Review question

# Are any of these four trig expressions equal to each other? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7973

## Solution

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

### 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\label{eq:2}\\ \cos \beta < \sin \beta\label{eq:4}\\ \end{align}

Now if we start with $\eqref{eq:4}$ and multiply it by $\sin\alpha$, we get

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

If we start with $\eqref{eq:2}$ 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.