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 denominator √3
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 = ∞.