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

Answer 1

Isaiah Allen

#-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#

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Answer 2

Luke Alvarez

To find the slope of the line normal to the tangent line of ( f(x) = \tan(x) + \sin(x - \frac{\pi}{4}) ) at ( x = \frac{5\pi}{8} ), first, find the derivative of ( f(x) ) using the sum rule and chain rule. Then evaluate the derivative at ( x = \frac{5\pi}{8} ) to get the slope of the tangent line. The slope of the normal line will be the negative reciprocal of the slope of the tangent line. Finally, compute the slope of the normal line.

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Answer from HIX Tutor

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