Find an equation of the tangent to the curve at the given point.
y=4(sinx)^2 point: pi/6,1)
So I took the derivative of the original function to get:
y' = 8cosx*sinx
I then chose a point to plug in to find a point for the slope. i picked pi/6 because i thought it would be easier with the cos and the sin but this is as far as I got.
I'm not sure if i've been doing the steps right, or if that's how to find the tangent of a curve.
I know the pointslop form is y-1=m(x+b)
your derivative will be the slope
and since (π/6,1) lies on the curve,
slope of tangent = 8cosπ/6 sinπ/6
= 8(√3/2)(1/2) = 2√3
( you might recall that sin 2x = 2sinxcosx
so 8sinxcosx = 4(2sinxcosx) = 4 sin2x
then slope = 4sin(π/3) = 4√3/2 = 2√3 )
now we have m and a point
y - 1 = 2√3(x - π/6)
y -1 = 2√3 x - π/3
y = 2√3x + 1-π/3
To find the equation of the tangent line to the curve y = 4(sin(x))^2 at the point (pi/6, 1), you are on the right track.
First, finding the derivative of the function is correct. The derivative of y = 4(sin(x))^2 is y' = 8cos(x)sin(x).
To find the slope of the tangent line at the point (pi/6, 1), substitute x = pi/6 into the derivative function: y' = 8cos(pi/6)sin(pi/6).
Recall that cos(pi/6) = sqrt(3)/2 and sin(pi/6) = 1/2. Plugging in the values, you get y' = 8(sqrt(3)/2)(1/2) = 4(sqrt(3))/2 = 2sqrt(3).
So the slope of the tangent line at the point (pi/6, 1) is 2sqrt(3).
Now, using the point-slope form, which is y - y1 = m(x - x1), where (x1, y1) is the point (pi/6, 1) and m is the slope (2sqrt(3)), we can plug in the values:
y - 1 = 2sqrt(3)(x - pi/6).
To make the equation look a bit cleaner, you can distribute the 2sqrt(3) to get:
y - 1 = 2sqrt(3)x - sqrt(3)pi/3.
Next, if you want to convert it to the slope-intercept form y = mx + b, where m is the slope and b is the y-intercept, you can rearrange the equation:
y = 2sqrt(3)x + (sqrt(3)pi/3) + 1.
Therefore, the equation of the tangent line to the curve y = 4(sin(x))^2 at the point (pi/6, 1) is y = 2sqrt(3)x + (sqrt(3)pi/3) + 1.