# Eq'n of Tangent Line to y=sinx and...

#### kablooey

I have a calc question I haven't been able to crack.

Here is the original question, almost verbatim. "A line passes through the point (-sqrt3, -pi/6) and is tangent to the graph of y=sinx at a point for which x is between 0 and pi/2. Determine an equation of the line."

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#### kablooey

Whoops! Sorry, I did a terrible job of transferring the original question to my picture. My sketch did correspond to the original question.

Here is the original question, almost verbatim. "A line passes through the point (-sqrt3, -pi/6) and is tangent to the graph of y=sinx at a point for which x is between 0 and pi/2. Determine an equation of the line."

Is there any way I can edit the original post so nobody else gets confused?

#### kablooey

Also, where might I learn how to insert math symbols directly into the post?

MHF Helper

#### skeeter

MHF Helper
Whoops! Sorry, I did a terrible job of transferring the original question to my picture. My sketch did correspond to the original question.

Here is the original question, almost verbatim. "A line passes through the point (-sqrt3, -pi/6) and is tangent to the graph of y=sinx at a point for which x is between 0 and pi/2. Determine an equation of the line."

Is there any way I can edit the original post so nobody else gets confused?
your original post has been edited ... I removed the attachment, you may want to correct it then repost.

Last edited:

#### skeeter

MHF Helper
corrected graph of tangent line attached ...

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#### kablooey

Here it is again.

#### skeeter

MHF Helper
solve the equation $\sin{a}+\dfrac{\pi}{6} = \cos{a}(a+\sqrt{3})$ for $0 < a < \dfrac{\pi}{2}$ ... I recommend you use technology.

tangent line will be $y = \cos{a}(x+\sqrt{3}) - \dfrac{\pi}{6}$

#### kablooey

I thought that a graphing calculator would be the right way to solve it, too.

But the answer key says that the equation of the line is
$$\displaystyle y=\frac{1}{2}x+\frac{\pi}{6}+\frac{\sqrt{3}}{2}.$$

That means the slope is
$$\displaystyle \frac{1}{2}$$
and so
$$\displaystyle a=\frac{\pi}{3}$$
. A slope of exactly
a half
suggests that there is an algebraic way of solving this and I do not believe graphing calculators are permitted for this question.

#### skeeter

MHF Helper
Check the tangent equation again ... should be $-\dfrac{\pi}{6}$.

I'll have to play with the problem to get $a=\dfrac{\pi}{3}$ ... I'm not seeing how to get there yet, but it does work.