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What is the equation of the normal line of #f(x)= sinx-cos2x# at #x = pi/8#?

Answer 1

#y_n = -(2x)/(sqrt(2+sqrt(2)) + 2sqrt(2)) + pi/(4(sqrt(2+sqrt(2)) + 2sqrt(2))) + ((sqrt(2-sqrt(2)))- sqrt(2))/2#

Find the first derivative #-> #the slope function: #f'(x) = cos x - (-sin 2x)(2) = cos x + 2 sin 2x#

Find the slope of the tangent line: #m_t = f'(pi/8) = cos(pi/8) + 2 sin(pi/4) = sqrt(2+sqrt(2))/2 + (2sqrt(2))/2# #m_t = (sqrt(2+sqrt(2)) + 2sqrt(2))/2#

The normal line is the perpendicular line which the negative reciprocal of the tangent line for its slope: #m_n = -2/(sqrt(2+sqrt(2)) + 2sqrt(2))#

#y_n -y_1= m_n(x - x_1)#

#f(pi/8) = sin(pi/8) - cos(pi/4) = (sqrt(2-sqrt(2)))/2 - sqrt(2)/2 = ((sqrt(2-sqrt(2)))- sqrt(2))/2#

#y_n -y_1= m_n(x - x_1):#

# y_n-((sqrt(2-sqrt(2)))- sqrt(2))/2 = -2/(sqrt(2+sqrt(2)) + 2sqrt(2))(x-pi/8)#

#y_n = -(2x)/(sqrt(2+sqrt(2)) + 2sqrt(2)) + pi/(4(sqrt(2+sqrt(2)) + 2sqrt(2))) + ((sqrt(2-sqrt(2)))- sqrt(2))/2#

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Answer 2

Axel Alvarez

The equation of the normal line of f(x) = sinx - cos2x at x = pi/8 is y = -8x + 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.