Introduction to Trigonometry

Stand at the base of a tall tower and look up at its top, or sit on a balcony and look down at a flower pot on a distant riverbank, or track a hot-air balloon rising into the sky — NCERT opens this chapter with exactly these three pictures, and in every one of them a right triangle appears in your imagination: a vertical height, a horizontal distance, and a line of sight joining them. The question each picture asks is the same: if you know one side and one angle of that triangle, can you find everything else without ever climbing the tower or crossing the river? Answering that question is the whole point of trigonometry, a word built from the Greek tri (three), gon (sides) and metron (measure) — literally, the measurement of triangles. This chapter builds the machinery from the ground up. It starts by defining six trigonometric ratios — sine, cosine, tangent, cosecant, secant, cotangent — as ratios of the two legs and the hypotenuse of a right triangle, taken with respect to one of its acute angles. A short but crucial argument, resting on similar triangles, shows these ratios depend only on the angle, never on how big or small the triangle happens to be. That table is then built by hand for the five angles $0°, 30°, 45°, 60°, 90°$ , using nothing more than an equilateral triangle, an isosceles right triangle, and the Pythagoras theorem you already know. Finally, dividing the Pythagoras theorem itself by each of its three terms in turn produces three trigonometric identities that hold for every acute angle, turning trigonometry from a table of numbers into an algebraic toolkit. By the end, you will define any of the six ratios from a triangle sketch, recover all six from just one of them, rebuild the standard-angle table in under a minute, and apply the three core identities confidently — the exact foundation the next chapter, on heights and distances, builds directly on top of.

Key takeaways

The trigonometric ratios of an acute angle $A$ in a right triangle are sin A, cos A, tan A, cosec A, sec A, cot A, each a ratio of two of the triangle's three sides — the side opposite to $A$ , the side adjacent to $A$ , and the hypotenuse.
Three ratios are primary: $\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}}$ , $\cos A = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan A = \dfrac{\text{opposite}}{\text{adjacent}}$ ; the other three, cosec A, sec A, cot A, are their exact reciprocals.
Also, $\tan A = \dfrac{\sin A}{\cos A}$ and $\cot A = \dfrac{\cos A}{\sin A}$ , linking all six ratios together.
The value of every trigonometric ratio of a given angle depends only on the angle, never on the size of the right triangle it is measured in — a direct consequence of AA similarity.
If just one trigonometric ratio of an acute angle is known, every other ratio can be recovered using a scale factor $k$ and the Pythagoras theorem.
The standard values at $A = 0°, 30°, 45°, 60°, 90°$ can be derived from an equilateral triangle (for $30°$ and $60°$ ) and an isosceles right triangle (for $45°$ ), and by a limiting argument (for $0°$ and $90°$ ); as $A$ increases from $0°$ to $90°$ , sin A rises from $0$ to $1$ while cos A falls from $1$ to $0$ .
tan A and sec A are not defined at $A = 90°$ ; cosec A and cot A are not defined at $A = 0°$ — because the ratio would need a division by zero.
Dividing the Pythagoras theorem $AB^2 + BC^2 = AC^2$ by $AC^2$ , $AB^2$ and $BC^2$ in turn produces the three trigonometric identities: $\sin^2 A + \cos^2 A = 1$ ; $1 + \tan^2 A = \sec^2 A$ (for $A \neq 90°$ ); $1 + \cot^2 A = \text{cosec}^2 A$ (for $A \neq 0°$ ).
Key idea: the six trigonometric ratios of an acute angle are fixed ratios of a right triangle's sides that depend only on the angle (proved by similar triangles), so a single table at 0°, 30°, 45°, 60°, 90° built once from an equilateral and an isosceles right triangle, together with the three identities obtained by dividing the Pythagoras theorem, solves every right triangle you will meet.

The core insight

In a right triangle, the six trigonometric ratios of an acute angle — sin, cos, tan and their reciprocals cosec, sec, cot — are fixed numbers that depend only on the angle (never on the triangle's size, by AA similarity), so one table built at $0°, 30°, 45°, 60°, 90°$ from an equilateral and an isosceles right triangle, together with the identities $\sin^2 A + \cos^2 A = 1$ , $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \text{cosec}^2 A$ obtained by dividing the Pythagoras theorem, is enough to solve any right triangle.

What this chapter covers

The chapter has three connected movements. It opens by defining the trigonometric ratios of an acute angle in a right triangle — sine, cosine, tangent, and their reciprocals cosecant, secant, cotangent — as ratios of the side opposite, the side adjacent and the hypotenuse, and proves, via AA similarity, that these values never change as long as the angle stays the same. It then works out, from first principles using an equilateral triangle and an isosceles right triangle, the exact values of all six ratios at the five standard angles $0°, 30°, 45°, 60°, 90°$ , and records what happens to sin A and cos A at the two extreme ends of the range. Finally it derives the three trigonometric identities — $\sin^2 A + \cos^2 A = 1$ , $1 + \tan^2 A = \sec^2 A$ , $1 + \cot^2 A = \text{cosec}^2 A$ — by dividing the Pythagoras theorem through by each of its terms in turn, and uses them to rewrite one ratio in terms of another.

The core ideas

Defining the trigonometric ratios of an acute angle

Take a right triangle $ABC$ , right-angled at $B$ , and focus on the acute angle $A$ . Relative to $A$ , the side $BC$ (which does not touch $A$ ) is the side opposite to $A$ ; the side $AB$ (which is part of angle $A$ ) is the side adjacent to $A$ ; and $AC$ , opposite the right angle, is the hypotenuse, always the longest side. The six trigonometric ratios of $A$ are defined directly from these three lengths.

\sin A = \frac{BC}{AC}, \quad \cos A = \frac{AB}{AC}, \quad \tan A = \frac{BC}{AB}, \quad \text{cosec}\,A = \frac{AC}{BC}, \quad \sec A = \frac{AC}{AB}, \quad \cot A = \frac{AB}{BC}. \qquad (1)

where: $BC$ is the side opposite $A$ , $AB$ is the side adjacent to $A$ , and $AC$ is the hypotenuse. Notice at once that cosec A, sec A and cot A are exactly the reciprocals of sin A, cos A and tan A, so only three ratios ever need to be found independently. It also follows directly that $\tan A = \dfrac{BC/AC}{AB/AC} = \dfrac{\sin A}{\cos A}$ and, by the same cancellation, $\cot A = \dfrac{\cos A}{\sin A}$ . One notational warning from the start: sin A is a single symbol for "the sine of angle A," not the product of "sin" and "A." "Sin" written on its own, separated from an angle, means nothing.

Figure

A right triangle A B C right angled at B with angle A marked at the vertex, side BC labelled opposite A, side AB labelled adjacent to A, and hypotenuse AC labelled, alongside the six ratio definitions

The moment you switch attention from angle $A$ to angle $C$ in the same triangle, the labels swap: now $AB$ is opposite $C$ and $BC$ is adjacent to $C$ , while $AC$ stays the hypotenuse for both angles. So $\sin C = AB/AC$ and $\cos C = BC/AC$ — the ratios are defined relative to whichever acute angle you are looking from, not fixed to one particular side.

Ratios do not change with the size of the triangle

A natural worry is whether these ratios are trustworthy: if you draw the same angle $A$ inside a bigger or smaller right triangle, could sin A come out different each time? It cannot, and the reason is similar triangles — any two right triangles sharing the acute angle $A$ are similar by the AA criterion, and similar triangles have proportional sides. The full argument, with points $P$ and $M$ marked on a second triangle sharing angle $A$ , is carried out step by step in the first derivation below. The upshot is what makes trigonometry usable at all: the value of a trigonometric ratio depends only on the angle, never on the lengths of the sides.

Trigonometric ratios of the standard angles

For the angles $0°, 30°, 45°, 60°$ and $90°$ — the ones you already know how to construct with a compass — the trigonometric ratios can be worked out exactly, without any measurement, using an equilateral triangle (for $30°$ and $60°$ ) and an isosceles right triangle (for $45°$ ). The two limiting angles $0°$ and $90°$ are found by imagining the angle $A$ in a right triangle shrinking towards $0°$ or growing towards $90°$ and watching what the ratios approach. The full derivation appears below; the results are collected once and for all in Table 1.

A	0°	30°	45°	60°	90°
sin A	0	1/2	1/root 2	root 3 / 2	1
cos A	1	root 3 / 2	1/root 2	1/2	0
tan A	0	1/root 3	1	root 3	not defined
cosec A	not defined	2	root 2	2/root 3	1
sec A	1	2/root 3	root 2	2	not defined
cot A	not defined	root 3	1	1/root 3	0

where: every entry follows from dividing two sides of the $30°$ - $60°$ - $90°$ triangle (sides in ratio $1 : \sqrt{3} : 2$ ) or the $45°$ - $45°$ - $90°$ triangle (sides in ratio $1 : 1 : \sqrt{2}$ ), except the $0°$ and $90°$ columns, which come from the limiting argument. Read the table both ways: given the angle, read off the ratio, or given a ratio's value, read backwards to identify the angle — a trick used constantly in board-level problems. Two features of Table 1 are worth remembering as patterns, not just numbers. As $A$ increases from $0°$ to $90°$ , $\sin A$ rises from $0$ to $1$ while $\cos A$ falls from $1$ to $0$ — mirror images, meeting at $A = 45°$ where both equal $1/\sqrt{2}$ . And $\tan A = \sin A/\cos A$ is undefined at $A = 90°$ because $\cos 90° = 0$ ; symmetrically $\cot A = \cos A/\sin A$ is undefined at $A = 0°$ because $\sin 0° = 0$ .

Figure

A graph with angle A from 0 to 90 degrees on the horizontal axis and ratio value from 0 to 1 on the vertical axis, showing sin A rising steadily from 0 to 1 and cos A falling steadily from 1 to 0, the two curves crossing at 45 degrees where both equal 1 over root 2

Trigonometric identities from the Pythagoras theorem

The last core idea turns the numeric table into algebra that works for every acute angle at once, not just the five standard ones. Start again from the right triangle $ABC$ , right-angled at $B$ , so that $AB^2 + BC^2 = AC^2$ by the Pythagoras theorem. Dividing this single equation through by $AC^2$ , by $AB^2$ , and by $BC^2$ in turn — each time rewriting the resulting fractions as trigonometric ratios of angle $A$ — produces three identities that hold for every acute angle:

\sin^2 A + \cos^2 A = 1, \qquad 1 + \tan^2 A = \sec^2 A, \qquad \cot^2 A + 1 = \text{cosec}^2 A. \qquad (2)

where: the first identity holds for all $A$ with $0° \le A \le 90°$ ; the second holds only for $0° \le A < 90°$ , since tan A and sec A are undefined at $90°$ ; the third holds only for $0° < A \le 90°$ , since cot A and cosec A are undefined at $0°$ . These are trigonometric identities, true for every angle in their domain, not just one triangle — the full derivation is below. Their main use is conversion: given any single ratio of an angle, these three identities (with the reciprocal relations) let every other ratio be written down algebraically, without redrawing a triangle.

Two to three worked derivations

Derivation AWhy trigonometric ratios do not depend on the size of the triangle

The goal is to prove rigorously the claim used everywhere else in this chapter: that the trigonometric ratios of an angle $A$ are the same number no matter which right triangle containing $A$ you measure them in. The proof needs nothing beyond the AA similarity criterion, applied to a second triangle sitting inside the first.

Step 1. Start with the right triangle $ABC$ , right-angled at $B$ , with acute angle $A$ . Choose any point $P$ on the hypotenuse $AC$ (other than $A$ or $C$ ), and draw $PM$ perpendicular to $AB$ , meeting $AB$ at $M$ . This creates a second, smaller right triangle $APM$ , right-angled at $M$ , sitting inside the first one and sharing the same angle $A$ .

Step 2. Compare triangles $APM$ and $ACB$ . They share angle $A$ (it is literally the same angle, common to both triangles), and each has a right angle — $\angle AMP = \angle ABC = 90°$ . Two pairs of equal angles force the third pair to be equal too, so by the AA similarity criterion, $\triangle APM \sim \triangle ACB$ .

Step 3. Since the triangles are similar, their corresponding sides are in the same proportion: $\dfrac{AM}{AB} = \dfrac{AP}{AC} = \dfrac{MP}{CB}$ . Rearranging the first and second ratios as a single proportion between $MP/AP$ and $CB/AC$ gives $\dfrac{MP}{AP} = \dfrac{CB}{AC}$ , because both equal the same constant ratio of similarity.

Step 4. But $\dfrac{MP}{AP}$ is exactly $\sin A$ computed inside the small triangle $APM$ (opposite over hypotenuse for that triangle), and $\dfrac{CB}{AC}$ is exactly $\sin A$ computed inside the original triangle $ACB$ . Step 3 says these are equal. Identical reasoning with the other two proportions shows $\cos A$ and $\tan A$ also agree between the two triangles. Since $P$ was an arbitrary point on the hypotenuse, this holds for every right triangle containing angle $A$ : the ratios are a property of the angle alone, never of the particular triangle's size.

Go deeper · Derivation B · 1 minDeriving the trigonometric ratios of 30 degrees, 45 degrees and 60 degrees from first principles

Derivation BDeriving the trigonometric ratios of 30 degrees, 45 degrees and 60 degrees from first principles

The goal is to compute every entry of Table 1 for the three "constructed" angles — building each value from a triangle whose side lengths are known exactly, rather than quoting the table from memory. Two triangles do all the work: an isosceles right triangle supplies $45°$ , and half of an equilateral triangle, cut by an altitude, supplies both $30°$ and $60°$ .

Step 1 — the 45 degree triangle. In right triangle $ABC$ , right-angled at $B$ , let $\angle A = 45°$ . Since the angles of a triangle sum to $180°$ , $\angle C = 180° - 90° - 45° = 45°$ as well, so the triangle is isosceles with $BC = AB$ . Let $BC = AB = a$ . By the Pythagoras theorem, $AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2$ , so $AC = a\sqrt{2}$ .

Step 2. Read off the ratios directly: $\sin 45° = \dfrac{BC}{AC} = \dfrac{a}{a\sqrt{2}} = \dfrac{1}{\sqrt{2}}$ , and by the same substitution $\cos 45° = \dfrac{AB}{AC} = \dfrac{1}{\sqrt{2}}$ , and $\tan 45° = \dfrac{BC}{AB} = \dfrac{a}{a} = 1$ . The reciprocals follow at once: $\text{cosec}\,45° = \sqrt{2}$ , $\sec 45° = \sqrt{2}$ , $\cot 45° = 1$ .

Step 3 — the 30 and 60 degree triangle. Take an equilateral triangle $ABC$ with every angle $60°$ , and drop the altitude $AD$ from $A$ to $BC$ . Because $ABC$ is equilateral (hence isosceles from $A$ ), the altitude $AD$ bisects the base, so $BD = DC$ , and it also bisects angle $A$ , so $\angle BAD = 30°$ . This makes $\triangle ABD$ a right triangle, right-angled at $D$ , with $\angle ABD = 60°$ and $\angle BAD = 30°$ . Let $AB = 2a$ , the side of the equilateral triangle; then $BD = \tfrac{1}{2}BC = a$ , and by the Pythagoras theorem, $AD^2 = AB^2 - BD^2 = (2a)^2 - a^2 = 3a^2$ , so $AD = a\sqrt{3}$ .

Step 4. Now read off both angles from the same right triangle $ABD$ . For $\angle BAD = 30°$ : opposite side is $BD = a$ , adjacent side is $AD = a\sqrt{3}$ , hypotenuse is $AB = 2a$ , giving $\sin 30° = \dfrac{a}{2a} = \dfrac{1}{2}$ , $\cos 30° = \dfrac{a\sqrt{3}}{2a} = \dfrac{\sqrt{3}}{2}$ , $\tan 30° = \dfrac{a}{a\sqrt{3}} = \dfrac{1}{\sqrt{3}}$ . For $\angle ABD = 60°$ , opposite and adjacent swap: $\sin 60° = \dfrac{AD}{AB} = \dfrac{\sqrt{3}}{2}$ , $\cos 60° = \dfrac{BD}{AB} = \dfrac{1}{2}$ , $\tan 60° = \dfrac{AD}{BD} = \sqrt{3}$ . Taking reciprocals fills in cosec, sec and cot for both angles, completing every non-limiting entry of Table 1 from two triangles alone.

Go deeper · Derivation C · 1 minProving the three Pythagorean trigonometric identities

Derivation CProving the three Pythagorean trigonometric identities

The goal is to derive, not merely state, all three identities of equation $(2)$ from the Pythagoras theorem, and to pin down why each one carries the angle restriction it does. The technique is one idea applied three times: take the equation the Pythagoras theorem hands us and divide it through by a different squared side each time.

Step 1. Start from the Pythagoras theorem in right triangle $ABC$ , right-angled at $B$ : $AB^2 + BC^2 = AC^2$ . This one equation is the source of all three identities; the only thing that changes below is which squared length we divide by.

Step 2 — divide by $AC^2$ . $\dfrac{AB^2}{AC^2} + \dfrac{BC^2}{AC^2} = \dfrac{AC^2}{AC^2} = 1$ , which is $\left(\dfrac{AB}{AC}\right)^2 + \left(\dfrac{BC}{AC}\right)^2 = 1$ , that is, $\cos^2 A + \sin^2 A = 1$ . Every term here is defined for every acute angle including $0°$ and $90°$ themselves, since sin and cos are always defined, so this identity has no restriction: it holds for $0° \le A \le 90°$ .

Step 3 — divide by $AB^2$ . $\dfrac{AB^2}{AB^2} + \dfrac{BC^2}{AB^2} = \dfrac{AC^2}{AB^2}$ , that is, $1 + \left(\dfrac{BC}{AB}\right)^2 = \left(\dfrac{AC}{AB}\right)^2$ , which is $1 + \tan^2 A = \sec^2 A$ . But $\tan A$ and $\sec A$ both have $AB$ in a denominator, and $AB \to 0$ precisely as $A \to 90°$ (the adjacent side shrinks to nothing) — so this identity is only meaningful for $0° \le A < 90°$ .

Step 4 — divide by $BC^2$ . $\dfrac{AB^2}{BC^2} + \dfrac{BC^2}{BC^2} = \dfrac{AC^2}{BC^2}$ , that is, $\left(\dfrac{AB}{BC}\right)^2 + 1 = \left(\dfrac{AC}{BC}\right)^2$ , which is $\cot^2 A + 1 = \text{cosec}^2 A$ . Here cot A and cosec A both have $BC$ in a denominator, and $BC \to 0$ as $A \to 0°$ (the opposite side shrinks to nothing), so this identity holds only for $0° < A \le 90°$ . Collecting all three completes equation $(2)$ , and each restriction is not a technicality to memorise blindly — it falls directly out of watching which side of the triangle is disappearing.

Worked examples

Board level - finding every ratio when tangent is given (Example 1 pattern)

Given $\tan A = \dfrac{4}{3}$ , find the other trigonometric ratios of angle $A$ .

Step 1. Draw a right triangle $ABC$ , right-angled at $B$ , so that $\tan A = \dfrac{BC}{AB} = \dfrac{4}{3}$ . Since only the ratio of the sides is fixed, write $BC = 4k$ and $AB = 3k$ for some positive number $k$ .

Step 2. Find the hypotenuse using the Pythagoras theorem: $AC^2 = AB^2 + BC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2$ , so $AC = 5k$ .

Step 3. Read off sin A and cos A directly from the three side lengths: $\sin A = \dfrac{BC}{AC} = \dfrac{4k}{5k} = \dfrac{4}{5}$ , and $\cos A = \dfrac{AB}{AC} = \dfrac{3k}{5k} = \dfrac{3}{5}$ .

Step 4. Take reciprocals for the remaining three ratios: $\cot A = \dfrac{1}{\tan A} = \dfrac{3}{4}$ , $\text{cosec}\,A = \dfrac{1}{\sin A} = \dfrac{5}{4}$ , $\sec A = \dfrac{1}{\cos A} = \dfrac{5}{3}$ . Answer: sin A = 4/5, cos A = 3/5, cot A = 3/4, cosec A = 5/4, sec A = 5/3.

Board level - finding two sides from one side and one standard angle (Example 6 pattern)

In triangle $ABC$ , right-angled at $B$ , $AB = 5$ cm and $\angle ACB = 30°$ . Determine the lengths of the sides $BC$ and $AC$ .

Step 1. Relative to $\angle C = 30°$ , the side $AB$ is opposite $C$ and the side $BC$ is adjacent to $C$ , so the ratio linking the known side $AB$ to the unknown side $BC$ is tangent: $\dfrac{AB}{BC} = \tan C$ .

Step 2. Substitute the known values: $\dfrac{5}{BC} = \tan 30° = \dfrac{1}{\sqrt{3}}$ . Cross-multiplying gives $BC = 5\sqrt{3}$ cm.

Step 3. For the hypotenuse $AC$ , use the ratio that links $AB$ (opposite $C$ ) directly to $AC$ : $\sin C = \dfrac{AB}{AC}$ , so $\dfrac{1}{2} = \dfrac{5}{AC}$ , giving $AC = 10$ cm.

Step 4. Check with the Pythagoras theorem instead of the trigonometric route: $AC = \sqrt{AB^2 + BC^2} = \sqrt{5^2 + (5\sqrt{3})^2} = \sqrt{25 + 75} = \sqrt{100} = 10$ cm, which matches Step 3 exactly. Answer: BC = 5 root 3 cm and AC = 10 cm.

Board HOTS / foundation JEE-NEET - solving a pair of trigonometric equations for two angles (Example 8 pattern)

If $\sin(A - B) = \dfrac{1}{2}$ and $\cos(A + B) = \dfrac{1}{2}$ , where $0° < A + B \le 90°$ and $A > B$ , find $A$ and $B$ .

Step 1. Recognise both equations as inverse look-ups in Table 1, not as algebra to expand blindly — the expressions $(A-B)$ and $(A+B)$ are simply being treated as single unknown angles for now. Since $\sin(A-B) = \dfrac{1}{2}$ and Table 1 shows $\sin 30° = \dfrac{1}{2}$ is the only angle in the valid range with that sine, conclude $A - B = 30°$ .

Step 2. Similarly, $\cos(A+B) = \dfrac{1}{2}$ matches $\cos 60° = \dfrac{1}{2}$ from Table 1, so $A + B = 60°$ . This step depends on knowing the standard-angle table well enough to read it in reverse, which is exactly the extra layer that makes this problem harder than a direct lookup.

Step 3. Now solve the two linear equations $A - B = 30°$ and $A + B = 60°$ together. Adding them eliminates $B$ : $2A = 90°$ , so $A = 45°$ .

Step 4. Substitute back into $A + B = 60°$ : $45° + B = 60°$ , so $B = 15°$ . Check both original conditions: $A > B$ since $45° > 15°$ , and $0° < A+B = 60° \le 90°$ , both satisfied. Answer: A = 45 degrees and B = 15 degrees. Timing note: budget about two minutes. The whole problem collapses once you stop trying to expand $\sin(A-B)$ or $\cos(A+B)$ with a compound-angle formula (not part of this chapter) and instead read $(A-B)$ and $(A+B)$ straight off Table 1 as single angles — that recognition is the one real hurdle, and after it the problem is just a pair of simultaneous linear equations. Always check the given range restrictions at the end; they exist precisely to rule out other angles (like $A - B = 150°$ ) that also satisfy the sine or cosine value.

Common traps

The single most common notational slip is reading sin A as "sin times A" — the six trigonometric symbols are indivisible names for a ratio; "sin" on its own has no meaning. A closely related error is confusing $\sin^{-1} A$ with $(\sin A)^{-1}$ : $\sin^2 A$ genuinely does mean $(\sin A)^2$ , but $\sin^{-1} A$ is not $1/\sin A$ (that quantity is cosec A) — it denotes the inverse sine function, a different idea met only in higher classes. Another frequent mistake is forgetting the angle restrictions on the second and third identities of equation $(2)$ : $1 + \tan^2 A = \sec^2 A$ silently breaks at $A = 90°$ , and $1 + \cot^2 A = \text{cosec}^2 A$ breaks at $A = 0°$ , for the same reason each time — a side has shrunk to zero. Students also sometimes assume sin A or cos A can exceed $1$ , forgetting that the hypotenuse is always the longest side, so sin A and cos A can never be greater than $1$ , whereas cosec A and sec A (their reciprocals) can never be less than $1$ . Finally, when only one ratio is given, skipping the scale factor $k$ and treating the ratio's numerator and denominator as literal side lengths usually still gives correct final ratios by luck, but it drops the Pythagoras-theorem step needed for an actual length rather than just a ratio.

Common trapTrigonometric ratios depend only on the angle, never on the size of the particular triangle you happen to draw

The single biggest misconception in this chapter is treating a trigonometric ratio as though it depends on the actual side lengths of one specific triangle, rather than purely on the angle. It shows up in subtle ways: assuming that scaling a triangle up or down changes sin A, or being surprised that $\sin 30° = 1/2$ works equally well for a triangle with sides $1, \sqrt{3}, 2$ centimetres and one with sides $100, 100\sqrt{3}, 200$ metres. The justification is the similar-triangles argument proved in the first derivation above: any two right triangles sharing angle $A$ are similar by AA, so their sides are proportional, and a ratio of proportional sides is always the same number. Whenever a problem gives only one trigonometric ratio, the correct first move is to introduce a scale factor $k$ (writing the two known-ratio sides as $4k$ and $3k$ , say) and find the third side using Pythagoras — never treat the ratio's numerator and denominator as literal, fixed lengths.

You should be able to…

Define all six trigonometric ratios of an acute angle using opposite, adjacent and hypotenuse, and state the three reciprocal relations.
Explain, using AA similarity, why a trigonometric ratio depends only on the angle, never on the triangle's size.
Given one trigonometric ratio, recover all five others using a scale factor $k$ and the Pythagoras theorem.
Derive sin, cos and tan at $30°, 45°$ and $60°$ from an equilateral triangle and an isosceles right triangle, without relying on memory.
State all six ratios at $0°$ and $90°$ , and explain why tan A, sec A are undefined at $90°$ and cosec A, cot A at $0°$ .
Derive the three trigonometric identities from the Pythagoras theorem and state each one's exact domain restriction.
Use the identities to rewrite one trigonometric ratio in terms of another, without redrawing a triangle.
Solve for an unknown side or angle in a right triangle given one side and one acute angle.
Avoid the notational traps: sin A is not a product, $\sin^{-1} A$ is not cosec A, and sin A, cos A can never exceed $1$ .

Ask the tutor: Want to work through deriving the 30-60-90 and 45-45-90 triangles from scratch yourself, or practise recovering all six ratios from a single given value?

Why is sin A not the same thing as sin multiplied by A?

Because sin A is a single, indivisible symbol meaning "the sine ratio of angle A," not two separate quantities being multiplied. "Sin" by itself is not a number or a variable — it only has meaning when it is attached to an angle, exactly the way "square root of nine" cannot be split into "square root" times "nine." This is worth stating explicitly because the notation genuinely looks like a product at first glance, especially once you also learn that $\sin^2 A$ really does mean $(\sin A)^2$ . Treat every one of the six ratio names — sin, cos, tan, cosec, sec, cot — as glued permanently to the angle that follows it.

Why do the trigonometric ratios of an angle stay exactly the same even if I draw a much bigger right triangle with that same angle?

Because a trigonometric ratio is, by definition, a ratio of two sides, not a length by itself, and any two right triangles that share an acute angle are similar by the AA criterion — matching the right angle and the shared acute angle automatically matches the third angle too. Similar triangles always have their corresponding sides in the same proportion, so the ratio of any two sides — which is exactly what sin, cos or tan measures — comes out identical whether the triangle has sides measured in centimetres or in kilometres. This is precisely why a single standard-angle table can be reused for every right triangle you will ever meet, instead of needing a new table for every possible triangle size.

Why are tan 90 degrees and cot 0 degrees not defined, when sin and cos are defined everywhere?

Because tan A and cot A are built as ratios that put a side length, not the hypotenuse, in the denominator, and that side shrinks to zero exactly at the angle in question. $\tan A = BC/AB$ has the adjacent side $AB$ in the denominator, and as $A \to 90°$ that side shrinks toward zero, so tan A grows without bound and is undefined exactly there (sec A fails for the same reason). Symmetrically, $\cot A = AB/BC$ has the opposite side $BC$ in the denominator, and $BC \to 0$ as $A \to 0°$ , so cot A and cosec A are undefined at $0°$ . Sine and cosine never hit this problem because their denominator is always the hypotenuse, which never shrinks to zero for any angle between $0°$ and $90°$ .

What is the fastest reliable way to remember or rebuild the whole standard-angle table under exam pressure?

Do not memorise thirty separate numbers — rebuild the two triangles instead, which takes under a minute and cannot be misremembered. Draw an isosceles right triangle with legs $1, 1$ and hypotenuse $\sqrt{2}$ to read off every ratio at $45°$ . Then draw an equilateral triangle of side $2$ , drop the altitude to get a $1, \sqrt{3}, 2$ right triangle, and read $30°$ and $60°$ straight off it, swapping which leg is "opposite" for each angle. Fill in $0°$ and $90°$ last, using the two patterns "sin A rises from 0 to 1, cos A falls from 1 to 0" and "tan A and sec A break at 90°, cosec A and cot A break at 0°" as a final consistency check rather than as independent facts to memorise.