Take the second derivative of 2sin2(x) and evaluate it at x=π/3.
So I got the second derivative, -8sin(2x)
How do I solve plug in x=π/3 and solve? Do I use the special triangle?
So if I am supposed to use special triangles, would the final answer be -3.64
Your second derivative is correct.
Yes, use your special triangles
especially the 30-60-90 triangle, where sin 60° = √3/2 and sin 120° = √3/2 by the CAST rule
think π/3 ----> 60°
so -8sin(2x) = -8(√3/2) = -4√3 = appr 6.928
how did you get -3.64 ?
To evaluate the expression for x = π/3, you can use the special triangle and the unit circle to find the value of sin(2x).
First, let's find the value of x = π/3 in terms of sin and cos values. In the special triangle, the angles are 30°, 60°, and 90°, and the side opposite the 30° angle is 1, the side opposite the 60° angle is √3, and the hypotenuse is 2. Since π/3 is equivalent to 60° in radians, we can see that sin(π/3) = √3/2 and cos(π/3) = 1/2.
Now let's find the value of sin(2x), where x = π/3. Since sin(2x) = 2sin(x)cos(x), we can substitute the values we found earlier:
sin(2(π/3)) = 2sin(π/3)cos(π/3)
= 2(√3/2)(1/2)
= √3/2
Now, substitute the value of sin(2x) = √3/2 back into the expression for the second derivative:
-8sin(2x) = -8(√3/2)
= -4√3
So, the second derivative of 2sin^2(x) evaluated at x = π/3 is -4√3.