A quicker generalized method to finding a curve tangent to another curve? Let's say we have a curve of $\sin(x)$ and we have to find a curve tangent to this in form of $c(x-d)^{1/3}$. This curve should have the same tangent line as $\sin(x)$ at any point around $(a,\sin(a))$. ($c$ and $d$ should in a closed form with no decimals).
I thought the quickest and most elegant method would be taking
$$\left((\sin(x))^{3}\right)^{1/3}$$
Then taking the tangent inside the parenthesis...
$$\left(3\sin(a)^{2}\cos(a)(x-a)+{\sin(a)}^{3}\right)^{1/3}$$
Finally I did some further factorization...
$${(3)^{1/3}\sin(a)}^{2/3}{\cos(a)}^{1/3}\left(x-a+\frac{\sin(a)^3}{3\sin(a)^{2}{\cos(a)}^{}}\right)^{1/3}$$
$${{3}^{1/3}\sin(a)}^{2/3}{\cos(a)}^{1/3}\left(x-a+\frac{\tan(a)}{3}\right)^{1/3}$$
I still think there is a quicker and more elegant method to doing this? What could it be? 
 A: May be, you took a long way. You have two functions $y_1(x)$ and $y_2(x)$ (the last one containing two parameters to be identified). What is required in your problem is to solve for the parameters from two equations $$y_1(a)=y_2(a)$$ $$y'_1(a)=y'_2(a)$$ So, considering your specific problem, then write $$\sin (a)=c \sqrt[3]{a-d}$$ $$\cos (a)=\frac{c}{3 (a-d)^{2/3}}$$ We can eliminate $c$ from the first equation (it becomes a function of $a$ and $d$), plug it into the second equation to be solved for $d$. This leads to $$d=a-\frac{\tan (a)}{3}$$ $$c=  \sqrt[3]{3\cos (a)\sin^2(a)}$$ My final word, if I may suggest : build the equations in the simplest manner; only the solution would require some gymnastics.
A: I really liked Claude's answer but every one has their quickest technique. Using Claude's vs my technique I felt like mine was the quickest
The reason is because for my method is one has to only take one derivative and solve for the tangent inside the equation. Using Claude's method you have to find two derivatives and solve for parametric variables of $c$ and $d$.
I think one should not immediately upvote anyone's post and try both techniques.
