Question

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)?

Answer 1

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

Explanation:

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!