2!}+\frac{\lvert z\rvert^3}{3!}+\frac{\...
5
Solve $\cos x=\frac{\sqrt 3}2$ and recall that $\cos(\pi-x)=-\cos x$.
4
Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
Then, from Maclaurin expansion, we have
$$ \sin 1 = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \ldots $$
Observe that the absolute value of each of these terms is ...
4
Hint:
$$\cos(\arcsin(x))=\sqrt{1-x^2},$$ for all $x\in (-\pi/2,\pi/2)$.
4
You can do this purely with (trigonometric) formulas, but here's a more geometric approach.
Since $\sin^{-1}(-a)=-\sin^{-1}(a)$ and $\cos(-a)=\cos(a)$, you have:
$$\cos\left(\sin^{-1}\left(-\tfrac{3}{5}\right)\right)=\cos\left(\sin^{-1}\left(\tfrac{3}{5}\right)\right)$$
Now imagine (or draw!) a right triangle with sides 3, 4 and 5:
$\sin^{-1}\left(\tfrac{3}...
4
The hint.
Use $$\arctan(n+1)-\arctan{n}=arccot(n^2+n+1)$$ and the telescopic summation.
4
The case of multiples of $\pi/11$ actually involves five "symmetrically equivalent" forms:
$4\sin(5\pi/11)-\tan(2\pi/11)=\sqrt{11}$
$4\sin(\pi/11)+\tan(4\pi/11)=\sqrt{11}$
$4\sin(4\pi/11)-\tan(5\pi/11)=-\sqrt{11}$
$4\sin(2\pi/11)+\tan(3\pi/11)=\sqrt{11}$
$4\sin(3\pi/11)+\tan(\pi/11)=\sqrt{11}$
All are derivable from the quadratic Gauss sum ...
4
Advanced Calculus
This is the well-known Laplace integral.
WLOG, assume $a,c > 0$
Let the integral be $I (a,c)$, then
$$
\newcommand{\abs}[1]{\left\vert #1 \right\vert}
\newcommand\rme{\mathrm e}
\newcommand\imu{\mathrm i}
\DeclareMathOperator{\diff}{\mathrm d\!}
\DeclareMathOperator\sgn{sgn}
I (a, 0) = a\int_0^{+\infty} \frac 1{x^2 + a^2} \mathrm ...
3
It's actually wrong. The correct answer is actually
$$\arctan(\tan(y))=y+k\pi\quad k\in\mathbb{Z}\text{ s.t. } |y+k\pi|<\pi/2$$
if the standard $\arctan:\mathbb{R}\to(-\pi/2,\pi/2)$ function is used as an inverse.
3
Note that
$$
2\sin\alpha\sec^2\alpha=\frac{2\sin\alpha}{\cos^2\alpha}=\frac{2\sin\alpha}{1-\sin^2\alpha}
$$
If
$$
f(x)=\arctan\frac{2x}{1-x^2}
$$
then
$$
f'(x)=\frac{2}{1+x^2}
$$
so $f(x)$ differs from $2\arctan x$ by a constant over $(-1,1)$. Since $f(0)=0=2\arctan0$, we can say that
$$
\arctan\frac{2\sin\alpha}{1-\sin^2\alpha}=2\arctan\sin\alpha
$$
For $x&...
3
You're making the mistake of not converting everything into degrees and just the $\pi$ into degrees.
$$(3\pi-10) rad = (540-10\times 57.296)^{\circ}$$
$$= -32.96^{\circ} \in [-90^{\circ},90^{\circ}]$$
3
$\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$.
$(\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta=1$
$\dfrac32-\sqrt2 k=1$
$k=\dfrac 1{2\sqrt2}$
3
This requires the use of the Laplace transform.
We set
$$J(t;q)=q\int_0^\infty \frac{\cos(tx)dx}{x^2+q^2}.$$
Then we lake the Laplace transform of it:
$$\begin{align}
\mathcal{L}\{J(t;q)\}(s)&=\int_0^\infty e^{-st}J(t;q)dt\\
&=q\int_0^\infty \int_0^\infty \frac{e^{-st}\cos(tx)}{q^2+x^2}dxdt\\
&=q\int_0^\infty \frac{1}{x^2+q^2}\int_0^\infty e^{-...
3
I'll leave it to you to prove by induction that the partial sum is $\arctan\left(1-\frac{1}{n^2+n+1}\right)$, so the limit is $\pi/4$. One approach to obtaining this partial sum is that of @achillehui's telescope, viz. $$\left[\arctan(2r^2+2r+1)\right]_{-1}^n=\arctan\frac{n^2+n}{n^2+n+1}.$$
Edit: just to spell it out, the definition $f(r):=\arctan(2r^2+2r+1)...
2
For the minimum, note that since $x,y\ge0\implies y\le\dfrac{2\pi}3$, we have $$\dfrac{x-y}2=\dfrac{x+y-2y}2=\dfrac{\dfrac{2\pi}3-2y}2=\frac\pi3-y$$ so $$\sin x+\sin y = \sqrt{3}\cos\frac{x-y}{2}=\sqrt3\cos\left(\frac\pi3-y\right)\ge\begin{cases}\sqrt3\cos\left(\frac\pi3-0\right)\\\sqrt3\cos\left(\frac\pi3-\frac{2\pi}3\right)\end{cases}=\frac{\sqrt3}2.$$
2
Your mistake is at this step:
$\sin(2\theta)=-\frac{1}{\sqrt{2}}$, hence $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$.
The correct step is:
$\sin(2\theta)=-\frac{1}{\sqrt{2}}$, hence $2\theta = \frac{5\pi}{4}+2k\pi$ or $\frac{7\pi}{4}+2k\pi$, where $k$ is any integer. So dividing by 2 we get
$\theta = \frac{5\pi}{8}+k\pi$ or $\frac{7\pi}{8}+k\pi$, where $...
2
Hence, $2\theta = \frac{5\pi}{4}$ or $\frac{7\pi}{4}$
Don't forget $+2\pi n$.
Since the signs of $\cos\theta$ and $\sin\theta$ place $\theta$ in the fourth quadrant, only two candidates hold up. You can choose between them by testing whether $\cos\theta > \cos\frac{7\pi}{4}$
2
If $0\le \theta<2\pi$, then $0\le 2\theta<4\pi$. You should have $4$ possible values of $\theta$.
$\cos (\theta) = \sqrt{\frac{1}{2}+\frac{1}{2\sqrt{2}}}$ and $\sin
(\theta) = -\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{2}}}$ imply that $\sin(2\theta)=-\dfrac1{\sqrt2}$, but not the other way round.
Note that $\displaystyle \tan\theta=\frac{-\sqrt{\frac{1}{2}-\...
2
Remember that if the range of $\theta$ is $0 \leq \theta \lt 2 \pi$ then the range od $2\theta$ will be $0\leq\theta \lt 4\pi$. So $2\theta = \ldots$
2
Suppose $\theta$ is an angle in radians. Let's create a new unit for angles called the $\mathrm{Moytaba}$, or $\mathrm{Moy}$ for short. Let us define it by $1 \,\,\mathrm{Moy} = c\,\,\mathrm{rad}$ where $c$ is some constant scaling factor (of course, any transformation between units of the same dimension has to be a constant scaling factor, for obvious ...
2
The fact that $\cos \frac{\pi}{3} = \frac{1}{2}$ is not something that requires memorization, but like many identities in mathematics, it is convenient and efficient to memorize because the proof is more sophisticated than the result.
In an equilateral triangle $\triangle ABC$, draw the altitude $\overline{AD}$ from $A$ to $\overline{BC}$. Since $\angle A =...
2
Simply use that $\sin\frac{\pi}3=\frac{\sqrt3}2$ and $\cos\frac{\pi}3=\frac12$.
Then $\tan \frac{\pi}3=\dfrac{\frac{\sqrt3}2}{\frac12}=\sqrt3$.
Thus $\arctan \sqrt3=\frac{\pi}3$.
Adjust for the sign: $\arctan -\sqrt3=-\frac{\pi}3$.
2
I agree with one of the comments to your question that it’s worth memorizing some of the basic triangles and the associated sines and cosines. However, a 30-60-90 triangle can be quickly reverse-engineered from $\tan\theta = -\sqrt3$: we have $$\frac yx = -\sqrt3 \\ x^2+y^2=1$$ from which $4x^2=1$, so $x=\pm\frac12$ and $y=\mp\frac12\sqrt3$. The hypotenuse ...
2
$$\log\frac{1+\cos x}{1-\cos x}=\log\frac{(1+\cos x)^2}{\sin^2 x}=2\log(1+\cos x)-2\log|\sin x|.$$
The first term is constant, while the sine is asymptotic to $|x|$ (for $x\to0$).
2
I think, it's better
$$\left|\sin\frac{\alpha}{2}\right|=\sqrt{\frac{1-\cos\alpha}{2}}$$ and
$$\left|\cos\frac{\alpha}{2}\right|=\sqrt{\frac{1+\cos\alpha}{2}}$$
Your mistake is that $\sqrt{x^2}=\pm x$ is wrong.
The right identity it's:
$$\sqrt{x^2}=|x|.$$
For example, after
$$\sin^4\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\frac{\sin^2\alpha}{4}=0$$ we ...
2
Use Prosthaphaeresis Formulas,
$$0=\sin x-\sin y=2\sin\dfrac{x-y}2\cos\dfrac{x+y}2$$
If $\sin\dfrac{x-y}2=0,\dfrac{x-y}2=m\pi$
$\implies x-y=2m\pi$
As $0\le x+y\le\pi, -\pi\le x-y\le\pi\implies m=0$
If $\cos\dfrac{x+y}2=0,\dfrac{x+y}2=(2r+1)\dfrac\pi2\implies x+y=(2r+1)\pi$
But $0\le x+y\le\pi,r=0$
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