Are any of these four trig expressions equal to each other?

Are any of these four trig expressions equal to each other?

Question

If $\alpha<\beta$, how many different values are there among the following expressions? $\sin\alpha\sin\beta \quad \sin\alpha\cos\beta \quad \cos\alpha\sin\beta \quad \cos\alpha\cos\beta$

Solution

Method 1

We have

$$\sin\alpha\sin\beta = \frac{b}{c}\frac{a}{c} = \frac{ab}{c^2},$$ $$\sin\alpha\cos\beta = \frac{b}{c}\frac{b}{c} = \frac{b^2}{c^2},$$ $$\cos\alpha\sin\beta = \frac{a}{c}\frac{a}{c} = \frac{a^2}{c^2},$$

and

$$\cos\alpha\cos\beta = \frac{a}{c}\frac{b}{c} = \frac{ab}{c^2}.$$

Now since $a > b, a^2 > ab > b^2$, and clearly the expressions take exactly three distinct values.

Method 2

We know that $0<\alpha<45^{\circ}<\beta<90^{\circ}$. Using the properties of $\cos$ and $\sin$ (which which are illustrated in the graph above), we have

$$\begin{align} \sin \alpha < \cos \alpha\\ \cos \beta < \sin \beta\\ \end{align}$$

Now if we start with $(4)$ and multiply it by $\sin\alpha$, we get

$$\sin\alpha\cos \beta < \sin\alpha\sin \beta.$$

If we start with $(2)$ and multiply by $\sin \beta$, we get

$$\sin \beta \sin \alpha <\sin \beta \cos \alpha.$$

Now $\alpha$ and $\beta$ are complementary angles, so we can write

$$\cos\alpha\cos\beta=\cos(90^{\circ}-\beta)\cos(90^{\circ}-\alpha)=\sin\beta\sin\alpha.$$

So we can order the four expressions as

$$\sin\alpha\cos\beta<\sin\alpha\sin\beta=\cos\alpha\cos\beta<\cos\alpha\sin\beta$$

and conclude that we have only three different values.