What is the slope of the line normal to the tangent line of $f(x) = tanx+sin(x-pi/4)$ at $x= (5pi)/8$ ?

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-29T03:23:48

$-0.1290$ , nearly.

At #x=5/8pi,

y= f(5/8pi)=tan(5/8pi)+sin(5/8pi-pi/4)#

$=tan(pi/2+pi/8)+sin(-pi/8)$

$=-cot(pi/8)-sin(pi/8)=-2.797$ , nearly

The slope of the normal at this point

$P(5/8pi, 2.0315)=P(1.9635,2.0315)$

is -1/y/ at P

$=-1/(tan x+sin(x-pi/4))'$ at $x =5/8pi$

$=-1/(sec^2(5/8pi)+cos(pi/8)$

$=-1/(csc^2(pi/8)+cos(pi/8))$

$=-0.1290$ , nearly.

The equation to the normal at P(1.9635,2.0315)# is

y-2.0315=-0.1290(x-1.9635)#. Simplifying,

$y=-0.1290 x+2.057$

What is the slope of the line normal to the tangent line of f(x) = tanx+sin(x-pi/4) f(x) = tanx+sin(x-pi/4) at x= (5pi)/8 x= (5pi)/8 ?