What is the slope of the line normal to the tangent line of f(x) = xcotx+2xsin(x-pi/3) at x= (15pi)/8 ?

1 Answer
Aug 12, 2017

-2/(15sqrt(3)pi + 4)

Explanation:

rArr xcotx + 2xsin(x - pi/3)

For reference :-
sin ((15pi)/8) = 1/2
cos ((15pi)/8) = -sqrt3/2
cosec ((15pi)/8) = 2
cot ((15pi)/8) = -sqrt3

As slope of any line is the derivative of the equation, I will differentiate the equation of tangent.

rArr dy/dx = -xcosec^2xcotx + 2[sin(x - pi/3) + cos(x - pi/3)]

clearly,
Here I can see the trigonometric formula of sin(A - B) and cos(A - B)
Solving,
sin(x - pi/3) + cos(x - pi/3)
by above 2 formulas, I get,
(sinx + cosx) + sqrt3(sinx - cosx)

So, at x = (15pi)/8,
The value of above equation becomes 2

Now, at x = (15pi)/8,
dy/dx = (15sqrt(3)pi + 4)/2 = m_"1"

Now, when a line is normal to the tangent line, it means that the line is perpendicular to the tangent line.

Let the slope of perpendicular line be m_"2"

If the lines are perpendicular to each other then,
m_"1"m_"2" = -1

So, using above equation the slope of line normal to the tangent comes out to be,

m_"2" = -2/(15sqrt(3)pi + 4)

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