Tangent: Value

How to find the value of sine (trigonometry): 1 example and its solution. + Tangent values of commonly used angles.

Example

Example

Solution

First, find the reference angle of 11π/6.

11π/6 is between 3π/2 and 2π.
So draw the terminal side on quadrant IV.

Then the reference angle is
2π - 11π/6.

2π = 12π/6

12π/6 - 11π/6 = π/6

So π/6 is the reference angle.

Draw a right triangle like this.

The central angle is π/6: 30º.
So this is a 30-60-90 triangle.

So the base is √3.
The height is -1.
And the hypotenuse is 2.

tan 11π/6 is the tangent of the right triangle.

Tangent is TOA:
Tangent,
Opposite side (-1),
Hypotenuse (√3).

So tan 11π/6 = -1/√3.

Rationalize the denominator3
by multiplying √3/√3.

Then -√3/3.

So tan 11π/6 = -√3/3.

Tangent Values of Commonly Used Angles

Table

These are the sine values
of commonly used angles.

Detail

θ = 0

The adjacent side (base) is 1.
The opposite side (height) is 0.
The hypotenuse is 1.

TOA:
Tangent,
Opposite side (0),
Adjacent side (1).

So tan 0 = 0/1 = 0.

θ = π/6

This is a 30-60-90 triangle.

The adjacent side (base) is √3.
The opposite side (height) is 1.
The hypotenuse is 2.

TOA:
Tangent,
Opposite side (1),
Adjacent side (√3).

So tan π/6 = 1/√3 (= √3/3).

θ = π/4

This is a 45-45-90 triangle.

The adjacent side (base) is 1.
The opposite side (height) is 1.
The hypotenuse is √2.

TOA:
Tangent,
Opposite side (1)
Adjacent side (1).

So tan π/4 = 1/1 = 1.

θ = π/3

This is a 30-60-90 triangle.

The adjacent side (base) is 1.
The opposite side (height) is √3.
The hypotenuse is 2.

TOA:
Tangent,
Opposite side (√3),
Adjacent side (1).

So tan π/3 = √3/1 = √3.

θ = π/2

The adjacent side (base) is 0.
The opposite side (height) is 1.
The hypotenuse is 1.

TOA:
Tangent,
Adjacent side (0)
Opposite side (1).

So tan π/2 = 1/0 = ∞.