Sine, Cosine, and Tangent of 30º and 60º

In an equilateral triangle, all of the angles measures 60º. The height, h, divides the equilateral triangle into two equal right angles. Using the Pythagorean Theorem, the height is:

h = \sqrt { { I }^{ 2 } - { (\frac { I }{ 2 }) }^{ 2 } } = \sqrt { \frac { 3 { I }^{ 2 } }{ 4 } } = \frac { \sqrt { 3 } }{ 2 } I

\sin { 30 } = \frac { \frac { I }{ 2 } }{ I } = \frac { 1 }{ 2 } \qquad \sin { 60\circ } = \frac { \frac { \sqrt { 3 }}{ 2 } I}{ I } = \frac { \sqrt { 3 }}{ 2 }

\cos { 30 } = \frac { \frac { \sqrt { 3 }}{ 2 } I}{ I } = \frac { \sqrt { 3 }}{ 2 } \qquad \cos { 60 \circ } =  \frac { \frac { I }{ 2 } }{ I } = \frac { 1 }{ 2 }

\tan { 30 } = \frac { \frac { 1 }{ 2 } }{ \frac { \sqrt { 3 }}{ 2 } } = \frac { 1 }{ \sqrt { 3 }} = \frac { \sqrt { 3 }}{ 3 } \qquad \tan { 60 \circ } = \frac { \frac { \sqrt { 3 } }{ 2 } }{ \frac { 1 }{ 2 }} = \sqrt { 3 }

 

Sine, Cosine, and Tangent of 45º

d = \sqrt { { I }^{ 2 } + { I }^{ 2 } } = \sqrt { 2 { I }^{ 2 } } = I \sqrt { 2 }

\sin { 45 } = \frac { I }{ I \sqrt { 2 } } = \frac { 1 }{ \sqrt { 2 } } = \frac { \sqrt { 2 }}{ 2 }

\cos { 45 } = \frac { I }{ I \sqrt { 2 } } = \frac { 1 }{ \sqrt { 2 }} = \frac { \sqrt { 2 } }{ 2 }

\tan { 45 } = \frac { \frac { \sqrt { 2 }}{ 2 }}{ \frac { \sqrt { 2 }}{ 2 }}

Table of Trigonometric Values

Angle 0 30 45 60 90 180 270
\sin {} 0 \frac { 1 }{ 2 } \frac { \sqrt { 2 }}{ 2 } \frac { \sqrt { 3 }}{ 2 } 1 0 -1
\cos {} 1 \frac { \sqrt { 3 }}{ 2 } \frac { \sqrt { 2 }}{ 2 } \frac { 1 }{ 2 } 0 -1 0
\tan {} 0 \frac { \sqrt { 3 }}{ 3 } 1 \sqrt { 3 } \rightarrow \infty 0 \rightarrow \infty

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 3.00/5 - 2 vote(s)
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.