Evaluating a trigonometric rational function
- Thread starter Shawnster
- Start date
- Joined
- Nov 24, 2012
- Messages
- 3,021
Hello, and welcome to FMH! 
I would first simplify the given function:
[MATH]f(x)=\frac{5\cos(9x)}{2\sin(x)}[/MATH]
And so:
[MATH]f\left(\frac{\pi}{6}\right)=\frac{5\cos\left(9\cdot\dfrac{\pi}{6}\right)}{2\sin\left(\dfrac{\pi}{6}\right)}[/MATH]
Or:
[MATH]f\left(\frac{\pi}{6}\right)=\frac{5\cos\left(\dfrac{3\pi}{2}\right)}{2\sin\left(\dfrac{\pi}{6}\right)}[/MATH]
Now the arguments for the trig. functions are special angles which allow us to immediately evaluate the trig. functions. Can you proceed?
I would first simplify the given function:
[MATH]f(x)=\frac{5\cos(9x)}{2\sin(x)}[/MATH]
And so:
[MATH]f\left(\frac{\pi}{6}\right)=\frac{5\cos\left(9\cdot\dfrac{\pi}{6}\right)}{2\sin\left(\dfrac{\pi}{6}\right)}[/MATH]
Or:
[MATH]f\left(\frac{\pi}{6}\right)=\frac{5\cos\left(\dfrac{3\pi}{2}\right)}{2\sin\left(\dfrac{\pi}{6}\right)}[/MATH]
Now the arguments for the trig. functions are special angles which allow us to immediately evaluate the trig. functions. Can you proceed?