Trigonometry is an important branch of Mathematics. It mainly deals with triangles and their angles. It provides the relationships between the lengths and angles of triangles. Also, it covers many other geometrical shapes like circles. In science for many complex derivations of the formulae and equations, we need trigonometry formula. There are a large number of uses of trigonometry and the formula of trigonometry. In this student will learn the Trigonometry Formula with examples. Let us learn it.

**Trigonometry Formula**

**What is Trigonometry?**

It is the study of the relationships which involve angles, lengths, and heights of triangles. It also relates to the different parts of circles as well as other geometrical figures. Trigonometry has many trigonometric ratios which are very fundamental in mathematics.

It has many identities that are very useful for learning and deriving the many equations and formulas in science. There are various fields where these identities of trigonometry and formula of trigonometry are used. Here we may see many useful trigonometric identities and formulas.

Trigonometric formulas involve many trigonometric functions. These formulas and identities are true for all possible values of the variables. Trigonometric Ratios are also very basic to provide the relationship between the measurement of the angles and the length of the side of the right-angled triangle.

We will consider the right-angled triangle. In these, we have three sides namely â€“ Hypotenuse, the opposite side (Perpendicular) and Adjacent side (Height). The largest 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 six ratios which are the core of trigonometry. These are,

- Sine (sin)
- Cosine (cos)
- Tangent (tan)
- Secant (sec)
- Cosecant (csc)
- Cotangent (cot)

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.

**Fundamental Trigonometric Ratios:**

In trigonometry, there are six possible ratios. A ratio is a comparison of two numbers (or sides of a triangle) by division. The Greek letter, Î¸, will be used to represent the reference angle in the right triangle.

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

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

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

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

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

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

These six ratios are used in different ways to compare two sides of a right triangle. We may notice here that cosecant is the reciprocal of sine, secant is the reciprocal of cosine and cotangent is the reciprocal of tangent.

**Some Important Formulas:**

**Trigonometry Formulas involving Product identities:**

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

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

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

**Trigonometry Formulas involving Sum to Product Identities:**

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

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

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

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

**Trigonometry Formulas involving Double Angle Identities:**

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

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

**Square Law:**

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

\(Sec ^2 x â€“ tan ^2 x will be always 1\)

\(Cosec ^2 x â€“ cot ^2 x will be always 1\)

**Solved Examples**

Q.1: Find the value of Calculate \(sin75 ^{\circ}sin15 ^{\circ}\)

Solution: As given,

\(sin75 ^{\circ}Â sin15 ^{\circ}\)

= \(sin(90 ^{\circ}-15 ^{\circ}) sin15 ^{\circ}\)

= \(cos15 ^{\circ} sin15 ^{\circ}\)

= \(\frac{1}{2} sin 30^{\circ}Â [ applying sin 2x = 2 sin (x) cos(x) ] \)

= \(\frac {1}{2} \times \frac{1}{2}\)

= \(\frac {1}{4}\)

Thus\( sin75 ^{\circ}Â sin15 ^{\circ} will be \frac{1}{4}\)

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