What is the equation of the normal line of f(x)=cos⁡(2x−π2) at x=π6?

f(x)=cos⁡(2x−π2);x=π6≈0.524∴f(π6)=cos⁡(2⋅π6−π2); or:f(π6)=cos⁡(π3−π2)=cos⁡(−π6)≈0.866;So the point is (0.524, 0.866) at which normal is drawn.f(x)=cos⁡(2x−π2)∴f′(x)=−sin⁡(2x−π2)⋅2 orf′(x)=−2sin⁡(2x−π2)∴f′(π6)=−2sin⁡(2⋅π6−π2) or∴f′(π6)=−2sin⁡(−π6)=−2⋅(−12)=1The slope of tangent at the point is mt=1, soslope of normal at the point is mn=−1∴ Equation ofnormal having slope -1 at point (0.524, 0.866) isy−0.866=−1(x−0.524)[y−y1=m(x−x1)] orx+y=0.524+0.866 or x+y=1.39Equation of normal is x+y=1.39 [Ans]

Best solutions to - "What is the equation of the normal line..."

Christopher Stone

Answered question

2022-04-07

What is the equation of the normal line of

f (x) = \cos (2 x - \frac{π}{2})

x = \frac{π}{6}

Answer & Explanation

Austin Sherman

Beginner2022-04-08Added 12 answers

f (x) = \cos (2 x - \frac{π}{2}); x = \frac{π}{6} \approx 0.524

∴ f (\frac{π}{6}) = \cos (2 \cdot \frac{π}{6} - \frac{π}{2});

: f (\frac{π}{6}) = \cos (\frac{π}{3} - \frac{π}{2}) = \cos (- \frac{π}{6}) \approx 0.866;

So the point is (0.524, 0.866) at which normal is drawn.

f (x) = \cos (2 x - \frac{π}{2}) ∴ f^{'} (x) = - \sin (2 x - \frac{π}{2}) \cdot 2

f^{'} (x) = - 2 \sin (2 x - \frac{π}{2})

∴ f^{'} (\frac{π}{6}) = - 2 \sin (2 \cdot \frac{π}{6} - \frac{π}{2})

∴ f^{'} (\frac{π}{6}) = - 2 \sin (- \frac{π}{6}) = - 2 \cdot (- \frac{1}{2}) = 1

The slope of tangent at the point is

m_{t} = 1

, so
slope of normal at the point is

m_{n} = - 1 ∴

Equation of
normal having slope -1 at point (0.524, 0.866) is

y - 0.866 = - 1 (x - 0.524) [y - y_{1} = m (x - x_{1})]

or
x+y=0.524+0.866 or x+y=1.39
Equation of normal is x+y=1.39 [Ans]

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