Tag Info

Hot answers tagged

13

Answer: $\displaystyle \int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}\,dx=\frac{\pi^2}{24}$ Proof: We are making use of $3$ Lemmas which are (quite ) easy to prove: 1. $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )}=\frac{\pi}{6} $ Proof: $$\begin{align*} \int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )} ...


11

Proposition : \begin{equation} \int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx=\frac{\pi}{2}\left[\arctan\sqrt{2\mu}-\arctan\sqrt{\frac{\mu}{\mu+1}}\right]\quad,\quad\text{for }\,\mu\ge0 \end{equation} Proof : Let \begin{equation} I(\mu)=\int_0^{\Large\frac{\pi}{4}}\arctan\sqrt{\frac{\mu\cos2x}{\cos^2x}}\,dx \end{equation} then ...


6

Notice, you can put $ t = \frac{1}{x} $. Hence $$ \lim_{x \to \infty} x - x \cos(\frac{4}{x} ) = \lim_{t \to 0} \frac{1}{t} - \frac{1}{t} \cos(4t) = \lim_{t \to 0} \frac{ 1 - \cos (4t) }{t} $$ Now, multiply this by $\frac{4}{4} $ and remember the following well known limit: $$ \lim_{\alpha \to 0 } \frac{ 1 - \cos \alpha}{\alpha} = 0 $$


5

Let $z=\cos\frac{a\pi}b+i\sin\frac{a\pi}b=e^{\frac{a\pi i}{b}}$. Then $z^{2b}=1$ and hence $z$ is algebraic. Finally $\cos\frac{a\pi}b=\frac12(z+z^{-1})$ is also algebraic.


5

My $100^{\mbox{th}}$ answer posted on Math S.E $\,$ (っ^▿^)۶🍸🌟٩(˘◡˘ ) Wolfram Mathematica $9.0$ is able to evaluate this indefinite integral. Here is the output \begin{equation} \sqrt{\frac{5+\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}-3\right)\tan ...


4

Short answer: The sine function takes values from $-1$ to $1$ because that's just the nature of the function. Longer answer: The elementary definition of the sine function for angles from $0$ to $\pi/2$ is that it is the ratio of the length of one (opposite to the corner of which the sine is calculated) catete of a right angle triangle to the hypothenuse ...


4

$\cfrac{3\pi}{8}-\left(-\cfrac{\pi}{8}\right)$ is half of the period. Hence, the period is $\pi$ i.e. $b=2$ and therefore, $c=\cfrac{\pi}{4}(+n\pi)$.


4

$\sin(x)$ is concave on that range and the line pass through $(0,0)$ and $(\pi/2,1)$ is $y=\frac{2}{\pi}x$


4

L'Hospital rule says that if $\lim_{x \to \infty} f(x)=\lim_{x \to \infty}g(x)= \infty$ and $\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ exists then $\lim_{x \to \infty} \frac{f(x)}{g(x)}$ exists and is the same thing. L'Hospital doesn't say anything about what happens if $\lim_{x \to \infty} \frac{f'(x)}{g'(x)}$ doesn't exists, and the converse of L'H is not ...


4

Use the recursive relation involving Chebychev polynomials: $$ T_{n+1}(X) = 2XT_n(X) - T_{n-1}(X) \\ T_n(\cos \theta) = \cos n\theta $$ Proof: first prove existence of such polynomials. Then the identity $$ \cos ((n+1)\theta) + \cos ((n-1)\theta) = 2\cos \theta \cos n\theta $$


4

Putting $x=\frac{1}{t}$ it becomes: $$\ \lim_{t\to0}\frac{1-\cos(4t)}{t}=\lim_{t\to0}\frac{1-\cos(4t)}{16t^2}\cdot\frac{16t^2}{t}=\lim_{t\to0}8\cdot t=0$$


3

We can write $$x-x\cos\frac4x=x\frac{1-\cos^2\frac4x}{1+\cos\frac4x}=\left(\frac{\sin\frac4x}{\frac4x}\right)^2\cdot \frac{16}{1+\cos\frac4x}\cdot\frac1x.$$ Now as $x\to\infty$, the limit of the first factor is $1$, the limit of the second is $8$, and the limit of the third is $0$. I let you conclude.


3

Add $$\sin (x+y) = \sin x\cos y + \cos x\sin y $$ and $$\sin (x-y) = \sin x\cos y - \cos x\sin y. $$ See what you get and apply it to your problem.


3

$$ \underbrace{\frac{\cos x - \dfrac 1 {\sin x}}{\sin x - \dfrac 1 {\cos x}} = \frac{(\cos x) (\sin x \cos x - 1)}{(\sin x)(\sin x \cos x - 1)}}_{\begin{smallmatrix} \text{Multiply both the numerator} \\ \text{and the denominator by $\sin x \cos x$.} \end{smallmatrix}} = \frac {\cos x}{\sin x} = \cdots $$


3

To get the first equality $$\frac {1- \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} {1+ \frac {sin^2(\frac x2)} {cos^2(\frac x2)}} \underbrace{=}_{(1)} \frac {cos^2(\frac x2) - sin^2(\frac x2)} {cos^2(\frac x2) + sin^2(\frac x2)}\underbrace{=}_{(2)} cos^2(\frac x2) - sin^2(\frac x2)$$ in step $(1)$ you multiply numerator and denominator by $\cos^2\frac x2$ ...


3

It is simply a trigonometric identity: $$\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$$ where $\theta = 2x$ in your case.


3

Your approach is correct, but you should probably say that you let $\theta=\frac{\pi}{8}$ before you introduce $x$. Continuing after your third line, you could write something like this: Let $\theta = \frac{\pi}{8}$, and denote $x=\cos^2\left(\frac{\pi}{8}\right)$. This gives $$ \frac{\cos\pi-1}{32}=4x^4-8x^3+5x^2-x$$ so $4x^4-8x^3+5x^2-x=-\frac{1}{16}$.


3

Try this and, if you do not understand any particular part, then ask in the comments


3

$$\displaystyle \overline{AB}=2rcos(x)=R$$, where R is radius of the shaded circle. We have $2R+R(2x)=\frac{2\pi r}{2}$ by the condition given, rearrange to get the equality you want to prove.


3

If you know that the tangent function is convex on $[0,90^\circ)$, you can do this: \begin{align*} \tan(55^\circ) = \tan(\tfrac13\cdot 45^\circ + \tfrac23\cdot 60^\circ) \le \tfrac13\tan(45^\circ) + \tfrac23\tan(60^\circ) = \frac{1+2\sqrt3}{3} < \frac32 < \frac\pi2 \end{align*} To prove the convexity, the simplest thing would be to calculate the second ...


3

This is a variant on AsdrubalBeltran's answer, starting from the equation $$-2+\sqrt{3}=\tan(165^\circ)=\tan(3\cdot55^\circ)=\frac{3\tan(55^\circ)-\tan^3(55^\circ)}{1-3\tan^2(55^\circ)}$$ This implies that $\tan(55^\circ)$ is a root of the cubic polynomial $$P(t)=t^3+(6-3\sqrt3)t^2-3t+(\sqrt3-2)$$ Note that $$P(0)=\sqrt3-2\lt0$$ and ...


3

Imagine a right-angled triangle with one leg $k$ and hypotenuse $4k$ and angle $\theta$ between them. Then $\cos \theta = \frac{k}{4k}= \frac14$ and $\sec \theta = 4$, making $\sec^{-1}4 = \theta$. The opposite leg is $\sqrt{(4k)^2-k^2}=\sqrt{15}k$ and so $\tan(\sec^{-1}4) = \tan \theta = \frac{\sqrt{15}k}{k}=\sqrt{15}$. Now you may need a calculator. ...


3

No. In general, $\cos(a-b)\neq \cos a-\cos b$. Example: $a=b=0$.


3

Maple gives one of the roots as $$ 1/20\,\sqrt [5]{31}{4}^{4/5}\sqrt [5]{{\frac {125\,\sqrt {5}\sqrt {-65 -22\,\sqrt {5}}-409\,\sqrt {-65-22\,\sqrt {5}}-1575\,\sqrt {5}-3725}{ \sqrt {-65-22\,\sqrt {5}}}}}+{\frac { \left( 1577\,{\frac {\sqrt {5}}{ \sqrt {-65-22\,\sqrt {5}}}}-65\,\sqrt {5}+181\,\sqrt {-65-22\,\sqrt {5 }}-13 \right) {31}^{2/5}{4}^{3/5}}{260\, ...


3

I would also have a problem proving this: setting $b=1$, we get $\arctan a$ on the left and $\arctan a-\frac{\pi}{4}$ on the right.


3

Then $\tan$ on both sides of the equation and use $\tan(x-y) = \frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$ to obtain $$\frac{a}{b} = \frac{a-b}{1+ab}$$ Thus the equation can only be satisfied if $a$ and $b$ satisfy the equation above.


3

To find the amplitude, take half the difference of the maximum and minimum values. In this case, the amplitude is $$a = \frac{1 - (-5)}{2} = \frac{6}{2} = 3$$ If you subtract the amplitude from the maximum value, you will find the average value. In this case, the average value of the function is $1 - 3 = -2$. If there were no vertical shift, the average ...


3

Hint \begin{equation} \begin{cases} S = {S_1} + {S_2} = {1 \over 2}Rr\sin {\theta _1} + {1 \over 2}R{l_2}sin\theta = {1 \over 2}r{l_2} \\ \\ {\cos }{\theta _1} = {{{r^2} + {R^2} - l_1^2} \over {2Rr}} \\ \\ {\cos}\theta = {{l_2^2 + {R^2} - {l^2}} \over {2R{l_2}}} \\ \\ {\cos}{\theta _1} = {\sin}\theta , {\sin}{\theta _1} = {\cos}\theta ...


3

\begin{align} \sin^2C &= \sin^2(A+B) = (\sin A \cos B + \cos A \sin B)^2 \\ &= \sin^2A \cos^2B + \sin^2B \cos^2A + 2\sin A \sin B \cos A \cos B \\ &\leqslant \sin^2A \cos^2B + \sin^2B \cos^2A + \sin A \sin B (cos^2A + cos^2B) \\ &= \sin^2A(1-\sin^2B)+\sin^2B(1-\sin^2A)+\sin A \sin B(2-\sin^2A - ...


3

Both are correct. In fact, they are the same. $$ 2\sqrt 2 (\cos\frac{-7\pi}{8} + i\sin\frac{-7\pi}{8}) = -2 \sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) $$ Moreover, $-2\sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8})$ looks better.



Only top voted, non community-wiki answers of a minimum length are eligible