Trigonometric Ratios: Quick Revision

A brief note on Trigonometric Ratios. Read it before a test. Try each question at the end before you open its answer.

The sides

• The hypotenuse faces the right angle and is the longest side.
• From an acute angle, the opposite side faces it and the adjacent side touches it.
• Change the angle and opposite and adjacent swap; the hypotenuse stays.

The six ratios

• $\sin A = \dfrac{\text{opp}}{\text{hyp}}$, $\cos A = \dfrac{\text{adj}}{\text{hyp}}$, $\tan A = \dfrac{\text{opp}}{\text{adj}}$.
• $\operatorname{cosec} A = \dfrac{1}{\sin A}$, $\sec A = \dfrac{1}{\cos A}$, $\cot A = \dfrac{1}{\tan A}$.
• $\tan A = \dfrac{\sin A}{\cos A}$ and $\cot A = \dfrac{\cos A}{\sin A}$.
• A ratio depends only on the angle, not on the size of the triangle.
• For an acute angle, $\sin A$ and $\cos A$ are less than $1$; $\sec A$ and $\operatorname{cosec} A$ are greater than $1$.

From one ratio to all six

1. Take the two sides of the given ratio as multiples of $k$.
2. Find the third side with Pythagoras.
3. Read off the other ratios; the $k$ cancels.

A worked example

Given $\tan A = \dfrac{7}{24}$, find $\cos A$ and $\operatorname{cosec} A$.

1. Sides. Opposite $7k$, adjacent $24k$.
2. Hypotenuse. $\sqrt{49k^2 + 576k^2} = 25k$.
3. Ratios. $\cos A = \dfrac{24}{25}$ and $\operatorname{cosec} A = \dfrac{25}{7}$.

Practice with new numbers

A program to try

The program starts from $\tan A = \dfrac{7}{24}$ and prints all six ratios.

Mistakes to avoid

• Naming sides before choosing the angle. Opposite and adjacent depend on it.
• Pairing reciprocals wrongly: $\operatorname{cosec}$ goes with $\sin$, $\sec$ with $\cos$.
• Adding sides instead of squares in Pythagoras.
• Reading $\sin A$ as $\sin \times A$. It is one number.

Check yourself

1. In triangle $JKL$ the right angle is at $K$. Which side is opposite angle $J$?

2. If $\sin A = \dfrac{12}{13}$, what is $\tan A$?

3. Can $\cos A$ equal $\dfrac{9}{8}$ for an acute angle?

4. If $\sec A = \dfrac{5}{3}$, what is $\cos A$?