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What are the six trigonometry functions?

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Trigonometry can be defined as the branch of mathematics that determines and studies the relationships between the sides of a triangle and angles subtended by them. Trigonometry is basically used in the case of right-angled triangles. Trigonometric functions define the relationships between the 3 sides and the angles of a triangle. There are 6 trigonometric functions mainly. Before going into the study of the trigonometric functions we will learn about the 3 sides of a right-angled triangle.

The three sides of a right-angled triangle are as follows,

  • Base The side on which the angle θ lies is known as the base.
  • Perpendicular It is the side opposite to the angle θ  in consideration.
  • Hypotenuse It is the longest side in a right-angled triangle and opposite to the 90° angle.

Trigonometric Functions

Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,

  • sine It is represented as sin θ and is defined as the ratio of perpendicular and hypotenuse.
  • cosine It is represented as cos θ and is defined as the ratio of base and hypotenuse.
  • tangent It is represented as tan θ and is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base.
  • cosecant It is the reciprocal of sin θ and is represented as cosec θ.
  • secant It is the reciprocal of cos θ and is represented as sec θ.
  • cotangent It is the reciprocal of tan θ and is represented as cot θ.

What are the six trigonometry functions?

The six trigonometric functions have formulae for the right-angled triangles, the formulae help in identifying the lengths of the sides of a right-angled triangle, lets take a look at all those formulae,

Trigonometric Functions Formulae
sin θ \frac{P}{H}
cos θ \frac{B}{H}
tan θ \frac{P}{B}
cosec θ \frac{H}{P}
sec θ \frac{H}{B}
cot θ \frac{B}{P}

The below table shows the values of these functions at some standard angles,

Function 30° 45° 60° 90°
sin\theta = \frac{P}{H} 0 \frac{1}{2} \frac{1}{\sqrt2} \frac{\sqrt3}{2} 1
cos\theta = \frac{B}{H} 1 \frac{\sqrt3}{2} \frac{1}{\sqrt2} \frac{1}{2} 0
tan\theta = \frac{sin\theta}{cos\theta}=\frac{P}{B} 0 \frac{1}{\sqrt3} 1 \sqrt3
cosec\theta = \frac{H}{P} 2 \sqrt2 \frac{2}{\sqrt3} 1
sec\theta = \frac{H}{B} 1 \frac{2}{\sqrt3} \sqrt2 2
cot\theta = \frac{B}{P} \sqrt3 1 \frac{1}{\sqrt3} 0

Note: It is advised to remember the first 3 trigonometric functions and their values at these standard angles for ease of calculations.

Sample Problems

Question 1: Evaluate sine, cosine, and tangent in the following figure.

Solution: 

Given P=3, B=4, H=5

Using the trigonometric formulas for sine, cosine and tangent,

sin\theta=\frac{P}{H}=\frac{3}{5}

cos\theta=\frac{B}{H}=\frac{4}{5}

tan\theta=\frac{P}{B}=\frac{3}{4}

Question 2: In the same triangle evaluate secant, cosecant, and cotangent. 

Solution: 

As it is known the values of sine, cosine and tangent, we can easily calculate the required ratios.

cosec\theta=\frac{1}{sin\theta}=\frac{5}{3}

sec\theta=\frac{1}{cos\theta}=\frac{5}{4}

cot\theta=\frac{1}{tan\theta}=\frac{4}{3}

Question 3: Given tan\theta=\frac{6}{8}, evaluate sin θ.cos θ.

Solution: 

tan\theta=\frac{P}{B}

Thus P=6, B=8

Using Pythagoras theorem,

H2=P2+B2

H2=36+64=100

Therefore, H =10

Now, sin\theta= \frac{6}{10}

cos\theta=\frac{8}{10}

Question 4: If cot\theta = \frac{12}{13}, evaluate tan2θ.

Solution: 

Given cot\theta=\frac{12}{13}

Thus tan\theta=\frac{1}{cot\theta}=\frac{13}{12}

\therefore tan^2\theta=\frac{169}{144}

Question 5: In the given triangle, verify sin2θ+cos2θ = 1

Solution: 

Given P=12, B=5, H=13

Thus sin\theta=\frac{12}{13}

cos\theta=\frac{5}{13}

sin^2\theta=144/169

cos^2\theta=25/169

sin^2\theta+cos^2\theta=\frac{169}{169}=1

Hence verified.


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Last Updated : 03 Sep, 2021
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