What is the slope of the line normal to the tangent line of $f (x) = x cot x + 2 x sin (x - \frac{π}{3})$ $f(x) = xcotx+2xsin(x-pi/3)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?

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What is the slope of the line normal to the tangent line of $f (x) = x cot x + 2 x sin (x - \frac{π}{3})$ $f(x) = xcotx+2xsin(x-pi/3)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?

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Answer 1 · 2017-08-12T15:05:29

$- \frac{2}{15 \sqrt{3} π + 4}$ $-2/(15sqrt(3)pi + 4)$

Explanation:

$\Rightarrow x cot x + 2 x sin (x - \frac{π}{3})$ $rArr xcotx + 2xsin(x - pi/3)$

For reference :-
$sin (\frac{15 π}{8}) = \frac{1}{2}$ $sin ((15pi)/8) = 1/2$
$cos (\frac{15 π}{8}) = - \frac{\sqrt{3}}{2}$ $cos ((15pi)/8) = -sqrt3/2$
$cos e c (\frac{15 π}{8}) = 2$ $cosec ((15pi)/8) = 2$
$cot (\frac{15 π}{8}) = - \sqrt{3}$ $cot ((15pi)/8) = -sqrt3$

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

$\Rightarrow \frac{d y}{d x} = - x cos e c^{2} x cot x + 2 [sin (x - \frac{π}{3}) + cos (x - \frac{π}{3})]$ $rArr dy/dx = -xcosec^2xcotx + 2[sin(x - pi/3) + cos(x - pi/3)]$

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

So, at $x = \frac{15 π}{8}$ $x = (15pi)/8$ ,
The value of above equation becomes $2$ $2$

Now, at $x = \frac{15 π}{8}$ $x = (15pi)/8$ ,
$\frac{d y}{d x} = \frac{15 \sqrt{3} π + 4}{2} = m_{1}$ $dy/dx = (15sqrt(3)pi + 4)/2 = m_"1"$

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 $m_{2}$ $m_"2"$

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

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

$m_{2} = - \frac{2}{15 \sqrt{3} π + 4}$ $m_"2" = -2/(15sqrt(3)pi + 4)$

ENJOY MATHS !!!!!!!!

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What is the slope of the line normal to the tangent line of $f (x) = x cot x + 2 x sin (x - \frac{π}{3})$ $f(x) = xcotx+2xsin(x-pi/3)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?

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What is the slope of the line normal to the tangent line of f(x)=xcotx+2xsin(x−π3)f(x) = xcotx+2xsin(x-pi/3) at x=15π8 x= (15pi)/8 ?

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Explanation:

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What is the slope of the line normal to the tangent line of $f (x) = x cot x + 2 x sin (x - \frac{π}{3})$ $f(x) = xcotx+2xsin(x-pi/3)$ at $x = \frac{15 π}{8}$ $x= (15pi)/8$ ?