I want to find the intersection of two functions. One is $f(x) = cos(x)$. The other one is $g(x) = x$. When I do the normal steps like equating both functions and finding the roots, I get something like this $\cos(x) = x$ $\cos(x) - x = 0$ How do I proceed further? Is there any argument to cosine such that it would give back the same argument as output? If yes, how do I find it analytically without graphing the function?
7892
There is an unique solution with $x$ being approximately $0.739085$. But is there also a closed form solution?
40603
I am asked to prove that $\cos(\arcsin(x)) = \sqrt{1-x^2}$  I have used the trig identity to show that $\cos^2(x) = 1 - x^2$  Therefore why isn't the answer denoted with the plus-or-minus sign?  as in $\pm \sqrt{1-x^2}$.  Thank you!
28096
How does one solve this equation?     $$\cos {x}+\sin {x}-1=0$$   I have no idea how to start it.  Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?  Thanks in advance!
58386
I have a function $F(θ) = \sin^{−1} \sqrt{\sin(11θ)}$  I derived the following answer using basic trigonometric and quotient rules.  $\dfrac{11\csc \left(\left(11\theta \right)^{\left(\frac{1}{2}\right)}\right)\cos \left(\left(11\theta \right)\right)}{2\sqrt{1-\sin \left(11\theta \right)}}$  My answer however is wrong.  Can anyone outline how to go about getting the correct solution to a problem such as this?
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I found this problem in a collection of contest problems of a Russian competition in 1995 and wasn't able to solve it.     Solve for real $x$:   $$ \cos (\cos (\cos (\cos(x))))=\sin (\sin (\sin (\sin (x)))) $$   My guess is that there is no solution, but how do I prove it? I tried to estimate  $LHS\ge \cos (1) \ge \cos(\pi/3)=1/2 $  and RHS similarly but the ranges overlap..  Do you have a better idea?
152
The "" formulas often come in handy, but it's not immediately obvious that they would be true.  \begin{align} \sin(\alpha \pm \beta) &amp;= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\ \cos(\alpha \pm \beta) &amp;= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \end{align}  So what I want to know is,    How can I prove that these formulas are correct? More importantly, how can I understand these formulas intuitively?   Ideally, I'm looking for answers that make no reference to Calculus, or to , although such answers are still encouraged, for completeness.
106132
For a hobby project gone too far, I'm trying to find the maximum separation between two ellipses. I have only a high-school-graduate-level grasp of algebra and calculus, and am happy to dig deeper than that... but I need some guidance.  I have 3 functions of the same form, $X(p, q)$, $Y(p, q)$, and $Z(p, q)$ for which I am trying to evaluate $p$ and $q$. I know that $X+Y+Z=0$. I know I need to express p and q in terms of each other, but as you can see below, the three functions together are quite complex. I've been futzing around with trig identities to try and isolate either variable, but am stuck. Perhaps fresh eyes will help; can anyone here see a way forward?  All values are known, except $p$ and $q$.  $X(p,q) = [b_1V_{x1}\cos(p)-a_1U_{x1}\sin(p)][(C_{x2}+a_2U_{x2}\cos(q)+b_2V_{x2}\sin(q))-(C_{x1}+a_1U_{x1}\cos(p)+b_1V_{x1}\sin(p))]$  $Y(p,q) = [b_1V_{y1}\cos(p)-a_1U_{y1}\sin(p)][(C_{y2}+a_2U_{y2}\cos(q)+b_2V_{y2}\sin(q))-(C_{y1}+a_1U_{y1}\cos(p)+b_1V_{y1}\sin(p))]$  $Z(p,q) = [b_1V_{z1}\cos(p)-a_1U_{z1}\sin(p)][(C_{z2}+a_2U_{z2}\cos(q)+b_2V_{z2}\sin(q))-(C_{z1}+a_1U_{z1}\cos(p)+b_1V_{z1}\sin(p))]$  Edit - A couple threads I've been trying to springboard off of:        I arrived here by taking the partial derivatives of the distance function for two points, one on each ellipse. Once I can find the critical points from that, I should have my solution(s).   The distance function is from the ellipses' parametric equations:  $x(θ)=C_x+aU_x\cos(θ)+bV_x\sin(θ)$  $y(θ)=C_y+aU_y\cos(θ)+bV_y\sin(θ)$  $z(θ)=C_z+aU_z\cos(θ)+bV_z\sin(θ)$  (Where $C$ is the centrepoint of the ellipse, $a$ and $b$ are its semimajor and semiminor axes, respectively, and $U$ and $V$ are unit vectors defining the plane of the ellipse. Angle $θ$ describes position on the ellipse)  So the distance function is:  $$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$  As the derivative of $\sqrt{n}$ is ${n'\over2\sqrt{n}}$, I only need concern myself with where the numerator ($n'$) is equal to $0$:  $$n'=-2f'(p)[-f(p)+j(q)]-2g'(p)[-g(p)+k(q)]-2h'(p)[-h(p)+l(q)]$$  (Where functions $f$, $g$, and $h$ refer to the parametrics for the 1st ellipse, and functions $j$, $k$, and  $l$ to those for the 2nd. $p$ and $q$ are used instead of $θ$)  $n'=0$ is essentially of the form $0=X+Y+Z$ as outlined above.
96430
I'm trying to comprehend the following: If I choose any starting value (e.g. 1) and keep clicking on cosine on the calculator (in radian mode), it gives me a result of about 0.739085...(I believe it's the result of cos(x) = x), but when I repeat the same procedure using sin and tan, I get something completely different (looks like for sin it's converging to 0 while for tan I get very wild results).  Thanks for your help.
8702
Find all roots in $\mathbb{C}$ of the equation $\cos(z)=2$ and plot them on an Argand diagram.  My solution: $\cos(z)=2$, $z=x+iy$ so we have that $\cos(x)\cos(iy)-\sin(x)\sin(iy)=2$ so $\cos(x)\cosh(y)=2$ and $\sin(x)\sinh(y)=0$ must hold. So if $\sinh(y)=0$ then $y=0$ so $\cosh(y)=1$ and so $\cos(x)=2$ which is a contradiction, therefore $x=n\pi$, for some $n\in \mathbb{Z}$ and so $\cos(x)=\pm 1$. However, $\cosh(y)&gt;0$, so if $\cos(x)=-1$ it follows that $\cos(x)\cosh(y)&lt;0$ and cannot equal $2$. Hence $\cos(x)=1$ so $x=2n\pi$ for some $n\in \mathbb{Z}$ and $y=\pm\cosh^{-1}(2)$ since $\cosh(y)=2$. Therefore since $z=x+iy$ $\implies z=2n\pi \pm i\cosh^{-1}(2)$, $n\in \mathbb{Z}$.  My question is, is this sufficent to answer the question $\cos(z)=2$? And if not how should I structure it using this method? Also why is it $\pm\cosh^{-1}(2)$? I don't understand this sutble point. And if I was to draw an Argand diagram would it just be on the imaginary axis at $\cosh^{-1}(2)$ every $2n\pi$?
35069
Given the function: $$f(x)=\cos(\sin(x))-\sin(\cos(x))$$ it has absolute maxima at $x=(2k+1)\pi$ with $k=0,1,..N$ and relative maxima at $x=2k\pi$. It is not clear where are the minima. Putting the derivative to zero doesn't help. Any suggestion on how to find the minimum value and where is it? Thanks.
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I know that if $\sin x=\sin y$ then $x=n\pi+(-1)^ny$  and $\cos x=\cos y$ implies that $x=y+2k\pi$ or $x=-y+2l\pi$. Since the latter does not fit the form above $x=y+2k\pi$ which implies that $k=0,1,-1$. Hence, we have x=y or $x=2\pi,y=0$ or $x=0,y=2\pi$.
30871
Is it possible to find $\displaystyle{\cos x \over \cos y}$ if $\displaystyle{\sin x\over \sin y}$ is given? If so, how would one approach this problem? Thank you in advance!
81246
I have the following functional equation. Find all continuous functions $f:(-1,1) \to \mathbb R$ such that $$ f(x+y)=\frac{f(x) + f(y)}{1 - f(x)f(y)} $$ The first obvious solution is $f(x) \equiv 0$. Another one I guessed, it is $f(x) = \pm \tan x$. I suspect, that there are no more solutions. The problem is that I don't know how to prove that.  Since we have a rational equation, I have no idea how to make any substitutions in order to get expression for $\tan x$ (as it is not a rational expression).  P.S. I can also show that it is true that (a) $f(0) = 0$ (take $x=y=0$), and (b) $f(-x) = -f(x)$ (take $y=-x$).  Update: there is a solution of this problem also here:
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I came up with this function: $$2\left(\frac{1}{1+e^{\textstyle\frac{-6\sin^{-1}(\cos(x))}{\pi/2}}}-\frac12\right)$$ to mimic a 'cosine'-esque function with flat peaks and valleys. Here it is as plotted by Wolfram Alpha:    What I was wondering is, is there a more elegant way to achieve this effect? (The values the function outputs need not be the same as those of this function - it only needs to look cosine-esque and have flat peaks and valleys).
119123
Find the value of   $$1\sin 2^{\circ} +2\sin 4^{\circ} + 3\sin 6^{\circ}+\cdots+ 90\sin180^{\circ}$$   My attempt  I converted the $\sin$ functions which have arguments greater than $90^\circ$ to $\cos$ but I have gone no where with it!  I also tried using double angle formula for the angles which are even.
59265
I've been wondering, what is the general form for the integral of $\sin(\cos x)$? I tried integrating it myself and couldn't, tried with Wolfram Alpha and it couldn't either. If you think about it conceptually, it should have a general form as it's graph if just like sin and cos, which both have general forms. If it doesn't have a general form, I'd like to know why. Thanks!
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Let $S$ := {$(x, y) ∈ \mathbb R^2: x^2 + y^2 = 1$} be the unit circle. Let $p = (0, -1)$ ∈ $S$. Define a map $F$ from $\mathbb R$ × {$0$} to $S$ as follows: Given $(t, 0)$, let $F((t, 0)) = (x, y)$ ∈ $S$ be the intersection of $S$ with the line through $p$ and $(t, 0)$. Derive a formula for $F$. Show that $t$ → $F((t, 0))$ is a bijective map from $\mathbb R$ to $S$ \ {$p$}, and also derive a formula for its inverse.  If we express $S$ in polar coordinates: $$S:=\{(r,\varphi):r=1,\varphi∈(-\frac\pi{2},\frac{3\pi}{2})\}$$ then, from simple trigonometric relations, the inverse function $\Phi$ is: $$\Phi(\varphi)=\tan\frac{\frac\pi2-\varphi}2$$ and its inverse $F$ is: $$F(t)=\frac\pi{2}-2\arctan t$$ After changing coordinates back to cartesian, $F$ will look like: $$F(t)=(\cos(\frac\pi{2}-2\arctan t),\sin(\frac\pi{2}-2\arctan t))$$ which can be further simplified to: $$F(t)=(\frac{2 t}{t^2 + 1},\frac{1 - t^2}{t^2 + 1})$$ Is this the formula of $F$? Is it correct? And I know that to show that $\phi$ : $\mathbb R$ →  $S$ \ {$p$} is a bijective map, I have to show that $\phi$ is injective and surjective. But how do I do that? Can someone please help me show that $\phi$ is bijective? and how can I derive a formula for its inverse?