Thread: 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.

Hello killer786

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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:

$\displaystyle f(x) = 3\cos\left(\frac{\pi}{4}x -\frac{3\pi}{2}\right) + 5$

$\displaystyle = 3\cos\frac{\pi}{4}\Big(x-6\Big) + 5$

Grandad