Hello everyone I have a tangent line with slope $= a$ to $y = \sin x$ in 2 points $(u_1,v_1) , (u_2 ,v_2)$
How can I prove that $\tan a = a$?
My direction was to express $a$ with the points by $a = \frac{u_1-u_2}{v_1-v_2}$
and the tangent line equation is $y = ax -au_1 +v_1$
$f(x) = \sin x \iff f'(u) = \cos u$
so $\cos u_1 = a$ and $\cos u_2 = a$?
Best Answer
You can't, as there are more solutions. Let's $(u,\sin(u))$ and $(v,\sin(v))$ be two points. (We exclude the trivial case $v=u+2k\pi$.) The equations of their tangents are $$ \begin{align} y-\sin(u)&=\cos(u)(x-u)\\ y-\sin(v)&=\cos(v)(x-v). \end{align} $$ We have $\cos(u)=\cos(v)$, which gives $v=2k\pi-u$ for some $k\in\mathbb Z$ Furthermore we must have equal $y$-intercepts, that is $$\sin(u)-u\cos(u)=\sin(v)-v\cos(v).$$ Plugging in $v=2k\pi-u$ and solving for $\tan(u)=\sin(u)/\cos(u)$ finally gives $$\tan(u)=u+k\pi.$$
Here's an example where $k=5$.