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

May 2011
28
0
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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May 2011
28
0
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?
 
May 2011
28
0
Also, where might I learn how to insert math symbols directly into the post?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
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
Jun 2008
16,217
6,765
North Texas
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 $y = \cos{a}(x+\sqrt{3}) - \dfrac{\pi}{6}$
 
May 2011
28
0
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
Jun 2008
16,217
6,765
North Texas
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.