Trigonometric ratios

Take a right triangle and fix your attention on one acute angle $\theta$. Relative to $\theta$, the three sides earn names: the hypotenuse (opposite the right angle, always longest), the opposite side (facing $\theta$), and the adjacent side (next to $\theta$, but not the hypotenuse). Every trigonometric ratio is just one of these three lengths divided by another. The three primary ratios are defined directly from the sides, and the other three are their reciprocals:

where: opposite, adjacent and hypotenuse are the three side lengths of the right triangle measured relative to the chosen angle $\theta$.

Two consequences follow immediately and are worth internalising before doing any problems. First, since the hypotenuse is always the longest side of a right triangle, both $\sin\theta$ and $\cos\theta$ are always less than or equal to 1 — a ratio like $\sin\theta = 1.2$ is simply impossible and signals an arithmetic slip. Second, the ratios pair up as reciprocals: $\sin\theta$ with $\csc\theta$, $\cos\theta$ with $\sec\theta$, and $\tan\theta$ with $\cot\theta$. Given any one ratio, its partner is available instantly by flipping the fraction, with no triangle redrawing needed. The figure below makes the labelling concrete: the same right triangle, with its three sides named relative to angle $\theta$.

Why do the ratios not care about the triangle's size? Any two right triangles built on the same angle $\theta$ are similar — they have identical corresponding angles (the right angle, $\theta$, and the third angle $90^\circ-\theta$), and similar triangles always have proportional corresponding sides. So the fraction opposite-over-hypotenuse comes out to the same number whether the triangle has sides of a few centimetres or a few kilometres. That single fact is the reason a triangle's shape — not its size — is what determines every trigonometric ratio.

• Mixing up "opposite" and "adjacent" when the chosen angle changes — the same side can be opposite to one angle and adjacent to the other angle in the same triangle, so always ask "opposite or adjacent to which angle?"
• Writing $\sin\theta$ as opposite over adjacent (that is actually $\tan\theta$) instead of opposite over hypotenuse — a very common slip under exam pressure.
• Thinking $\sin\theta$ or $\cos\theta$ can exceed 1, or reading $\sin A$ as "sin times A" rather than as one inseparable symbol naming a ratio.
• Forgetting that $\csc\theta$, $\sec\theta$, $\cot\theta$ are undefined whenever the matching denominator ratio is zero — for example $\tan 0^\circ = 0$, so $\cot 0^\circ$ is undefined.

Problem. In right triangle $ABC$, right-angled at $B$, the side opposite to angle $A$ is $BC = 3$ cm and the hypotenuse $AC = 5$ cm. Find $\sin A$, $\cos A$, $\tan A$, and then find all three reciprocal ratios $\csc A$, $\sec A$, $\cot A$.

Step 1. Find the third side using the Pythagoras theorem. Here $AB$ is the side adjacent to angle $A$.

Step 2. Identify the sides relative to angle $A$: opposite $= BC = 3$, adjacent $= AB = 4$, hypotenuse $= AC = 5$ — the familiar 3-4-5 triangle.

Step 3. Apply the three primary definitions.

Step 4. Take reciprocals to get the remaining three ratios — no new triangle work needed, just flip each fraction.

Step 5. Check consistency using $\tan A = \sin A/\cos A$: $(3/5)\div(4/5) = 3/4$. It matches. All six ratios of angle $A$ are: $\sin A=\tfrac{3}{5}$, $\cos A=\tfrac{4}{5}$, $\tan A=\tfrac{3}{4}$, $\csc A=\tfrac{5}{3}$, $\sec A=\tfrac{5}{4}$, $\cot A=\tfrac{4}{3}$.

Competitive stretch

Problem. If $\cot\theta = \dfrac{7}{8}$, find $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$.

Step 1. Since $\cot\theta = \text{adjacent}/\text{opposite} = 7/8$, build a right triangle with adjacent $=7$, opposite $=8$, relative to $\theta$.

Step 2. Hypotenuse from Pythagoras: $\sqrt{7^2+8^2} = \sqrt{113}$, so $\sin\theta = 8/\sqrt{113}$ and $\cos\theta = 7/\sqrt{113}$.

Step 3. Expand using $(1-x)(1+x)=1-x^2$: the expression becomes $\dfrac{1-\sin^2\theta}{1-\cos^2\theta} = \dfrac{1-64/113}{1-49/113} = \dfrac{49/113}{64/113} = \dfrac{49}{64}$. The value of the expression is $\dfrac{49}{64}$, which also equals $\cot^2\theta$ directly — a shortcut using the identity $\sin^2\theta+\cos^2\theta=1$ from the Trigonometric Identities topic.

Ask the tutor: Want to practise going from one given trigonometric ratio to all six, or need help spotting opposite versus adjacent in a tricky diagram — ask the AI tutor to walk through a few more examples with you.