Question

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

Answer 1

`-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)`

ENJOY MATHS !!!!!!!!