How do you find an equation of the tangent line to the curve at the given point: #y = csc (x) - 2 sin (x)# and P = (pi/6, 1)?

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Answer 1 · 2016-11-15T14:19:06

The equation is #y = -3sqrt(3)x + (pisqrt(3) + 1)/2#

Start by differentiating the function.

#y= 1/sinx - 2sinx#

#dy/dx= (0 xx sinx - cosx xx 1)/(sinx)^2 - 2cosx#

#dy/dx= -cosx/sin^2x - 2cosx#

#dy/dx = -cotxcscx - 2cosx#

Now, find the slope of the tangent.

#dy/dx = -cot(pi/6)csc(pi/6) - 2cos(pi/6)#

#dy/dx = -1/tan(pi/6) xx 1/sin(pi/6) - 2cos(pi/6)#

#dy/dx = -1/(1/sqrt(3)) xx 1/(1/2) - 2(sqrt(3)/2)#

#dy/dx= -sqrt(3) xx 2 - sqrt(3)#

#dy/dx = -2sqrt(3) - sqrt(3)#

#dy/dx= -3sqrt(3)#

We now find the equation of the tangent line with our point and our slope.

#y_ y_1 = m(x- x_1)#

#y - 1 = -3sqrt(3)(x - pi/6)#

#y - 1 = -3sqrt(3)x + (pisqrt(3))/2#

#y = -3sqrt(3)x + (pisqrt(3))/2 + 1#

#y = -3sqrt(3)x + (pisqrt(3) + 1)/2#

Hopefully this helps!