# Tag Info

7

As you know $$\matrix{ {0 \le \sin (x) \le 1,} & {0 \le x \le \pi } \cr }$$ and hence $$0 \le \sin(x) + 2\sin(2x) + 3\sin(3x) + 4\sin(4x) \le 10$$ and we can have your equality only when all of the $\sin$'s are equal to $1$, i.e, $$\matrix{ {\sin (x) = 1} \hfill & \to \hfill & {x = {\pi \over 2}} \hfill \cr {\sin (2x) = ... 5 Notice that we can write$$ \frac{2n}{n^4 - n^2 + 1} = \frac{(n^2+n) - (n^2-n)}{1 + (n^2+n)(n^2-n)}. $$In view of the addition formula for the tangent, we find that$$ \arctan \bigg( \frac{x-y}{1+xy} \bigg) = \arctan x - \arctan y $$for any x > y > 0. Thus we have$$ \arctan\bigg( \frac{2n}{n^4 - n^2 + 1} \bigg) = \arctan(n^2+n) - ...

5

The statemnet $\sqrt{\cos^2(x)}=|\cos(x)|$ is true for all real $x$. Wolfram|Alpha, on the other hand, has the often annoying habit of considering all complex numbers by default. So, it's not willing to apply the identity $$\sqrt{x^2}=|x|$$ unless it knows that $x$ is real. Given that $\cos(x)$ is real for all real $x$, the identity you cite is true for any ...

4

$\sin \theta + \cos \theta = \sqrt 2 \sin (\theta + \frac{\pi}{4})$ Sine is positive in the first and second quadrants. So the problem reduces to finding $\theta$ such that $0 < (\theta + \frac{\pi}{4}) < \pi$, giving $-\frac{\pi}{4} <\theta < \frac{3\pi}{4}$

4

Well, to think about the equation $$y=\arccos(\sin(x))$$ let's first look at an easier one, obtained by taking the cosine of both sides (noting that $\cos(\arccos(x))=x$ - that is $\arccos$ is a right-inverse of $\cos$): $$\cos(y)=\sin(x).$$ If we plot this, we get the following graph: $\hskip1.5in$ which is just a series of diagonal lines going in two ...

4

There are six equally likely outcomes: $ABP>BCP>ACP$ $ABP>ACP>BCP$ $BCP>ACP>ABP$ $BCP>ABP>ACP$ $ACP>BCP>ABP$ $ACP>ABP>BCP$ Each of them are mutually exclusive and their sum is $1$. Hence $BCP$ being the largest has probability $1\over3$. (The equal area case has probability $0$ and can be ignored because we have a ...

4

As already said in comments and answers, you cannot do much with embedded trigonometirc functions except developing as Taylor series. Built around $x=0$ (remember that, for any $x$, $|\cos(x)| \leq 1$), you should get \sin(\cos(x))=\sin (1)-\frac{1}{2} x^2 \cos (1)+x^4 \left(\frac{\cos (1)}{24}-\frac{\sin (1)}{8}\right)+x^6 \left(\frac{\sin ... 3 You know that $$\textrm{sin}^2x + \textrm{sin}^2 2x + \textrm{sin}^2 3x = 1 \qquad (\textrm{Eq.} 1)$$ Notice that \begin{aligned} \textrm{sin}^2x + \textrm{cos}^2 x = 1 \Rightarrow \textrm{sin}^2x = 1 - \textrm{cos}^2 x \end{aligned} Substituting the identity in Equation (1): ... 3 It's true for 0 \le \theta < \pi/2, but not in general. For \pi/2 < \theta < 3 \pi/2, \sec(\theta) < 0 so \cosh^{-1}(\sec(\theta)) is not real, although \tan^{-1}(\sin(\theta)) is. For 3 \pi/2 < \theta < 2 \pi, \sin(\theta) < 0 so \tanh^{-1}(\sin(\theta)) < 0, while \sec(\theta) > 1 and \cosh^{-1}(\sec(\theta)) ... 3 (This is not a comment, but an addendum to hbp's answer.) I'm glad I raised a bounty for this question because, thanks to hbp, we can show that the general quintic can be solved in terms of trigonometric and hyperbolic functions. The general quintic can be reduced to the one-parameter Brioschi form,w^5-10cw^3+45c^2w-c^2=0\tag1$$with solution (see ... 3 You can certainly use a double angle formula - as long as you use it correctly... Since \sin2x=2\sin x\cos x, the second term in your expression is$$(2\sin2x\cos2x)\sin3x\ ,$$and then the bit in brackets is...? 3 Steps To Carry Out 1) First take a look at this link which is a guide for DeMoivre's formula. 2) Using step 1 show that$$\sin (7x) = 64\sin \left( x \right)\cos {\left( x \right)^6} - 80\sin \left( x \right)\cos {\left( x \right)^4} + 24\sin \left( x \right)\cos {\left( x \right)^2} - \sin \left( x \right)$$3) Replace \cos^2(x)=1-\sin^2(x) and obtain ... 3$$2\cos(t)\cos(2t)-\sin(t)\sin(2t)=\frac{\cos(t)+3\cos(3t)}{2}$$Proof:$$2\cos(t)\cos(2t)-\sin(t)\sin(2t)=\frac{\cos(t)+3\cos(3t)}{2}\Longleftrightarrow2\left(2\cos(t)\cos(2t)-\sin(t)\sin(2t)\right)=\cos(t)+3\cos(3t)\Longleftrightarrow2\left(\cos(-t)+\cos(3t)-\sin(t)\sin(2t)\right)=\cos(t)+3\cos(3t)\Longleftrightarrow$$... 3 A right triangle with a 60 degree angle would be half an equilateral triangle. Hence BC would be half as long as AC, so AC=8. But 7^2+4^2\ne 8^2 so indeed Pythagoras tells us that this is not the case. 3 let \frac{\pi}{16}=A then tan13A=-tan3A and tan9A=-tan7A so$$S=tanA-tan3A+tan5A-tan7A=\frac{-sin2A}{cosAcos3A}+\frac{-sin2A}{cos5Acos7A}S=-2sin2A\left(\frac{1}{2cosAcos3A}+\frac{1}{2cos5Acos7A}\right)S=-2sin2A\left(\frac{1}{cos4A+cos2A}+\frac{1}{cos12A+cos2A}\right)$$... 3$$ 2^6 \cos^6 x = (e^{ix}+e^{-ix})^6=e^{6ix}+6e^{4ix}+15e^{2ix} +20+15e^{-2ix}+6e^{-4x}+e^{-6ix} \\ = 2(\cos 6x + 6 \cos 4x +15 \cos 2x +10) $$and so forth 3 First we prove \tan A + \tan B + \tan C = \tan A \tan B \tan C. We know that A + B + C = 180 from the angle addition formula, so \tan((A + B) + C) = \frac{\tan(A + B) + \tan C}{1−\tan(A + B) \tan C} = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan A \tan C - \tan B \tan C} We know that tan 180 = 0, ... 2 Hint: The area is 20\left(\frac{AD+BC}{2}\right). So all we need to know is AD+BC. Claim: This is equal to AB+CD. To show that AD+BC=AB+CD, use repeatedly the fact that if the line segments PX and PY are tangent to a circle, where X and Y are the points of tangency, then PX=PY. 2 The inverse of the cosine is defined on [-1, 1] and maps to [0, \pi]. The sine function on [-\pi/2, \pi/2] maps to [-1,1].$$ \arccos(\sin(x)) = \arccos(\cos(\pi/2 - x)) = -x + \pi/2 $$The sine funcion on [\pi/2, \pi/2 + \pi] is$$ \sin(x) = -\sin(x-\pi) = \sin(\pi - x) $$and maps to [-1,1]. We get$$ \arccos(\sin(x)) = \arccos(\cos(\pi/2 - ...

2

Just for some new ideas! I would reccomend a completely different method. This method uses the Gudermannian $\text{gd}$ function. So you would substitute $x=\text{gd}(a);\text{d}x=\text{sech}\space a\text{d}a$ That transforms the integral into: $$\int \frac{\tanh a}{\tanh a+\text{sech}\space a}(\text{sech}\space a)\mathrm da$$ Through some hyperbolic trig ...

2

In $1^{st}$ quadrant both $\sin \theta$ and $\cos \theta$ are positive. In $2^{nd}$ quadrant from $90^{\circ}$ to $135^{\circ}$ degrees $\sin \theta > \cos \theta$ and $\sin\theta$ is positive while $\cos\theta$ is negative. In $3^{rd}$ quadrant both are negative. In $4^{th}$ quadrant from $315^{\circ}$ to $360^{\circ}$ $\cos \theta > \sin \theta$ ...

2

You want to read the alternate form assuming $x$ is real. (The one about $x>0$ is a red herring, just written there in case that was what a user wanted to know about.) If $x$ is not real but rather complex then the square root is a more complicated animal (see https://en.wikipedia.org/wiki/Square_root for more info).

2

This doesn't hold. Try $A=B=C=\pi/3$.

2

HINT: $$\int\frac{1}{\left(t+\cos(a)\right)^2+\sin^2(a)}\space\text{d}t=$$ Substitute $u=\cos(a)+t$ and $\text{d}u=\text{d}t$: $$\int\frac{1}{\sin^2(a)+u^2}\space\text{d}u=$$ $$\int\frac{\csc^2(a)}{u^2\csc^2(a)+1}\space\text{d}u=$$ $$\csc^2(a)\int\frac{1}{u^2\csc^2(a)+1}\space\text{d}u=$$ Substitute $s=u\csc(a)$ and ...

2

$$\frac{1+\tan^2 (x/2)}{1-\tan^2 (x/2)}=\frac{\cos^2(x/2)+\sin^2 (x/2)}{\cos^2(x/2)-\sin^2 (x/2)}=\frac{1}{\cos^2(x/2)-\sin^2 (x/2)}=\frac{1}{\cos x}=\sec x$$

2

Thanks to Tito for the nice question. Here is a solution in terms of hyperbolic sine, which may not be what you want. $$\sinh(5t) = 5 \sinh t+ 20 \sinh^3 t + 16 \sinh^5 t.$$ With $x = 2\sinh t$ and $b = -2\sinh 5t$, we have $$x^5+ 5 x^3 + 5 x + b = 0\tag1,$$ which is the $a = 1$ case. So the solution is $$x = 2\sinh t = 2\sinh\left( ... 2 Translate your circle to the origin. The coordinates of (x_{i},y_{i}) become (x_{i}-a,y_{i}-b) for i=1,2. You know that if X_{1} and X_{2} are two vectors of the euclidean space \mathbb{R}^{2}, the angle \theta formed by X_{1} and X_{2} is obtained as follows (see the remark at the bottom of this answer):$$\langle ...

2

$$\begin{cases} \displaystyle r_1=\frac{4a\cos\theta}{\sin^2\theta}\\ \displaystyle r_2=\frac{4a\sin\theta}{\cos^2\theta}\\ \end{cases}$$ Try to find $\sin\theta$ and $\cos\theta$ as following: $$r_1\sin^2\theta=4a\cos\theta\tag{1}$$ $$r_2\cos^2\theta=4a\sin\theta\tag{2}$$ To equation $(1)$ plug computed $\sin\theta$ from equation $(2)$: ...

2

Mathematica's FullSimplify (and Simplify) functions have an option called ComplexityFunction. This is used to determine which form (of the ones it tries) is the simplest. In this situation the default measure wasn't sufficient to achieve the simplification you wanted. Try: FullSimplify[-2 h*Cos[t] Sin[t], ComplexityFunction -> LeafCount]

2

$AB = 0$ does not exclude $A = B = 0$. You can simply rearrange the equations: $$\begin{vmatrix} 1 & 1 & 1 \\ \cos\theta_2 & \cos\theta_3 & \cos\theta_1 \\ \sin\theta_2 & \sin\theta_3 & \sin\theta_1 \\ \end{vmatrix}=0$$ And use an analogous proof to show that ${\beta = \alpha} \vee {\beta = \gamma}$. And rearrange a final time to ...

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