# Trigonometry Formulas

## Trigonometry Formulas

Trigonometry is a branch of Mathematics that deals explicitly with the angles of a triangle and tries to see the relationship between each element of the triangle which is three sides and three angles. Trigonometry also deals with the relationship of each angle of triangles with circles and is very specifically used in the branch of science and engineering.

The best and the worst part about trigonometry is that one question has multiple answers. One ratio has multiple formulas, and most importantly, knowing where to use each formula can be really tiresome. It requires a lot of problem solving and memorisation of the formulas. And this is because the whole study is relational. We can find a lot of relations among each element of the triangle.

So the question lies, why must a high school student learn trigonometry and trigonometry formulas? The question is a great one, actually. So if we look at our world, our world consists of a lot of shapes. One of the major ones is a circle, and the other one is a triangle. There are also other shapes like curves and, very specifically, waves. We have tons of places that have waves. Waves from the sea, sound or light, or mechanical waves.

Before we get into this rabbit hole, let us imagine a right angle triangle. Why this one specifically? So the answer is to make our lives simpler. A Right Angle Triangle has a lot of element fixes, like one of the angles being strictly 90°, so the other angles have a range to fit in ( the sum of the other two must be 90°). Also, it has some sides defined to be the hypotenuse and the perpendicular sides let the perpendicular be y and the base be x. Let the angle marked be called the greek letter, so there are six standard ratios, each with their basic definition here. In this case would be

Sinθ=perpendicular/hypotenuse=y/r

Cosθ=base/hypotenuse=x/r

Tanθ=perpendicular/base=y/x=Sinθ/Cosθ

Cotθ=base/perpendicular=x/y=Cosθ/Sinθ

Secθ=hypotenuse/base=r/x=1/Cosθ

Cosecθ=hypotenuse/perpendicular=r/y=1/Sinθ

We can see the basic list itself is pretty extensive, so here we have curated all the possible basic formulas that one must know at the end of high school for easy retention of these definitions.

## Trigonometry Formulas

Trigonometry all formulas are available on the website. This is an extensive list of all the formulas. And guess what? This list has all the formulas that may use any of the ratios of trigonometry in the syllabus.

### Trigonometry Formulas List

The formulas or basically the relations that we are going to learn and understand are the followings

• Basic Formulas
• Reciprocal Identities
• Trigonometry Table
• Periodic Identities
• Cofunction Identities
• Sum and Difference Identities
• Double Angle Identities
• Triple Angle Identities
• Half Angle Identities
• Product Identities
• Sum to Product Identities
• Inverse Trigonometry Formulas

### Basic Trigonometric Function Formulas

In trigonometry, there are 6 basic relations defined with 3 sides and 3 angles. Let’s have a look at them

• sin θ = Opposite Side/Hypotenuse
• cos θ = Adjacent Side/Hypotenuse
• tan θ = Opposite Side/Adjacent Side
• sec θ = Hypotenuse/Adjacent Side
• cosec θ = Hypotenuse/Opposite Side
• cot θ = Adjacent Side/Opposite Side

Here the opposite is the side opposite to the angle taken into consideration and adjacent is the side adjacent to the same angle considered.

### Reciprocal Identities

All these relations are when taken reciprocals of produce a different set of results. Before moving further, try taking reciprocals of each and find out what they become.

For example

• sin θ = Opposite Side/Hypotenuse

1/sin θ= hypotenuse/ opposite

Which is following this same ratio? Cosecant, so let’s see the reciprocal identities for all the angles.

• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• cot θ = 1/tan θ
• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ

### Trigonometry Table

These are the common values of angles that are used for solving problems. Memorising these just saves some time, so here is a table.

 Angles (In Degrees) 0° 30° 45° 60° 90° 180° 270° 360° Angles (In Radians) 0° π/6 π/4 π/3 π/2 π 3π/2 2π sin 0 1/2 1/√2 √3/2 1 0 -1 0 cos 1 √3/2 1/√2 1/2 0 -1 0 1 tan 0 1/√3 1 √3 ∞ 0 ∞ 0 cot ∞ √3 1 1/√3 0 ∞ 0 ∞ cosec ∞ 2 √2 2/√3 1 ∞ -1 ∞ sec 1 2/√3 √2 2 ∞ -1 ∞ 1

### Values of Trigonometric Functions in Different Domain and Ranges

Here is a table of the domain and range of these ratios. Let’s understand what it means specifically.

All values are present in the number line. The angle can be any number. However, when we apply the sine function to it, the answer will be between -1 and 1 and will not be exactly equal to those two mentioned values.

Let’s understand why.

Sin is the opposite/hypotenuse. Now imagine a triangle, a RAT. Now can we have a RAT with a side exactly equal to the measurement of the hypotenuse? No never. So it can never be equal to 1.

Try to understand this table with all the ratios in a similar manner.

Let’s look at the above table of values and let’s see the values of Sin in 0 and 360. As we see they are the same. So basically, we see these values repeat after a particular set of intervals. Hence we call these functions periodic.

• sin (π/2 – A) = cos A & cos (π/2 – A) = sin A
• sin (π/2 + A) = cos A & cos (π/2 + A) = – sin A
• sin (3π/2 – A) = – cos A & cos (3π/2 – A) = – sin A
• sin (3π/2 + A) = – cos A & cos (3π/2 + A) = sin A
• sin (π – A) = sin A & cos (π – A) = –
•  cos A
• sin (π + A) = – sin A & cos (π + A) = – cos A
• sin (2π – A) = – sin A & cos (2π – A) = cos A
• sin (2π + A) = sin A & cos (2π + A) = cos A

### Cofunction Identities (in Degrees)

Co function is basically the relationship of one ratio with another one. Remember they’re in degrees.

• sin(90°−x) = cos x
• cos(90°−x) = sin x
• tan(90°−x) = cot x
• cot(90°−x) = tan x
• sec(90°−x) = cosec x
• cosec(90°−x) = sec x

### Sum & Difference Identities

• sin(x+y) = sin(x)cos(y)+cos(x)sin(y)
• cos(x+y) = cos(x)cos(y)–sin(x)sin(y)
• tan(x+y) = (tan x + tan y)/ (1−tan x •tan y)
• sin(x–y) = sin(x)cos(y)–cos(x)sin(y)
• cos(x–y) = cos(x)cos(y) + sin(x)sin(y)
• tan(x−y) = (tan x–tan y)/ (1+tan x • tan y)

These are used when we have an angle on which we have to use the function of Sin, which can be defined by two standard angles or if we can define any unknown angle by the sum or difference of any standard angle.

Double Angle Identities

These are used when the given angle is the double of any standard angle.

• sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan2 x)]
• cos(2x) = cos2(x)–sin2(x) = [(1-tan2 x)/(1+tan2 x)]
• cos(2x) = 2cos2(x)−1 = 1–2sin2(x)
• tan(2x) = [2tan(x)]/ [1−tan2(x)]
• sec (2x) = sec2 x/(2-sec2 x)
• csc (2x) = (sec x. csc x)/2

### Triple Angle Identities

• Sin 3x = 3sin x – 4sin3x
• Cos 3x = 4cos3x-3cos x
• Tan 3x = [3tanx-tan3x]/[1-3tan2x]

For other ratios, we just find the answer to these and invert them.

### Half Angle Identities

These are used when the given angle is half of the standard angle. Also for other ratios, we invert the answer found by these ratios.in these, as we see, we can also find the ratio of tan with sin and cos.

### Product identities

These are used when we need to find the product of functions of different angles.

### Sum to Product Identities

1. Sinα±sinβ=2sin12(α±β)Cos12(α∓β)
2.  Cosα+Cosβ=2Cos12(α+β)Cos12(α−β)
3. Cosα–Cosβ=−2sin(α+β)/2sin(α–β)/2

## Inverse Trigonometry Formulas with Their Domain and Range

Remember, inverse trigonometry is different from reciprocals. Reciprocal is when we put the entire function from numerator to denominator and vice versa, which is sin x and 1/sin x. On the other hand, the inverse is basically answering the question, let’s take the example of sin inverse, what angle of Sin is the given value of opposite/hypotenuse.

So the formulas are

• sin-1 (–x) = – sin-1 x
• cos-1 (–x) = π – cos-1 x
• tan-1 (–x) = – tan-1 x
• cosec-1 (–x) = – cosec-1 x
• sec-1 (–x) = π – sec-1 x
• cot-1 (–x) = π – cot-1

θ = sin−1

(x) is equivalent to x = sin θ

θ = cos−1

(x) is equivalent to x = cos θ

θ = tan−1

(x) is equivalent to x = tan θ

These properties hold for x in the domain and θ in

the range

sin(sin−1(x)) = x

cos(cos−1(x)) = x

tan(tan−1(x)) = x

sin−1(sin(θ)) = θ

cos−1(cos(θ)) = θ

tan−1(tan(θ)) = θ

Use the same logic as above to see why the domains and the ranges are there as these.

### What is the Sin 3x Formula?

Sine 3x is basically the sine of triple the angle that can be permissible in any RAT

Remember Sin 3x will also be between -1 and 1. The range is true regardless of any angle put into this function.

The formula for Sin 3x is

Sin 3x = 3sin x – 4sin3x

## Trigonometry Formulas From Class 10 to Class 12

We saw an extensive list of formulas, but all of them are not presented to students because it will be very overwhelming for any student. So the formulas are divided into three grades, 10th, 11th and 12th.

For Triple Angle, the Below Mentioned Trigonometric Functions are Used:

Triple angle formulas include the square of these ratios, so to solve them, we require

### Square Law Formulas

Sin2x+cos2x willbealways1

Sec2x–tan2x willbealways1

Cosec2x–cot2x willbealways1

sin(A+B+C)=sinAcosBcosC+cosAsinBcosC+cosAcosBsinC–sinAsinBsinC.

cos(A+B+C)=cosAcosBcosC–cosAsinBsinC–sinAcosBsinC–sinAsinBcosC.

tan(A+B+C)=tanA+tanB+tanC–tanAtanBtanC1–tanAtanB–tanBtanC–tanAtanC

cot(A+B+C)=cotAcotBcotC–cotA–cotB–cotCcotAcotB+cotBcotC+cotAcotC–1

### Trigonometry Formulas Major systems

All trigonometry formulas are divided into two parts

• Identities
• Ratios

Identities are basic definitions and ratios are the relations between each identity.

This extensive list of formulas is made because it is crucial for a student to know what to do and to make problem-solving easier.

The part that makes this branch beautiful and troublesome is that one question may have multiple answers. And all questions are asked in the pattern of proving the equation is equal to a certain value.

Taking these formulas into consideration will help students further to understand these problems and solve them in a time-bound fashion.

### Solution: As given,

sin75° sin15°

= sin(90° −15° )sin15°

= cos15° sin15°

= 12sin30° [applyingsin2x=2sin(x)cos(x)]

= 12×12

= 14

Thus sin75° sin15° willbe14

### tan2A = 9/16

tan A = ¾

Q.4: A person 100 metres from the base of a tree, observes that the angle between the ground and the top of the tree is 18 degrees. Estimate the height h of the tree to the nearest tenth of a metre.

Solution:

• Use the tangent
tan(18o) = h / 100
• Solve for h to obtain
h = 100 tan(18o) = 32.5 metres.

Q.5: The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation is situated 300 metres away from the take-off point. What is the upward speed, assumed constant, of the balloon? Give the answer in metres per second and round to two decimal places.

Solution:

• Use the tangent to write
tan(25o) = h1 / 300
and
tan(60o) = (h1 + h2) / 300
• Solve for h1 and h2
h1 = 300 tan(tan(25o))
and
h1 + h2 = 300 tan(60o)
• Use the last two equations to find h2
h2 = 300 [ tan(60o) – tan(25o) ]
• If it takes the balloon 2 minutes (10:00 to 10:02) to climb h2, the the upward speed S is given by
S = h2 / 2 minutes
= 300 [ tan(60o) – tan(25o) ] / (2 * 60) = 3.16 m/sec

Q.6: Point P has initial coordinates (x,y). It is then rotated by an angle about the origin to point P’ (the distance r from the origin is conserved). What are the new coordinates (x’,y’) of point P’.

Solution:

• Express x , y , x’ and y’ using angles b and a + b as follows
x = r cos b
y = r sin b
x’ = r cos(a + b)
y’ = r sin(a + b)
and
• Expand x’ and y’.
x’ = r cos(a + b)
= r cos a cos b – r sin a sin b
y’ = r sin(a + b)
= r sin a cos b + r cos a sin b
• We now use x = r cos b and y = r sin b in the above expressions to obtain
x’ = x cos a – y sin a
y’ = x sin a + y cos a
• The above relationships between x, y, x’ and y’ may be written in matrix form as follows

Q.7: When the top T of a mountain is viewed from point A, 2000 m from ground, the angle of depression a is equal to 15o and when it is viewed from point B on the ground the angle of elevation b is equal to 10o. If points A and B are on the same vertical line, find the height h of the mountain. (round answer to one decimal place).

Solution:

• Let h be the height of the mountain as shown in the figure below. Use the right triangles MTB and MTA to write

tan(10o) = h / d
tan(15o) = (2000 – h) / d

• Solve for d the last 2 equations as follows
d = h / tan(10o) and d = (2000 – h) / tan(15o)
• and eliminate d as follows
h / tan(10o) = (2000 – h) / tan(15o)
• Solve the above for h
h = 2000 tan(10o) / [ tan(15o) + tan(10o)]
= 793.8 m (rounded to 1 decimal place)