Finding slope of tangent line to $y = 2\sin^2(4x) + 3\cos^2(2x)$ at the point $x_0 = \frac{10\pi}{3}$

Question

This is more in regards to substitution. I understand how to find a point, taking the derivative to finding the slope and finding the equation $y=mx+b$. However, I seem to make a repeated algebraic mistake in substitution. I was just wondering if someone could outline this for me.

I attached a picture of the problem.

Thanks,

Answer 1

You got a bit off track from the start by substituting wrongly. Here's my attempt at the first part:

$2\sin^2(4x)+3\cos^2(2x)$ where $x_0=\frac {10\pi} 3$

$=2\sin(4*(\frac {10\pi} 3))^2+3\cos^2(2*(\frac {10\pi} 3))$ (replaced $x$ with $x_0$ here)

$=2\sin^2(\frac {40\pi} 3)+3\cos^2(\frac {20\pi} 3)$

$=2\sin^2(\frac {4\pi} 3)+3\cos^2(\frac {2\pi} 3)$

$=2(-\sqrt3/2)^2+3(-1/2))^2=2*(3/4)+3*(1/4)=9/4$

At the beginning you sort of got mixed up about multiples of pi and the sine of those multiples. $\frac {10\pi} 3$ is not equal to $\frac {4\pi} 3$, but $\sin(\frac {10\pi} 3)$ is equal to $\sin(\frac {4\pi} 3)$.

Answer 2

The source of your error is in substituting for $x$, you replaced $x$ with $\sin x$ in the first term and by $\cos x$ in the second term.

Let $f(x) = 2\sin^2(4x) + 3\cos^2(2x)$. Then at $x = \dfrac{10\pi}{3}$, we obtain \begin{align*} f\left(\frac{10\pi}{3}\right) & = 2\sin^2\left(4 \cdot \frac{10\pi}{3}\right) + 3\cos^2\left(2 \cdot \frac{10\pi}{3}\right)\\ & = 2\sin^2\left(\frac{40\pi}{3}\right) + 3\cos^2\left(\frac{20\pi}{3}\right)\\ & = 2\sin^2\left(\frac{4\pi}{3} + 12\pi\right) + 3\cos^2\left(\frac{2\pi}{3} + 6\pi\right)\\ & = 2\sin^2\left(\frac{4\pi}{3}\right) + 3\cos^2\left(\frac{2\pi}{3}\right)\\ & = 2\left(-\frac{\sqrt{3}}{2}\right)^2 + 3\left(-\frac{1}{2}\right)^2\\ & = 2 \cdot \frac{3}{4} + 3 \cdot \frac{1}{4}\\ & = \frac{6}{4} + \frac{3}{4}\\ & = \frac{9}{4} \end{align*}

Also, when you made an arithmetic mistake in taking the derivative. Applying the Chain Rule yields \begin{align*} f'(x) & = 2 \cdot 2\sin(4x)\cos(4x) \cdot 4 + 3 \cdot 2\cos(2x)[-\sin(2x)] \cdot 2\\ & = 16\sin(4x)\cos(4x) - 12\cos(2x)\sin(2x) \end{align*} Notice that I used brackets around the $-\sin(2x)$ term to indicate that we are multiplying $\cos(2x)$ by $-\sin(2x)$ rather than subtracting $\sin(2x)$ from $\cos(2x)$.