Introduction to Trigonometry
Why This Matters
How tall is the Qutub Minar? You can’t exactly drag a measuring tape up its 73 metres. But stand some distance away, measure that distance, and measure the angle at which your eyes look up to the top — and you can work out the height without climbing anything. How wide is a river you can’t cross? Same trick. How high is a hot-air balloon drifting in the sky? Same trick again.
The tool behind all of these is trigonometry — from the Greek tri (three), gon (sides), metron (measure): literally “measuring triangles”. Its big discovery is that in a right triangle, the angles and the sides are locked together by fixed ratios. Know an angle, and the shape of the triangle is decided — so the ratio of any two sides is decided too, no matter how big or small you draw it.
Early astronomers used exactly this to measure the distances to stars and planets they could never reach. Today the same ratios sit inside engineering, physics, GPS, and computer graphics. This chapter builds the foundation: six ratios, their exact values at the most useful angles, and three identities that connect them — all from one familiar friend, the Pythagoras theorem.
The Big Idea
In a right triangle, pick one acute angle. The three sides get names relative to that angle: the side facing it is the opposite, the side along it (touching it, not the hypotenuse) is the adjacent, and the longest slanting side is the hypotenuse. The six trigonometric ratios (sin, cos, tan, and their reciprocals cosec, sec, cot) are just ratios of these sides. Their key magic: they depend only on the angle, not on the triangle’s size — because all right triangles with the same acute angle are similar, so their sides stay in the same proportion.
Let’s Break It Down
The six trigonometric ratios
Take a right triangle and focus on one acute angle, call it θ (the Greek letter theta). Looking from θ, name the sides:
Now the definitions. The first three are the main ratios:
- sin θ = opposite / hypotenuse
- cos θ = adjacent / hypotenuse
- tan θ = opposite / adjacent
The other three are simply their reciprocals (flip them upside down):
- cosec θ = 1 / sin θ = hypotenuse / opposite
- sec θ = 1 / cos θ = hypotenuse / adjacent
- cot θ = 1 / tan θ = adjacent / opposite
Two more relationships worth burning into memory, straight from the definitions:
tan θ = sin θ / cos θ and cot θ = cos θ / sin θ
(Check: sin θ / cos θ = (opp/hyp) / (adj/hyp) = opp/adj = tan θ.)
A memory hook for the big three: “SOH-CAH-TOA” — Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj.
One important note on notation: sin θ is a single quantity — “the sine of θ”. It is not “sin” multiplied by θ; “sin” on its own means nothing. Also, we write sin²θ to mean (sin θ)², and the same for the others.
Why the ratios depend only on the angle
Here is the idea that makes trigonometry work at all. Draw two right triangles with the same acute angle θ — one tiny, one huge. By the AA similarity criterion (each has a right angle and the angle θ), the two triangles are similar. Similar triangles have proportional sides, so opp/hyp comes out the same in both, and so does every other ratio. That is why sin θ, cos θ, etc. are fixed numbers attached to the angle — the size of the triangle simply does not matter.
This also explains a quick fact: since the hypotenuse is the longest side, opposite/hypotenuse and adjacent/hypotenuse are always less than 1. So sin θ and cos θ can never exceed 1.
In a right triangle, tan A = 4/3. Find the other five trigonometric ratios of A.
- tan A = opposite/adjacent = 4/3. So the side opposite A is 4k and the adjacent side is 3k, for some positive number k (the angle, hence every ratio, is the same whatever k is).
- Find the hypotenuse by Pythagoras: hyp² = (4k)² + (3k)² = 16k² + 9k² = 25k², so hyp = 5k.
- Now read off the main ratios: sin A = opp/hyp = 4k/5k = 4/5, and cos A = adj/hyp = 3k/5k = 3/5.
- The reciprocals follow: cosec A = 1/sin A = 5/4, sec A = 1/cos A = 5/3, cot A = 1/tan A = 3/4. So sin A = 4/5, cos A = 3/5, tan A = 4/3, cosec A = 5/4, sec A = 5/3, cot A = 3/4.
If sin θ = 1/3 in a right triangle, what is cosec θ?
Trigonometric ratios of special angles
Some angles turn up constantly: 0°, 30°, 45°, 60° and 90°. For 30°, 45° and 60° we get exact values from two special triangles you can build with a ruler and compass.
The 45° angle. Take a right triangle whose two acute angles are both 45°. Then the two legs are equal — call each one 1.
By Pythagoras, hyp² = 1² + 1² = 2, so hyp = √2. Now read off, taking either 45° angle (opposite = 1, adjacent = 1, hyp = √2):
- sin 45° = opp/hyp = 1/√2
- cos 45° = adj/hyp = 1/√2
- tan 45° = opp/adj = 1/1 = 1
The 30° and 60° angles. Start with an equilateral triangle (all angles 60°) of side 2. Drop a perpendicular from the top vertex to the base; it cuts the base exactly in half and splits the triangle into two identical right triangles. Each has a 60° angle at the bottom, a 30° angle at the top, base = 1 (half of 2), and hypotenuse = 2 (the original side).
The vertical side: by Pythagoras it is √(2² − 1²) = √(4 − 1) = √3. Now read off both angles.
For the 60° angle (opposite = √3, adjacent = 1, hyp = 2):
- sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3/1 = √3.
For the 30° angle (now opposite = 1, adjacent = √3, hyp = 2):
- sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3.
The 0° and 90° angles. Imagine shrinking angle A towards 0°: the opposite side shrinks to nothing while the hypotenuse and adjacent become almost equal. So sin 0° = 0 and cos 0° = 1. Pushing A towards 90° does the reverse — the adjacent side shrinks away — giving sin 90° = 1 and cos 90° = 0. From these, tan 0° = 0/1 = 0, while tan 90° = 1/0 is not defined. (Any ratio that needs dividing by 0 is “not defined”.)
Here are all the values in one table — learn this, but notice the pattern: sin goes 0, 1/2, 1/√2, √3/2, 1 and cos is the same list reversed.
| ∠A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | not defined |
| cosec A | not defined | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | not defined |
| cot A | not defined | √3 | 1 | 1/√3 | 0 |
Evaluate sin 60° cos 30° + sin 30° cos 60°.
- Substitute from the table: sin 60° = √3/2, cos 30° = √3/2, sin 30° = 1/2, cos 60° = 1/2.
- First product: sin 60° cos 30° = (√3/2)(√3/2) = 3/4. Second product: sin 30° cos 60° = (1/2)(1/2) = 1/4.
- Add them: 3/4 + 1/4 = 4/4 = 1.
In a right triangle, the side opposite an acute angle is half the hypotenuse. What is that angle?
- The ratio opposite/hypotenuse is exactly sin of the angle. Here opposite/hypotenuse = 1/2, so sin (angle) = 1/2.
- From the table, the acute angle whose sine is 1/2 is 30°.
- So the angle is 30°.
Trigonometric identities — proved from Pythagoras
An identity is an equation that is true for every allowed value of the angle. There are three famous trigonometric identities, and remarkably all three come from a single source: the Pythagoras theorem. Take a right triangle ABC, right-angled at B. With a = adjacent (AB), b = opposite (BC) and h = hypotenuse (AC), Pythagoras says:
a² + b² = h² … (★)
Identity 1: sin²A + cos²A = 1. Divide every term of (★) by h²:
a²/h² + b²/h² = h²/h²
i.e. (a/h)² + (b/h)² = 1. But a/h = cos A and b/h = sin A, so
cos²A + sin²A = 1, that is sin²A + cos²A = 1.
This holds for all A with 0° ≤ A ≤ 90°.
Identity 2: 1 + tan²A = sec²A. This time divide (★) by a²:
a²/a² + b²/a² = h²/a²
i.e. 1 + (b/a)² = (h/a)². Now b/a = tan A and h/a = sec A, so
1 + tan²A = sec²A.
(Valid for 0° ≤ A < 90°; at 90° both tan and sec are undefined.)
Identity 3: 1 + cot²A = cosec²A. Divide (★) by b²:
a²/b² + b²/b² = h²/b²
i.e. (a/b)² + 1 = (h/b)². Now a/b = cot A and h/b = cosec A, so
1 + cot²A = cosec²A.
(Valid for 0° < A ≤ 90°; at 0° both cot and cosec are undefined.)
That’s the whole engine of the chapter: one Pythagoras equation, divided three different ways, gives three identities. They let you swap between ratios — know one, and you can find any other.
In a right triangle right-angled at C, AB = 29 and BC = 21, with angle B = θ. Show that cos²θ + sin²θ = 1.
- Find the third side AC by Pythagoras: AC = √(AB² − BC²) = √(29² − 21²) = √(841 − 441) = √400 = 20.
- For angle B (θ): opposite = AC = 20, adjacent = BC = 21, hypotenuse = AB = 29. So sin θ = 20/29 and cos θ = 21/29.
- Then sin²θ + cos²θ = (20/29)² + (21/29)² = (400 + 441)/29² = 841/841 = 1 — exactly as the identity promises.
Given sin A = 3/5, find cos A and tan A using the identities (A acute).
- By identity 1, cos²A = 1 − sin²A = 1 − (3/5)² = 1 − 9/25 = 16/25.
- Since A is acute, cos A is positive, so cos A = √(16/25) = 4/5.
- Then tan A = sin A / cos A = (3/5) / (4/5) = 3/4. So cos A = 4/5 and tan A = 3/4.
Prove that sec A (1 − sin A)(sec A + tan A) = 1.
- Write everything in sin and cos: sec A = 1/cos A and tan A = sin A/cos A. So the left side becomes (1/cos A)(1 − sin A)(1/cos A + sin A/cos A).
- Combine the last bracket: 1/cos A + sin A/cos A = (1 + sin A)/cos A. So the expression is (1/cos A)(1 − sin A)(1 + sin A)/cos A.
- Multiply the two brackets: (1 − sin A)(1 + sin A) = 1 − sin²A. So we have (1 − sin²A)/cos²A.
- By identity 1, 1 − sin²A = cos²A. So (1 − sin²A)/cos²A = cos²A/cos²A = 1, which is the right side. Proved.
Common Mistakes
sin A means 'sin' multiplied by A, so you can cancel the 'sin' from both sides of an equation.
It is written like a product, sin × A, and in algebra we cancel common factors all the time — so it feels like 'sin' is just a factor sitting next to A.
sin A is a single inseparable symbol meaning 'the sine of angle A'. 'sin' alone has no value and cannot be cancelled. You can only simplify using actual ratio values or identities, never by 'cancelling sin'.
The opposite and adjacent sides are fixed labels of the triangle, so they stay the same whichever angle you use.
Once you've labelled a triangle's sides for angle A, those labels look permanent — like the hypotenuse, which really is fixed.
Only the hypotenuse is fixed (it's always opposite the right angle). 'Opposite' and 'adjacent' are relative to the chosen angle: switch from angle A to angle C and they swap. The opposite of A is the adjacent of C.
The identity is sin²A + cos²A = 1, so writing sin A² + cos A² = 1 is the same thing.
The little 2 floats around the same symbols, so its exact position seems harmless.
sin²A means (sin A)² — you square the value of sin A. 'sin A²' reads as 'the sine of (A²)', the sine of a squared angle, which is a completely different and meaningless-here quantity. Always square the whole ratio: (sin A)².
Since sin θ and cos θ can be anything, sin θ = 3/2 is fine for some angle.
sin is 'just a ratio', and 3/2 is a perfectly good ratio, so it looks acceptable.
sin θ = opposite/hypotenuse, and the hypotenuse is the LONGEST side, so this ratio can never exceed 1. The same goes for cos θ. A value like 3/2 (> 1) is impossible for a sine or cosine.
Quick Check
In a right triangle, which ratio equals adjacent / hypotenuse?
What is the exact value of tan 60°?
Which of these is the identity that comes from dividing a² + b² = h² by the adjacent side squared?
If cos A = 4/5 for an acute angle A, what is sin A?
Practice Problems
Easy
In a right triangle ABC right-angled at B, AB = 24 cm and BC = 7 cm. Find sin A and cos A.
First the hypotenuse: AC = √(AB² + BC²) = √(24² + 7²) = √(576 + 49) = √625 = 25 cm.
For angle A: opposite = BC = 7, adjacent = AB = 24, hypotenuse = AC = 25.
So sin A = 7/25 and cos A = 24/25. (Check: (7/25)² + (24/25)² = (49 + 576)/625 = 625/625 = 1 ✓.)
Evaluate 2 tan²45° + cos²30° − sin²60°.
From the table: tan 45° = 1, cos 30° = √3/2, sin 60° = √3/2.
2 tan²45° = 2 × 1² = 2.
cos²30° = (√3/2)² = 3/4, and sin²60° = (√3/2)² = 3/4.
So cos²30° − sin²60° = 3/4 − 3/4 = 0.
Total = 2 + 0 = 2.
Medium
Given 15 cot A = 8, find sin A and sec A (A acute).
15 cot A = 8 means cot A = 8/15, i.e. adjacent/opposite = 8/15.
So take adjacent = 8k and opposite = 15k. Hypotenuse = √((8k)² + (15k)²) = √(64 + 225)k = √289 k = 17k.
Then sin A = opposite/hypotenuse = 15/17, and sec A = hypotenuse/adjacent = 17/8, so sec A = 17/8.
If sin(A − B) = 1/2 and cos(A + B) = 1/2, where 0° < A + B ≤ 90° and A > B, find A and B.
sin(A − B) = 1/2. The acute angle with sine 1/2 is 30°, so A − B = 30° … (1).
cos(A + B) = 1/2. The angle with cosine 1/2 is 60°, so A + B = 60° … (2).
Add (1) and (2): 2A = 90° → A = 45°. Subtract (1) from (2): 2B = 30° → B = 15°.
So A = 45° and B = 15°.
If 3 cot A = 4, find whether (1 − tan²A)/(1 + tan²A) equals cos²A − sin²A.
3 cot A = 4 → cot A = 4/3, so tan A = 3/4. Take opposite = 3k, adjacent = 4k; hypotenuse = √(9 + 16)k = 5k.
So sin A = 3/5, cos A = 4/5.
Left side: tan²A = 9/16, so (1 − 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25.
Right side: cos²A − sin²A = 16/25 − 9/25 = 7/25.
Both equal 7/25, so yes, the two expressions are equal here.
Challenge
In a right triangle OPQ, right-angled at P, OP = 7 cm and OQ − PQ = 1 cm. Find sin Q and cos Q.
OQ is the hypotenuse. By Pythagoras: OQ² = OP² + PQ². Write OQ = 1 + PQ (since OQ − PQ = 1).
So (1 + PQ)² = 7² + PQ² → 1 + 2PQ + PQ² = 49 + PQ² → 1 + 2PQ = 49 → 2PQ = 48 → PQ = 24 cm.
Then OQ = 1 + 24 = 25 cm.
For angle Q: opposite = OP = 7, adjacent = PQ = 24, hypotenuse = OQ = 25.
So sin Q = 7/25 and cos Q = 24/25.
Prove that (cot A − cos A)/(cot A + cos A) = (cosec A − 1)/(cosec A + 1).
Start with the left side and write cot A = cos A/sin A:
Numerator: cot A − cos A = cos A/sin A − cos A = cos A(1/sin A − 1) = cos A(1 − sin A)/sin A.
Denominator: cot A + cos A = cos A/sin A + cos A = cos A(1/sin A + 1) = cos A(1 + sin A)/sin A.
Divide — the common factor cos A/sin A cancels:
(cot A − cos A)/(cot A + cos A) = (1 − sin A)/(1 + sin A) = (1/sin A − 1)/(1/sin A + 1).
Since 1/sin A = cosec A, this is (cosec A − 1)/(cosec A + 1), the right side. Proved.
Summary
You should now be able to explain:
- In a right triangle, relative to an acute angle: sin = opposite/hypotenuse, cos = adjacent/hypotenuse, tan = opposite/adjacent, and the reciprocals cosec = 1/sin, sec = 1/cos, cot = 1/tan. Also tan = sin/cos and cot = cos/sin.
- The ratios depend only on the angle, not the triangle’s size, because same-angle right triangles are similar (AA), so their sides stay proportional.
- Since the hypotenuse is longest, sin and cos never exceed 1.
- The exact values for 0°, 30°, 45°, 60°, 90°, derived from the 45-45-90 triangle (sides 1, 1, √2) and the 30-60-90 triangle (sides 1, √3, 2).
- The three identities, all proved by dividing a² + b² = h²: sin²A + cos²A = 1 (÷h²), 1 + tan²A = sec²A (÷a²), 1 + cot²A = cosec²A (÷b²).
- How to find every ratio from any one, and how to prove identities by converting everything to sin and cos.
What’s Next
You now have the ratios and identities. Next, in Some Applications of Trigonometry, you put them to work on the very problems we opened with — heights and distances. You’ll meet the angle of elevation (looking up to the top of a tower) and the angle of depression (looking down to a boat from a cliff), and use a single trig ratio to compute a height or distance you could never measure directly.