Trigonometry formulas provided below can help students get acquainted with different formulas, which can be helpful in solving questions on trigonometric with ease. Trigonometry problems are very diverse and learning the below formulae help in solving them better. Multiple formulae could be required to solve the problem, so practice to make sure you know when to use each of these formulae.

Trigonometry is a branch of Mathematics that majorly deals with triangles. Trigonometry is also known as the study of relationships between lengths and angles of triangles; sometimes, it also deals with circles.

There is an enormous number of uses of trigonometry and formula of trigonometry. For example, the technique of triangulation is used in geography to measure the distance between landmarks; in astronomy, to measure the distance to nearby stars and also in satellite navigation systems.

## Trigonometry Formulas Major systems

All trigonometric formulas are divided into two major systems: The Trigonometric Identities and Trigonometric Ratios. Trigonometric Identities are some formulas that involve the trigonometric functions. These trigonometry 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.

Here we provide the students with all Trigonometry formula pdf that can be easily downloaded by them. These trigonometric formulas are helpful for the students in question-solving. Pdfs are easy to refer while solving a problem. Along with these trigonometric identities help us to derive the trigonometric formulas, which are sometimes asked in the examination.

We also provide the basic trigonometric table pdf that gives the relation of all trigonometric functions along with their standard value. These trigonometric formulae are helpful in determining the domain, range, and value of a compound trigonometric function. Students can refer to the formulas provided below or can also download the trigonometric formulas pdf that is provided above.

### Trigonometry Formulas From Class 10 to Class 12

Trigonometry Formulas For Class 12 |

Trigonometry Formulas For Class 11 |

Trigonometry Formulas For Class 10 |

## Trigonometry Functions Formulas

In a right-angled triangle, we have 3 sides namely â€“ Hypotenuse, Opposite side (Perpendicular) and Adjacent side(Height). The longest side is known as the hypotenuse, the side opposite to the angle is opposite and the side where both hypotenuse and opposite rests is the adjacent side.

There are basically 6 Laws used for finding the elements in Trigonometry. They are called trigonometric functions. The six trigonometric functions are sine, cosine, secant, co-secant, tangent and co-tangent.

By using a right angled triangle as reference, the trigonometric functions or trigonometric identities are derived:

\(\sin \theta = \frac{Opposite}{Hypotenuse}\)

\(\sec \theta = \frac{Hypotenuse}{Adjacent}\)

\(\cos\theta = \frac{Adjacent}{Hypotenuse}\)

\(\tan \theta =\frac{Opposite}{Adjacent}\)

\(cosec \theta = \frac{Hypotenuse}{Opposite}\)

\(cot \theta = \frac{Adjacent}{Opposite}\)

The **Reciprocal Identities** are given as:

\(cosec\theta =\frac{1}{\sin\theta }\)

\(sec\theta =\frac{1}{\cos\theta }\)

\(cot\theta =\frac{1}{\tan\theta }\)

\(sin\theta =\frac{1}{cosec\theta }\)

\(cos\theta =\frac{1}{\sec\theta }\)

\(tan\theta =\frac{1}{cot\theta }\)

All these are taken from a right angled triangle. With the length and base side of the right triangle given, we can find out the sine, cosine, tangent, secant, cosecant and cotangent values using trigonometric formulas. The reciprocal trigonometric identities are also derived by using the trigonometric functions.

## Trigonometry Formulas List

### A.Trigonometry Formulas involving Periodicity Identities:

- \(sin(x+2\pi )=sin\; x\)
- \(cos(x+2\pi )=cos\; x\)
- \(tan(x+\pi )=tan\; x\)
- \(cot(x+\pi )=cot\; x\)

All trigonometric identities are cyclic in nature. They repeat themselves after this periodicity constant. This periodicity constant is different for different trigonometric identity. tan 45 = tan 225 Â but this is true for cos 45 and cos 225. Refer to the above trigonometry table to verify the values.

### B.Trigonometry Formulas involving Cofunction Identities â€“ degree:

- \(sin(90^{\circ}-x)=cos\; x\)
- \(cos(90^{\circ}-x)=sin\; x\)
- \(tan(90^{\circ}-x)=cot\; x\)
- \(cot(90^{\circ}-x)=tan\; x\)

### C.Trigonometry Formulas involving Sum/Difference Identities:

- \( \sin (x + y) = \sin(x) \cos(y) + \cos(x) \sin(y)\)
- \(\cos(x + y) = \cos(x) \cos(y) â€“ \sin(x) \sin(y)\)
- \(\tan(x+y)=\frac{\tan\: x+\tan\: y}{1-\tan\: x\cdot \tan\: y}\)
- \(\sin(x â€“ y) = \sin(x) \cos(y) â€“ \cos(x) \sin(y)\)
- \(\cos(x – y) = \cos(x) \cos(y) + \sin(x) \sin(y)\)
- \(\tan(x-y)=\frac{\tan\: x – \tan\: y}{1+\tan\: x\cdot tan\: y}\)

### D.Trigonometry Formulas involving Double Angle Identities:

- \(\sin(2x) = 2\sin(x).\cos(x)\)
- \(\cos(2x) = \cos^{2}(x) – \sin^{2}(x)\)
- \(\cos(2x) = 2 \cos^{2}(x) -1\)
- \(\cos(2x) = 1 – 2 \sin^{2}(x)\)
- \(\tan(2x) = \frac{[2\: \tan(x)]}{[1 -\tan^{2}(x)]}\)

### E.Trigonometry Formulas involving Half Angle Identities:

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

Also, \(\tan(\frac{x}{2}) = \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}\\ \\ \\ =\sqrt{\frac{(1-\cos(x))(1-\cos(x))}{(1+\cos(x))(1-\cos(x))}}\\ \\ \\ =\sqrt{\frac{(1-\cos(x))^{2}}{1-\cos^{2}(x)}}\\ \\ \\ =\sqrt{\frac{(1-\cos(x))^{2}}{\sin^{2}(x)}}\\ \\ \\ =\frac{1-\cos(x)}{\sin(x)}\) So, \(\tan(\frac{x}{2}) =\frac{1-\cos(x)}{\sin(x)}\)

### F.Trigonometry Formulas involving Product identities:

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

### G.Trigonometry Formulas involving Sum to Product Identities:

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

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