Trigonometry Basics

Trigonometry is all about triangles. The word ‘Tri’ Greek word which defines as ‘Three’, ‘gon’ means ‘length’ and ‘metry’ is ‘measurement’. So basically in trigonometry basics, we will learn about the measurement of lengths and angles of triangles. Trigonometry is basically used for the right-angled triangle, where one of the angles is 90°.

Let us learn more about trigonometry formula and problems based on them in this article.

Trigonometry Formula

Let us learn, trigonometry basics formula here, taking an example of the right-angled triangle.

Consider ABC is a right angled triangle, right angled at B and have one hypotenuse AC and one adjacent AB adjacent to angle A and one perpendicular BC opposite to Angle A.

Trigonometry formulas are based on three major trigonometric ratios, Sine, Cosine and Tangent.

Sine or sin\(\angle\)A= \(\frac{Perpendicular}{Hypotenuse}\)=\(\frac{BC}{AC}\)

Cosines or cos\(\angle\)A= \(\frac{Adjacent}{Hypotenuse}\)=\(\frac{AB}{AC}\)

Tangent or tan\(\angle\)A= \(\frac{Perpendicular}{Base}\)=\(\frac{BC}{AB}\)

Similarly, we can write the formula for Reciprocal properties, Sec, Cosec and Cot ratios.

Sec\(\angle\)A =1/Cos\(\angle\)A=\(\frac{Hypotenuse}{Adjacent}\)=\(\frac{AC}{AB}\)

Cosec\(\angle\)A =1/Sin\(\angle\)A=\(\frac{Hypotenuse}{Perpendicular}\)=\(\frac{AC}{BC}\)

Cot\(\angle\)A =1/tan\(\angle\)A=\(\frac{Base}{Perpendicular}\)=\(\frac{AB}{BC}\)

Also,

Sec\(\angle\)A .Cos\(\angle\)A=1

Cosec\(\angle\)A .Sin\(\angle\)A=1

Cot\(\angle\)A .Tan\(\angle\)A=1

Quotient Properties in Trigonometry:

Tan\(\angle\)A=Sin\(\angle\)A/Cos\(\angle\)A

Cot\(\angle\)A =Cos\(\angle\)A/Sin\(\angle\)A

Sin\(\angle\)A=Tan\(\angle\)A/Cos\(\angle\)A

Cos\(\angle\)A=Sin\(\angle\)A/Tan\(\angle\)A

Sec\(\angle\)A=Tan\(\angle\)A/Sin\(\angle\)A

Cosec\(\angle\)A=Cos\(\angle\)A/Tan\(\angle\)A

Some more basic trigonometry formulas:

Sin(90-\(\angle\)A)=Cos\(\angle\)A

Cos(90-\(\angle\)A)=Sin\(\angle\)A

Tan(90-\(\angle\)A)=Cot\(\angle\)A

Cot(90-\(\angle\)A)=Tan\(\angle\)A

Sec(90-\(\angle\)A)=Cosec\(\angle\)A

Cosec(90-\(\angle\)A)=Sec\(\angle\)A

Pythagorean formulas for trigonometry:

sin^2\(\angle\)A+cos^2\(\angle\)A=1

tan^2\(\angle\)A+1=sec^2\(\angle\)A

cot^2\(\angle\)A+1=cosec^2\(\angle\)A

sin2\(\angle\)A=2sin\(\angle\)Acos\(\angle\)A

cos2\(\angle\)A=cos^2\(\angle\)A-sin^2\(\angle\)A

tan2\(\angle\)A=2tan\(\angle\)A/(1-tan^2\(\angle\)A)

cot2\(\angle\)A=(cot^2\(\angle\)A-1)/2cot\(\angle\)A

Trigonometry Table

Angle 30° 45° 60° 90°
Sin\(\angle\)A 0 1/2 \(1/\sqrt{2}\) \(\sqrt{3}/2\) 1
Cos\(\angle\)A 1 \(\sqrt{3}/2\) \(1/\sqrt{2}\) 1/2 0
Tan\(\angle\)A 0 1/\(\sqrt{3}\) 1 \(\sqrt{3}\) Undefined
Cot\(\angle\)A Undefined \(\sqrt{3}\) 1 \(\sqrt{3}\)/2 0
Sec\(\angle\)A 1 2/\(\sqrt{3}\)/2 \(\sqrt{2}\) 2 Undefined
Cosec\(\angle\)A Undefined 2 \(\sqrt{2}\) 2/\(\sqrt{3}\)/2 1

Trigonometry Problem

With the above-given formulas, we can solve trigonometry basics problems very easily.

For example: If a triangle ABC has sides, right angled at B, has perpendicular as 10cm and base as 5cm. Find the tangent and cotangent of the angle formed at point A.

Solution: we know,

Tan\(\angle\)A=Perpendicular/Base

Tan\(\angle\)A=10/5

=2 unit

And,

cot\(\angle\)A=1/Tan\(\angle\)A

=1/2

This way you can solve many trigonometric problems, based upon the trigonometric functions, identities and formulas. Also, learn trigonometry basics class 10th problems by clicking on the links given at the end of this article.

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Practise This Question

XY is drawn parallel to the base BC of a ΔABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY is :