Given
f(x)=tan x-sin 2xf(x)=tanx−sin2x at x_1=(4pi)/3x1=4π3
Solve for the point (x_1, y_1)(x1,y1) first
y_1=tan x_1-sin 2*x_1y1=tanx1−sin2⋅x1
y_1=tan ((4pi)/3)-sin (2*(4pi)/3)y1=tan(4π3)−sin(2⋅4π3)
y_1=sqrt3-sqrt3/2y1=√3−√32
y_1=sqrt3/2y1=√32
Our point (x_1, y_1)=((4pi)/3, sqrt3/2)(x1,y1)=(4π3,√32)
Solve for the slope mm
f(x)=tan x-sin 2xf(x)=tanx−sin2x
find the first derivative f' (x)=m
f' (x)=sec^2 x- 2*cos 2x
m=f' ((4pi)/3)=sec^2 ((4pi)/3)- 2*cos (2((4pi)/3))
m=(-2)^2-2(-1/2)
m=4+1=5
For the normal line
m_n=-1/m
m_n=-1/5
Solve for the normal line:
y-y_1=m_n(x-x_1)
y-sqrt3/2=-1/5(x-(4pi)/3)
y-sqrt3/2=-1/5x+(4pi)/15
color(blue)(y=-1/5x+(4pi)/15+sqrt3/2)
Kindly see the graph of f(x)=tan x-sin 2x and the normal line y=-1/5x+(4pi)/15+sqrt3/2 at the point ((4pi)/3, sqrt3/2)
![Desmos]()
God bless....I hope the explanation is useful.
