Question 889292
Take the derivative of the function and find its value at {{{x=pi/4}}}.
This value is the slope of the tangent line which is perpendicular to the normal line.
{{{y=6.5sin(2.5x)}}}
{{{dy/dx=2.5(6.5)cos(2.5x)}}}
{{{dy/dx(pi/4)=(65/4)cos((5pi)/8)}}}
{{{dy/dx(pi/4)=(65/4)(-0.3827)}}}
{{{m[t]=-6.2186}}}
Perpendicular line slopes are negative reciprocals.
{{{m[n]*m[t]=-1}}}
{{{m[n]=-1/(-6.2186)}}}
{{{m[n]=0.161}}}
Use the point slope form of a line,
{{{y-6.005=0.161(x-0.7854)}}}
{{{y-6.005=0.161x-0.1264}}}
{{{y=0.161x+5.879}}}
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{{{drawing(300,300,-5,5,-2,8,circle(0.7854,6.005,0.05),grid(1),graph(300,300,-5,5,-2,8,6.5sin(2.5x),0.161x+5.879,10.8891-6.21861x))}}}