# How to find $F=\left(\frac{B}{2}\right)^{2H}-1$ when $H$ is maximum of $W_{1}$ and $B$ is minimum of $W_{2}$?

The problem is as follows:

A pair of wavelenghts are $$W_{1}$$ and $$W_{2}$$:

$$W_{1}=5\sin\alpha-5+12\cos\alpha$$

$$W_{2}=3\sin\beta+\sqrt{3}\cos\beta$$

Assuming $$H$$ and $$B$$ represent the maximum and the minimum value for $$W_{1}$$ and $$W_{2}$$ respectively.

Find:

$$F=\left(\frac{B}{2}\right)^{2H}-1$$

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&\textrm{6560}\\ 2.&\textrm{5860}\\ 3.&\textrm{6562}\\ 4.&\textrm{5680}\\ \end{array}$$

I'm confused exactly how to tackle this problem from my precalculus workbook. What I attempted to do was to use the boundaries of the sine and the cosine function individually to see if I could obtain the maximum and the minimum values:

$$-1<\sin\alpha<1$$

$$-10<5\sin\alpha-5<0$$

Then for $$\cos\alpha$$:

$$-1<\cos\alpha<1$$

$$-12<12\cos\alpha<12$$

$$-22<5\sin\alpha-5+12\cos\alpha<12$$

Thus the maximum value for the earlier function would be $$12$$.

Whereas for the other would be:

$$-1<\sin\beta<1$$

$$-3<3\sin\beta<3$$

and for the other term:

$$-\sqrt{3}<\sqrt{3}\cos\beta<\sqrt{3}$$

$$-3-\sqrt{3}<3\sin\beta+\sqrt{3}\cos\beta<3+\sqrt{3}$$

Then the minimum value would be for the function would be $$-3-\sqrt{3}$$.

Hence:

$$B=12$$ and $$H=-3-\sqrt{3}$$

But the problemas has not ended yet, they are asking the result of $$F$$

Thus:

$$F=\left(\frac{-3-\sqrt{3}}{2}\right)^{2(2)}-1$$

Then this is where I got stuck. What could possibly be wrong here?. Which part went wrong?. Can someone help me here?. Could it be that the problem lies in the method to get the minimum and the maximum value for those functions?. The sort of approach which I'm looking to get is derivative-free in other words without using derivatives as this problem was intended to be solved relying only using precalculus.

• This is your first wave $W_{1}=5\sin\alpha-5+1$, but in your work, you said $-22<5\sin\alpha-5+12\cos\alpha<12$, which is different Feb 15 at 1:30
• @SomeGuy I'm sorry there was a typo in the original question, I've edited now to reflect the right question as it appears in my workbook. Feb 15 at 2:24

Let's try to make $$a\sin x+ b\cos x$$ of the form $$\boxed{r\sin(x+\theta)}$$ and $$r>0$$

\begin{align}&5\sin\alpha+12 \cos\alpha = r(\cos\theta\sin\alpha+\sin\theta\cos\alpha) = r\sin(\alpha+\theta) \\&\Rightarrow r^2 = 5^2+12^2 = 13^2 \\&\Rightarrow r = 13 \text{ and } \theta =\arctan\frac{12}{5} \end{align}

Now,

$$W_1= 5\sin\alpha+12 \cos\alpha - 5 = 13\sin(\alpha+\theta) - 5$$

Max. value $$13\sin(\alpha+\theta)$$ can attain is $$13$$.

So max. value of $$W_1$$ is $$\boxed{H = W_{1_{\max}} = 13-5 = 8}$$

Similarly

\begin{align}&3\sin\beta+ \sqrt3 \cos\beta = r\sin(\beta+\theta') = r(\sin\beta\cos\theta'+\cos\beta\sin\theta') \\\Rightarrow &r = \sqrt{9+3} = 2\sqrt3\text{ and }\theta' = \arctan\frac1{\sqrt3}\end{align}

So,

$$W_2 = 2\sqrt3\sin(\beta+\theta')$$

Min. value of $$\sin(\beta+\theta') = -1$$ and min. value of $$W_2$$ is $$\boxed{B = W_{2_{\min}} = 2\sqrt3 (-1) =-2\sqrt3}$$

So, $$F = \left(\frac{-2\sqrt3}{2}\right)^{2(8)}-1 = 3^8-1 = 6560$$