Answer
The equation for the tangent line at $x=-3$ is $ y=27\cos26(x+3)-\sin26.$
Work Step by Step
The equation for the tangent line at $x_0$ is:
$y-f(x_0)=f'(x_0)(x-x_0)$
Here is $x_0=-3$ and $f(x)=\sin(1+x^3)$
$$f'(x)=(\sin(1+x^3))'=\cos(1+x^3)(1+x^3)'=\cos(1+x^3)\cdot3x^2=
3x^2\cos(1+x^3)$$
Now we have:
$$f(-3)=\sin(1+(-3)^3)=\sin(1-27)=\sin(-26)=-\sin26$$
$$f'(-3)=3\cdot(-3)^2\cos(1+(-3)^3)=27\cos(1-27)=27\cos(-26)=27\cos26$$
So, the equation for the tangent line at $x=-3$ is:
$$y-(-\sin26)=27\cos26(x-(-3))\Rightarrow y=27\cos26(x+3)-\sin26$$
