# trig wave equations

• Jan 18th 2010, 07:58 AM
killer786
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.
• Jan 18th 2010, 08:20 AM
Hello killer786

Welcome to Math Help Forum!
Quote:

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 $\cos(px+q)$ has a maximum value $+1$ when $px +q = 0, 2\pi, 4\pi, ...$ and a minimum value $-1$ when $px+q = \pi, 3\pi, 5\pi, ...$ Since there are multiple maxima and minima, there will be multiple values of $p$ and $q$. Below, I've taken the first (and most obvious) values.

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

$B-A = 2$

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

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

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

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