Formulas · Mathematics

Trigonometry

3 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 10

sin²A + cos²A = 1

sin^{2} A + cos^{2} A = 1

What each letter means
$A$	an angle (in a right triangle, one of the acute angles)

For any angle, the square of its sine plus the square of its cosine is exactly 1. Two more identities follow from it: 1 + tan²A = sec²A and 1 + cot²A = cosec²A.

Why it works

In a right triangle with hypotenuse h, sin A = opposite ÷ h and cos A = adjacent ÷ h. So sin²A + cos²A = (opposite² + adjacent²) ÷ h², and by Pythagoras the top is h². Dividing that same Pythagoras equation by adjacent² or by opposite² gives the other two identities.

When to use it

Finding one ratio from another (cos A from sin A), simplifying trigonometric expressions, and proving identities.

How to use it

To find cos A from sin A: cos²A = 1 − sin²A, then take the positive square root for an acute angle.
To simplify: look for sin² + cos² (replace it by 1), or for 1 − sin² (replace it by cos²).
For tan and sec, or cot and cosec, use the two other forms.

To remember: Squares of sine and cosine always make one.

Other forms

Same formula, another form $1 + tan^{2} A = sec^{2} A$ Divide sin²A + cos²A = 1 by cos²A (for 0° ≤ A < 90°).
Same formula, another form $1 + cot^{2} A = cosec^{2} A$ Divide sin²A + cos²A = 1 by sin²A (for 0° < A ≤ 90°).
Rearranged $cos^{2} A = 1 - sin^{2} A$ The form for finding cosine from sine.

Worked examples

Story

A ramp makes an angle A with the ground, and sin A = 5/13: every 13 m along the ramp rises 5 m. How far across the ground do those 13 m go?

cos A = √(1 − 25/169) = √(144/169) = 12/13.
$13 \times \frac{12}{13} = 12$
They go 12 m across the ground.

Answer: 12 m

Picture

A point on a circle of radius 1 around the origin, at angle A from the x-axis, is (cos A, sin A). Check that (cos 30°, sin 30°) = (√3/2, 1/2) is on that circle.

Its distance from the origin is √(cos²30° + sin²30°).
$\frac{3}{4} + \frac{1}{4} = 1$
The distance is √1 = 1, so the point is on the circle.

Answer: Yes: cos²30° + sin²30° = 1

Direct

A is an acute angle and sin A = 3/5. Find cos A.

cos²A = 1 − sin²A
$1 - \frac{9}{25} = \frac{16}{25}$
$\frac{16}{25} = \frac{4}{5}$
A is acute, so cos A is positive: cos A = 4/5.

Answer: 4/5

Reverse

A is acute and cos A = 7/25. Find sin A and tan A.

Try it first, then show the working

sin A = √(1 − 49/625) = √(576/625) = 24/25.
$\frac{24}{25} \div \frac{7}{25} = \frac{24}{7}$
So tan A = 24/7.

Answer: sin A = 24/25, tan A = 24/7

Exam

Prove that sec A (1 − sin A)(sec A + tan A) = 1.

Try it first, then show the working

sec A + tan A = (1 + sin A) ÷ cos A, so the left side is (1 − sin A)(1 + sin A) ÷ cos²A.
(1 − sin A)(1 + sin A) = 1 − sin²A = cos²A.
So the left side is cos²A ÷ cos²A = 1.

Answer: Both sides are 1

Common mistake: Writing sin²A as sin(A²); sin²A means (sin A)². Also, sin A + cos A is not 1; only the squares add to 1.

Class 10

tan A = sin A ÷ cos A, and the reciprocal ratios

tan A = \frac{sin A}{cos A}

What each letter means
$A$	an angle (in a right triangle, one of the acute angles)

The six trigonometric ratios are linked: tan is sine divided by cosine, and cosec, sec and cot are the flips (reciprocals) of sin, cos and tan.

Why it works

In a right triangle, sin A = opposite ÷ hypotenuse and cos A = adjacent ÷ hypotenuse. Dividing them, the hypotenuse cancels, leaving opposite ÷ adjacent, which is tan A. Each reciprocal ratio is the same fraction turned upside down.

When to use it

Finding tan from sin and cos (or any ratio from the others), and rewriting an expression in sines and cosines to simplify or prove it.

How to use it

To get tan A, divide sin A by cos A.
cosec A = 1 ÷ sin A, sec A = 1 ÷ cos A, cot A = 1 ÷ tan A = cos A ÷ sin A.
To find all six ratios from one, draw a right triangle with that ratio's sides and find the third side by Pythagoras.

To remember: Tan is sine over cosine; each co-ratio flips its partner: cosec with sin, sec with cos, cot with tan.

Other forms

Same formula, another form $cosec A = \frac{1}{s i n A}$ The reciprocal of sine.
Same formula, another form $sec A = \frac{1}{c o s A}$ The reciprocal of cosine.
Same formula, another form $cot A = \frac{c o s A}{s i n A}$ The reciprocal of tan.

Worked examples

Story

A ladder's top rests 8 m up a wall and its foot is 6 m from the wall. For the angle A the ladder makes with the ground, find sin A, cos A and tan A, and check that tan A = sin A ÷ cos A.

The ladder is √(8² + 6²) = 10 m long.
sin A = 8/10 = 4/5, cos A = 6/10 = 3/5, tan A = 8/6 = 4/3.
$\frac{4}{5} \div \frac{3}{5} = \frac{4}{3}$

Answer: sin A = 4/5, cos A = 3/5, tan A = 4/3

Picture

In triangle ABC, right-angled at B, AB = 24 cm and BC = 7 cm. Find sin A, cos A and tan A.

The hypotenuse is AC = √(24² + 7²) = √625 = 25 cm.
sin A = BC/AC = 7/25 and cos A = AB/AC = 24/25.
$\frac{7}{25} \div \frac{24}{25} = \frac{7}{24}$
So tan A = 7/24.

Answer: sin A = 7/25, cos A = 24/25, tan A = 7/24

Direct

sin A = 3/5 and cos A = 4/5. Find tan A, cot A, sec A and cosec A.

$\frac{3}{5} \div \frac{4}{5} = \frac{3}{4}$
So tan A = 3/4.
cot A = 4/3, sec A = 5/4 and cosec A = 5/3 (each is a flip).

Answer: tan A = 3/4, cot A = 4/3, sec A = 5/4, cosec A = 5/3

Reverse

A is acute and tan A = 4/3. Find sin A and cos A.

Try it first, then show the working

Draw a right triangle with opposite side 4 and adjacent side 3: the hypotenuse is √(16 + 9) = 5.
sin A = 4/5 and cos A = 3/5. Check: (4/5) ÷ (3/5) = 4/3.

Answer: sin A = 4/5, cos A = 3/5

Exam

If 15 cot A = 8, find sin A and sec A.

Try it first, then show the working

cot A = 8/15, so tan A = 15/8: opposite 15, adjacent 8, hypotenuse √(225 + 64) = √289 = 17.
sin A = 15/17, and sec A = hypotenuse ÷ adjacent = 17/8.

Answer: sin A = 15/17, sec A = 17/8

Common mistake: Pairing the reciprocals wrongly: cosec goes with sin, and sec goes with cos (not the other way round).

Class 10

Heights and distances: h = b tan θ

h = b tan θ

What each letter means
$h$	the height (straight up from the ground) (length)
$b$	the distance along the ground to the foot of the height (length)
$θ$	the angle of elevation: how far you look up from the level (degrees)

Looking up at the top of a tower makes a right triangle: the height is opposite the angle of elevation and the ground distance is next to it, so height = ground distance × tan(angle).

Why it works

tan θ is opposite ÷ adjacent, and here the opposite side is the height h and the adjacent side is the ground distance b. So tan θ = h ÷ b, which rearranges to h = b tan θ.

When to use it

Finding a height you cannot climb (a tower, a tree, a building), or a distance you cannot walk (across a river), from an angle and one measured length.

How to use it

Draw the right triangle: the height is upright, the ground is flat, and your line of sight is the slanting side.
Mark the angle of elevation at your eye, between the ground and the line of sight.
Multiply the ground distance by tan of the angle (tan 30° = 1/√3, tan 45° = 1, tan 60° = √3).

To remember: Up over along is tan.

Other forms

Rearranged $b = \frac{h}{t a n θ}$ How far away, from a known height and the angle.
Special case $h = b$ When θ = 45°: the height equals the ground distance.

Worked examples

Story

When the sun is 30° above the horizon, a tower casts a shadow 40 m long. How tall is the tower?

The shadow is the ground distance and the sun's angle is the angle of elevation. tan 30° = 1/√3.
$h = 40 \times \frac{1}{3} = \frac{40}{3}$
That is 40√3/3 m, about 23.1 m.

Answer: 40/√3 m, about 23.1 m

Picture

A pole 10 m tall casts a shadow 10 m long. What is the angle of elevation of the sun?

tan θ = 10 ÷ 10 = 1.
The angle whose tan is 1 is 45°.

Answer: 45

Direct

From a point 15 m from the foot of a tree, the angle of elevation of its top is 45°. How tall is the tree?

tan 45° = 1.
$h = 15 \times 1 = 15$

Answer: 15

Reverse

From the top of a building 60 m high, the angle of depression of a car on the road is 30°. How far is the car from the foot of the building?

Try it first, then show the working

The angle of depression from the top equals the angle of elevation from the car, 30°.
$b = \frac{60}{t a n 30} = 603$
That is about 103.9 m.

Answer: 60√3 m, about 103.9 m

Exam

The angles of elevation of the top of a tower from two points on the same side of it, 20 m apart in a straight line with its foot, are 30° and 60°. Find the height of the tower.

Try it first, then show the working

Let the nearer point be x m from the foot. Then h = x tan 60° = √3 x, and h = (x + 20) tan 30° = (x + 20) ÷ √3.
So 3x = x + 20, which gives x = 10 m.
$h = 103$
The tower is 10√3 m, about 17.3 m, tall.

Answer: 10√3 m, about 17.3 m

Common mistake: Using sin or cos when the sides you have are the height and the ground distance (that pair needs tan), or measuring the angle from the upright instead of from the ground.