# At which points the tangent lines of the function $y=\cos x$ are parallel to $-\frac{1}{2}x+1$?

I did the following:

$$\cos' x=-1/2 \\ -\sin x=-1/2 \\ \sin x = 1/2$$

So, I guess that in this case I have to find values such that $\sin x = 1/2$, these values are $\pi/5$ and $5\pi /6$ (I know that there are more values than these, those are just some values that do have these property). Generally, I'd have:

Now I can write:

$$y=mx+b\\ 1/2=\pi/5\cdot -1/2+b \\1/2+\pi/10=b\\\frac{5+\pi}{10}=b$$

Then:

$$y=-1/2x+\frac{5+\pi}{10}$$

But this line doesn't seems to be tangent to $y=\cos x$, see:

I may have done some silly mistake, but I can't find it.

• First, you get two solutions $\pi/6$ and $5\pi/6$ modulo $2\pi$, why does it become $\pi/5$ and $5\pi/6$? – user37238 Apr 27 '15 at 10:55
• Secondly, how do you go from $y=mx+b$ to the next line? Especially, how do you get the value $1/2$ on the left hand side? – user37238 Apr 27 '15 at 10:56
• @user37238 When I evaluate $-\sin x=- 1/2$, I'd have the pair $(\pi/6,-1/2)$. The first item of the pair is $x$, the second one is $y$. I just made the substitution. – Billy Rubina Apr 27 '15 at 11:01
• You're looking for a line tangent to the curve $y=\cos x$. You have found the points where the slope is $-1/2$. Now that you know the slope (at the point $x=\pi/6$) of the tangent line you need to find the value of the function ($\cos$) at this point to completely know the tangent line. – user37238 Apr 27 '15 at 11:08
• @user37238 Yes! I guess you helped me to find the silly mistake. Let me write it. – Billy Rubina Apr 27 '15 at 11:12

Indeed the start is right. $$cos'x = {{-1}\over{2}} \\ sin(x) = {1 \over 2}$$ Giving you the condition which was asked. Solving for it gives you. $$x = (-1)^narcsin({1\over2}) + 2n\pi;\ n \ is\ integer \\ x = (-1)^n({\pi \over 6}) +2n\pi$$
Then if you wish to find the tangent line you should use $(y - y_0) = y' (x - x_0)$ Where $x_0$ is $(-1)^n({\pi \over 6}) +2n\pi$ and $y_0$ is $cos((-1)^n({\pi \over 6}) +2n\pi)$