1. ## trig wave equations

well im having quite alot of trouble doing questions like these could someone give me some insight into how they should be done thank you!

A function f(x) is known to be of the form F(x)=Acos(px+q) + B. It has a local maximum at (6, 8) and the next local minimum is at (10, 2). Find the constants A, p, q and B and hence give the formula for the function.

2. Hello killer786

Welcome to Math Help Forum!
Originally Posted by killer786
well im having quite alot of trouble doing questions like these could someone give me some insight into how they should be done thank you!

A function f(x) is known to be of the form F(x)=Acos(px+q) + B. It has a local maximum at (6, 8) and the next local minimum is at (10, 2). Find the constants A, p, q and B and hence give the formula for the function.
The function $\displaystyle \cos(px+q)$ has a maximum value $\displaystyle +1$ when $\displaystyle px +q = 0, 2\pi, 4\pi, ...$ and a minimum value $\displaystyle -1$ when $\displaystyle px+q = \pi, 3\pi, 5\pi, ...$ Since there are multiple maxima and minima, there will be multiple values of $\displaystyle p$ and $\displaystyle q$. Below, I've taken the first (and most obvious) values.

So $\displaystyle f(x)$ varies between $\displaystyle B+A$ and $\displaystyle B-A$. Since these maximum and minimum values are 8 and 2 respectively, we get:
$\displaystyle B+A = 8$

$\displaystyle B-A = 2$

$\displaystyle \Rightarrow B = 5, A = 3$
The maximum occurs when $\displaystyle x =6$. So, from what I said above, the simplest value is:
$\displaystyle 6p+q = 0$

$\displaystyle \Rightarrow q = -6p$
(1)
The minimum is when $\displaystyle x = 10$. So:
$\displaystyle 10p+q = \pi$

$\displaystyle \Rightarrow 4p = \pi$, from
(1)

$\displaystyle \Rightarrow p = \frac{\pi}{4}$ and $\displaystyle q = -\frac{3\pi}{2}$
So the formula we want is:
$\displaystyle f(x) = 3\cos\left(\frac{\pi}{4}x -\frac{3\pi}{2}\right) + 5$
$\displaystyle = 3\cos\frac{\pi}{4}\Big(x-6\Big) + 5$