Finding the equation of the tangent line at a point
With the equation $$f(x)=\sin^2(x)-\cos^2(x),$$ I need to find the equation of the tangent line at $(\frac{\pi}{4},0).$
I am stuck. I took the derivative of the function, but I don't know where to go from there if that's what I need to do.
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$\begingroup$First of all, we can actually simplify this a lot for ourselves. We actually have that $f(x) = \sin^2 x - \cos^2 x = -\cos(2x)$. (Check this yourself.) Since we want a tangent line, we need slope (derivative) and a point on the line. Our tangent line touches our function at the point $(\frac{\pi}{4},0)$, so this will be the point on our line. So let's calculate the derivative:
$$f'(x) = 2\sin(2x).$$
Thus the slope at $x=\frac{\pi}{4}$ is $f'\left(\frac{\pi}{4}\right) = 2\sin\left(2\frac{\pi}{4}\right) = 2$. The equation of a line in point slope form is given by $y-y_0=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line. But we know $m=2$ so we have $y-y_0=2(x-x_0)$. Do you see how to take it from here?
When doing tangent line problems, I always use point slope form, since it makes life much simpler. That way you avoid having to calculate the $y$-intercept.
$\endgroup$ $\begingroup$Evaluate the derivative at the given point and that becomes the slope of the tangent line passing through the point and use the formula for finding the equation of a straight line to find the required equation of the tangent line.
$f(x)=\sin^2x-\cos^2x=-(\cos^2x-\sin^x)=-\cos(2x)$ and so $f'(x)=2\sin(2x)$ Hence $f'(\pi/4)=2$
Thus the equation is $y=2(x-\frac{\pi}{4})$
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