Using Trigonometric Ratios

Video coming soon2 worked examples

Theory

From previous study of trigonometry, you should be familiar with the following ratios (SOH CAH TOA):

\sin x = \frac{O}{H} \quad \cos x = \frac{A}{H} \quad \tan x = \frac{O}{A}

Given one of these ratios as a fraction, we can draw a right-angled triangle and use Pythagoras' Theorem to find the third side. We can then state the other two trigonometric ratios.

⚠️ Common Examiner Traps

Draw the right-angled triangle: from $\sin A = \tfrac{3}{5}$ you can build a triangle and read off $\cos A$ and $\tan A$ directly. This is faster and safer than identities.
Use Pythagoras for the third side: and check the quadrant before deciding its sign.
The quadrant sets the signs: an acute angle gives everything positive, but if the question restricts the angle elsewhere some ratios turn negative.
Leave answers as exact fractions: these questions are almost always non-calculator.

Worked examples

Example 1

If $\tan p = \frac{7}{24}$ , state the values of $\sin p$ and $\cos p$ .

Since $\tan = \frac{O}{A}$ , we can imagine a right-angled triangle with Opposite = 7 and Adjacent = 24.

Use Pythagoras to find the Hypotenuse ( $H$ ):

\begin{aligned} H^2 &= 7^2 + 24^2 \\ H^2 &= 49 + 576 \\ H^2 &= 625 \\ H &= 25 \end{aligned}

Now we can write the other ratios:

\sin p = \frac{7}{25} \quad \text{and} \quad \cos p = \frac{24}{25}

Example 2

Acute angles $p$ and $q$ are such that $\sin p = \frac{4}{5}$ and $\cos q = \frac{12}{13}$ . Show that $\sin(p + q) = \frac{63}{65}$ .

First, find the missing sides for both angles using Pythagoras.

For angle p:

$\sin p = \frac{4}{5} \left(\frac{O}{H}\right)$

\begin{aligned} A^2 &= 5^2 - 4^2 \\ A^2 &= 25 - 16 = 9 \\ A &= 3 \end{aligned}

So, $\cos p = \frac{3}{5}$

For angle q:

$\cos q = \frac{12}{13} \left(\frac{A}{H}\right)$

\begin{aligned} O^2 &= 13^2 - 12^2 \\ O^2 &= 169 - 144 = 25 \\ O &= 5 \end{aligned}

So, $\sin q = \frac{5}{13}$

Now expand $\sin(p + q)$ using the compound angle formula:

\begin{aligned} \sin(p + q) &= \sin p \cos q + \cos p \sin q \\ &= \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) + \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \\ &= \frac{48}{65} + \frac{15}{65} \\ &= \frac{63}{65} \end{aligned}