In view of the coronavirus pandemic, we are making LIVE CLASSES and VIDEO CLASSES completely FREE to prevent interruption in studies
Home > Formulas > Maths Formulas > Trigonometric Functions Formula
Maths Formulas

Trigonometric Functions Formula

Trigonometry is a branch of Mathematics that deals mostly with triangles Trigonometry is also known as the study of relationships between the lengths and angles of triangles. Sometimes, it also dealing with circles. There are various uses of trigonometry and the formula of trigonometry. For example, the technique of triangulation is used in Geography to measure the distance between two landmarks. Also, in Astronomy, to measure the distance to nearby stars and many more. In this topic, we will discuss the Trigonometric Functions Formula with some examples.

Concept of Trigonometry

Trigonometry is the study of the relationships which involves angles, lengths, and heights of triangles. It also relations the different parts of circles and other geometrical figures. Applications of trigonometry are also found in engineering, astronomy, Physics and many other areas of study.

Trigonometric identities are very useful for learning and deriving the many equations and formulas in science. There are many and varied fields where these identities of trigonometry and formula of trigonometry are used.

Trigonometric Identities are popular formulas that involve trigonometric functions. These identities are true for all values of the variables. Trigonometric Ratio is known for the relationship between the measurement of the angles and the length of the side of the right triangle.

There are six functions which are the core of trigonometry. There are three primary ratios given as bellow:

  • Sine  (sin)
  • Cosine  (cos)
  • Tangent  (tan)

The other three are not used as often but can be derived from the three primary functions. These derived functions are:

  • Secant  (sec)
  • Cosecant  (csc)
  • Cotangent  (cot)

Consider the right-angled triangle. For each angle, there are six functions in trigonometry. Each function is the ratio of the two sides of the triangle. The only difference between the six functions is which pair of sides we are using.

Trigonometric Functions

Trigonometric Functions Formula

1] Fundamental Trigonometric Ratios

\(Sin\theta = \frac{{perpendicular}}{{hypotenuse}} = \frac{y}{r}\)

\(Cos\theta = \frac{{base}}{{hypotenuse}} = \frac{x}{r}\)

\(Tan\theta = \frac{{perpendicular}}{{base}} = \frac{y}{x} = \frac{{Sin\theta }}{{Cos\theta }}\)

\(Cot\theta = \frac{{base}}{{perpendicular}} = \frac{x}{y} = \frac{{Cos\theta }}{{Sin\theta }}\)

\(Sec\theta = \frac{{hypotenuse}}{{base}} = \frac{r}{x} = \frac{1}{{Cos\theta }}\)

\(Co\sec \theta = \frac{{hypotenuse}}{{perpendicular}} = \frac{r}{y} = \frac{1}{{Sin\theta }}\)

2] Trigonometry Formulas involving Half Angle Identities

\(\sin\frac{x}{2}=\pm \sqrt{\frac{1-\cos\: x}{2}}\)

\(\cos\frac{x}{2}=\pm \sqrt{\frac{1+\cos\: x}{2}}\)

\(\tan(\frac{x}{2}) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\)

\(\tan(\frac{x}{2}) =\frac{1-\cos(x)}{\sin(x)}\)

3] Trigonometry Formulas involving Product Identities

\(\sin\: x\cdot \cos\:y=\frac{\sin(x+y)+\sin(x-y)}{2}\)

\(\cos\: x\cdot \cos\:y=\frac{\cos(x+y)+\cos(x-y)}{2}\)

\(\sin\: x\cdot \sin\:y=\frac{\cos(x+y)-\cos(x-y)}{2}\)

4] Trigonometry Formulas involving Sum to Product Identities

\(\sin\: x+\sin\: y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}\)

\(\sin\: x-\sin\: y=2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\)

\(\cos\: x+\cos\: y=2\cos\frac{x+y}{2}\cos\frac{x-y}{2}\)

\(\cos\: x-\cos\: y=-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}\)

5] Trigonometry Formulas involving Double Angle Identities:

\(sin 2x = 2 sin (x cos(x)\)

\(tan 2x = \frac {sin 2x}{1-2 sin^2x}\)

6] Square Law

\(Sin ^2 x + cos ^2 x\) will be always 1

\(Sec ^2 x – tan ^2 x\) will be always 1

\(Cosec ^2 x – cot ^2 x\) will be always 1

Solved Examples for Trigonometric Functions Formula

Q.1: Find the value of cos x and tan x, if value of sin x = \(\frac{1}{2}\)

Solution: Since,

\(Sin\theta = \frac{{perpendicular}}{{hypotenuse}}\)

So perpendicular , p = 1

And hypotenuse h = 2


Base, \(b = \sqrt {2^2 – 1^2}\)

\(b = \sqrt{3}\)

\(Cos\theta = \frac{{base}}{{hypotenuse}}\)

i.e. \(cos x = \frac{\sqrt{3}}{2}\)


\(Tan\theta = \frac{{perpendicular}}{{base}}\)

So,\( tan x = \frac{1}{\sqrt{3}}\)

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

Leave a Reply

Notify of

Stuck with a

Question Mark?

Have a doubt at 3 am? Our experts are available 24x7. Connect with a tutor instantly and get your concepts cleared in less than 3 steps.
toppr Code

chance to win a

study tour

Download the App

Watch lectures, practise questions and take tests on the go.

Get Question Papers of Last 10 Years

Which class are you in?
No thanks.