Question

find an equation of the tangent line to the curve y = sin(5x) + cos(8x) at the point (π/6, y(π/6)). tangent line: y =

Explanation:

Step1: Find \( y(\frac{\pi}{6}) \)

Substitute \( x = \frac{\pi}{6} \) into \( y=\sin(5x)+\cos(8x) \).
\( y(\frac{\pi}{6})=\sin(5\times\frac{\pi}{6})+\cos(8\times\frac{\pi}{6})=\sin(\frac{5\pi}{6})+\cos(\frac{4\pi}{3}) \)
We know that \( \sin(\frac{5\pi}{6})=\frac{1}{2} \) and \( \cos(\frac{4\pi}{3})=-\frac{1}{2} \), so \( y(\frac{\pi}{6})=\frac{1}{2}+(-\frac{1}{2}) = 0 \).

Step2: Find the derivative \( y' \)

Using the chain rule, the derivative of \( \sin(5x) \) is \( 5\cos(5x) \) and the derivative of \( \cos(8x) \) is \( - 8\sin(8x) \). So \( y' = 5\cos(5x)-8\sin(8x) \).

Step3: Find the slope at \( x = \frac{\pi}{6} \)

Substitute \( x=\frac{\pi}{6} \) into \( y' \).
\( y'(\frac{\pi}{6})=5\cos(5\times\frac{\pi}{6})-8\sin(8\times\frac{\pi}{6})=5\cos(\frac{5\pi}{6})-8\sin(\frac{4\pi}{3}) \)
We know that \( \cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2} \) and \( \sin(\frac{4\pi}{3})=-\frac{\sqrt{3}}{2} \).
So \( y'(\frac{\pi}{6})=5\times(-\frac{\sqrt{3}}{2})-8\times(-\frac{\sqrt{3}}{2})=-\frac{5\sqrt{3}}{2}+4\sqrt{3}=\frac{3\sqrt{3}}{2} \).

Step4: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(\frac{\pi}{6},0) \) and \( m = \frac{3\sqrt{3}}{2} \).
\( y-0=\frac{3\sqrt{3}}{2}(x - \frac{\pi}{6}) \)
Simplify to get \( y=\frac{3\sqrt{3}}{2}x-\frac{3\sqrt{3}\pi}{12}=\frac{3\sqrt{3}}{2}x-\frac{\sqrt{3}\pi}{4} \)

Answer:

\( y=\frac{3\sqrt{3}}{2}x-\frac{\sqrt{3}\pi}{4} \)