What is the slope of the line normal to the tangent line of f(x) = cscx+sin(2x-(3pi)/8) f(x)=cscx+sin(2x−3π8) at x= (11pi)/8 x=11π8?
1 Answer
Explanation:
First, find the derivative of
f'(x) = d/dx(cscx)+d/dx(sin(2x-(3pi)/8))
f'(x) = -cscxcotx + cos(2x-(3pi)/8)*d/dx(2x-(3pi)/8)
f'(x) = -cscxcotx + 2cos(2x-(3pi)/8)
Now we need to plug in
f'((11pi)/8) = -csc((11pi)/8)cot((11pi)/8)+2cos(2(11pi)/8-(3pi)/8)
f'((11pi)/8) = (-1)/(sin((11pi)/8)tan((11pi)/8))+2cos((19pi)/8)
To evaluate these trig functions, use the half angle rules:
sin(theta/2)= +-sqrt((1-costheta)/2)
cos(theta/2) = +-sqrt((1+costheta)/2)
tan(theta/2) = (1-costheta)/sintheta
Using these rules, we can evaluate these trig functions like this:
sin((11pi)/8) = -sqrt((1-cos((11pi)/4))/2) = -sqrt((1-cos((3pi)/4))/2) = -sqrt((1+sqrt2/2)/2) = -sqrt((2+sqrt2)/4) = -sqrt(2+sqrt2)/2
tan((11pi)/8) = (1-cos((11pi)/4))/sin((11pi)/4) = (1+sqrt2/2)/(sqrt2/2) = 1/(sqrt2/2)+(sqrt2/2)/(sqrt2/2) = sqrt2+1
cos((19pi)/8) = cos((3pi)/8) = sqrt((1+cos((3pi)/4))/2) = sqrt((1-sqrt2/2)/2) = sqrt((2-sqrt2)/4) = sqrt(2-sqrt2)/2
Now we plug these back into our function and simplify.
f'((11pi)/8) = (-1)/((-sqrt(2+sqrt2)/2)(sqrt2+1))+2(sqrt(2-sqrt2)/2)
= 2/((sqrt(2+sqrt2))(sqrt2+1))+sqrt(2-sqrt2)
= 2/((sqrt(2+sqrt2))(sqrt2+1))+sqrt(2-sqrt2)(((sqrt(2+sqrt2))(sqrt2+1))/((sqrt(2+sqrt2))(sqrt2+1)))
= (2+sqrt((2-sqrt2)(2+sqrt2))(sqrt2+1))/((sqrt(2+sqrt2))(sqrt2+1))
= (2+sqrt(2)(sqrt2+1))/((sqrt(2+sqrt2))(sqrt2+1))
= (4 + sqrt2)/((sqrt(2+sqrt2))(sqrt2+1))
Finally, since we're looking for the slope of the normal line, we need to flip this fraction and make it negative.
"norm " = -((sqrt(2+sqrt2))(sqrt2+1))/(4+sqrt2)
And then rationalize the denominator by multiplying by the conjugate.
= -((sqrt(2+sqrt2))(sqrt2+1)(4-sqrt2))/(16 - 2)
= -((sqrt(2+sqrt2))(3sqrt2+2))/14
Final Answer
