Problem:
Consider the function $f(x) = –\sin(8x) + 6\cos(4x) – 8x$ where $–\pi/4 < x < \pi/2$. Find the exact $x$-coordinates of the points on the graph of f at which there is a horizontal tangent line.
This is my attempt.
Find derivative
$$ f’(x) = -\cos(8x)\cdot8 – 6\sin(4x)\cdot4 – 8 $$
$$f’(x) = -8\cos(8x) – 24\sin(4x) – 8$$
$$ f’(x) = -8(\cos(8x) + 3\sin(4x))$$
$$f’(x) = 0 $$
$$0 = -8(\cos(8x) + 3\sin(4x) + 1)$$
Use double angle formula $1-\sin(4x)$ to replace $\cos(8x)$
$$0 = -8(1 – \sin^2(4x) + 3\sin(4x) + 1)$$
Let $u = \sin(4x)$
$$ 0 = 1 – u^2 + 3u + 1$$
$$ 0 = -u^2 + 3u + 2$$
Use quadratic formula to solve for $u$.
$$\sin(x) = (3\pm \sqrt{17}) / 2$$
Answer can only be negative due to out of bounds so $\sin(x) = (3-\sqrt17) / 2$
Multiply equation by $4$ because $\cos(4x) $
$$\sin(4x) = [4\cdot(3-\sqrt{17})] / 2$$
I tried multiplying by $4$ but $\sin$ becomes out of bounds
So I just tried to get the inverse of $\sin(x)$ to find the reference angle.
$$\sin^{-1}(\frac{3–\sqrt{17}}{2}) = -0.596\text{rad} = -34^\circ$$
$$x = (\pi + 0.596), (2\pi-0.596), (-0.596), (-\pi+0.596)$$
After graphing these tangent lines in a calculator, the points are close but not exact and thus are not horizontal. I've done other horizontal tangent line problems but this one, it seems I have to use the quadratic formula to find $\sin(x)$, rather than factoring.
Any advice on what I'm doing wrong would be greatly appreciated.
Best Answer
You have used the wrong double angle formula, it should be $$ \cos 8x = 1 - 2 \sin^{2} (4x) $$ and corresponding quadratic equation would be $ 2u^2 - 3u - 2= 0$ and at last you would get $\sin 4x = - \frac{1}{2}$