Question

How do you find at equation of the tangent line to the graph `y=(cosx)/(sinx+2)` at x =pi/2?

Answer 1

`y = pi/6-1/3x`

Explanation:

Start by finding the y-coordinate of tangency.

`y = cos(pi/2)/(sin(pi/2) + 2)`

`y= 0/(1 + 2)`

`y = 0`

Now differentiate using the quotient rule.

`y' = (-sinx(sinx + 2) - cosx(sinx))/(2 + sinx)^2`

`y' = (-sin^2x - 2sinx - cosxsinx)/(2 + sinx)^2`

The slope of the tangent is given by evaluating your point `x= a` into the derivative.

`m_"tangent" = (-sin^2(pi/2) - 2sin(pi/2) - cos(pi/2)sin(pi/2))/(2+ sin(pi/2))^2`

`m_"tangent" = (-(1)^2 - 2(1) - 0(1))/(2 + 1)^2`

`m_"tangent" = (-1 - 2)/(2 + 1)^2`

`m_"tangent"= -3/9=-1/3`

We now find the equation of the tangent.

`y - y_1 = m(x - x_1)`

`y - 0 = -1/3(x - pi/2)`

`y = pi/6-1/3x`

Hopefully, this helps!