The functions sine, cosine, and tangent, are based on the relationships between triangle side lengths and the internal angles.

The functions may also be used to determine side lengths and angles of any triangle, by means of the sine rule and the cosine rule.

A circle is defined as the set of points which are the square root of the sum of the square of the sine and the square of the cosine of the angles of a 360° rotation.

$$sin^2θ + cos^2θ = r^2$$where r is the radius of the circle.

θ | sine | cosine | tan |
---|---|---|---|

0 | 0 | $$1$$ | $$0$$ |

30 (π/6) | $${1}/{2}$$ | $${√3}/2$$ | $${1}/{√3}$$ |

45 (π/4) | $${1}/{√2}$$ | $${1}/{√2}$$ | $$1$$ |

60 (π/3) | $${√3}/{2}$$ | $${1}/{2}$$ | $${√3}$$ |

90 (π/2) | $${0}$$ | $${1}$$ | $${-}$$ |

cosecant: csc$θ = 1/{sinθ}$

secant: sec$θ = 1/{cosθ}$

cotangent: cot$θ = 1/{tanθ}$

Trig identities are a list of useful equations involving cosine, sine, tangent, and their inverses.

sin$^2θ + $cos$^2θ = 1$

sin$^2θ = 1 - $cos$^2θ$

$cos^2θ = 1 - sin^2θ$

(cos$θ - $sin$θ)^2 = 1 - 2$cos$θ$sin$θ$

cos$(2θ) = $cos$^2θ - $sin$^2{θ} = 2$cos$^{2}θ -1= 1 - 2$sin$^{2}θ$

tan$(2θ) = {2tanθ}/{1-tan^2(θ)}$

sin$^2θ = 1 - $cos$^2θ$

$cos^2θ = 1 - sin^2θ$

(cos$θ - $sin$θ)^2 = 1 - 2$cos$θ$sin$θ$

Double angle formulae:

sin$(2θ) = 2$sin$θ$cos$θ$cos$(2θ) = $cos$^2θ - $sin$^2{θ} = 2$cos$^{2}θ -1= 1 - 2$sin$^{2}θ$

tan$(2θ) = {2tanθ}/{1-tan^2(θ)}$

cos$(α+β)= $cos$(α)$cos$(β) - $sin$(α)$sin$(β)$

cos$(α-β)= $cos$(α)$cos$(β) + $sin$(α)$sin$(β)$

sin$(α+β)= $sin$(α)$cos$(β) + $cos$(α)$sin$(β)$

sin$(α-β)= $sin$(α)$cos$(β) - $cos$(α)$sin$(β)$

tan$(α+β)= {tan(α)+tan(β)}/{1-tan(α)tan(β)}$

tan$(α-β)= {tan(α)-tan(β)}/{1+tan(α)tan(β)}$

cos$(3θ)= $cos$θ(1-4$sin$^2θ)= $cos$θ(4$cos$^2θ-3)$

sin$(3θ)= $sin$θ(4$cos$^2θ-1)= $sin$θ(3-4$sin$^2θ)$

tan$(3θ)= {tanθ(3-tan^2θ)}/{1-3tan^2θ}$

cos$^2(θ/2)={1+cosθ}/2$

sin$^2(θ/2)={1-cosθ}/2$

tan$^2(θ/2)={1-cosθ}/{1+cosθ}$

tan$(θ/2)={1-cosθ}/{sinθ}={sinθ}/{1+cosθ}$

cos$α$cos$β=1/2$(cos$(α+β)+$cos$(α-β)$)

cos$α$sin$β=1/2$(sin$(α+β)-$sin$(α-β)$)

sin$α$sin$β=1/2$(-cos$(α+β)+$cos$(α-β)$)

${sinA}/a={sinB}/b={sinC}/c$

or $a/{sinA} = b/{sinB}=c/{sinC}$

The sine rule can be used when 2 side lengths and 1 non-included angle (not between the two known sides) are known, or 2 angles and one side are known.

$a^2 = b^2+c^2-2bc$cos$A$

or $b^2 = a^2+c^2-2ac$cos$B$

or $c^2 = a^2+b^2-2ab$cos$C$

The cosine rule can be used when 2 side lengths and 1 included angle are known, or when 3 sides are known.

The area of any triangle is: $A = 1/2bc$sin$A$, or $A = 1/2ac$sin$B$, or $A = 1/2ab$sin$C$.

Arc length of sector = $r/θ$

Area of sector = ${θr^2}/2$

Length of chord: $2r$sin$(θ/2)$, where the chord is subtended by central angle $θ$ of a circle with radius $r$.

Sagitta = perpendicular line from the centre of a chord to the circumference of a circle. The length of the sagitta is $r - r$cos$C/2$, where $r$ is the radius of the circle and C the angle at the centre subtended by the chord.

The top of an 82m lighthouse, on top of a 114m cliff, is seen by a ship at two points, A and B, both on a straight line towards the lighthouse. At A, the angle of elevation is 6.4°, and at B the angle of elevation is 8.7°. What is the distance between A and B?

Let x be the distance AB, and y be the distance B to cliff base.

Using the sine rule: ${x+y}/{sin(90-6.4)}={82 + 114}/{sin(6.4)}$

∴ $x=196{sin83.6}/{sin(6.4)}-y = 1747.0-y$

And ${y}/{sin(90-8.7)}={82 + 114}/{sin(8.7)}$

∴ ${y}={196}{sin(81.3)}/{sin(8.7)}=1280.9$

$x= 1747-1280.9 = 466.1m$

Using trigonometry, the Indians around the 6th century CE were able to estimate the relative distances of Earth-Moon and Earth-Sun. This is how they did it:

In Jaipur, in the State of Rajasthan, north-west India, there is a beautifully preserved astronomical observatory. It contains many large instruments made of stone, which measure the equinoxes, declinations of planetary objects, and other astronomical phenomena.

In such a place, the movements of planetary objects could be observed and recorded, in the manner of Tycho Brahe, using non-optical enhancement techniques. The inclinations of the the elliptic, for example, would plot the angle of Earth's inclination, and hence the seasons. The Indians knew long before the Europeans that the Earth was a sphere and not the centre of all planetary orbits.

The Moon could be observed with great accuracy, allowing the marking of the exact time when the third quarter moon was precisely overhead. At this moment, the Sun is rising on the horizon. However, the angle between the vertical to the Moon and to the Sun is the hypotenuse of a right-angled triangle, and therefore necessarily less than 90°. But by how much?

Trigonometry determines that, if the distance to the Moon is d, then the distance to the Sun is a certain multiple of this distance, or x⋅d.

If the angle to the Sun from the vertical can be measured accurately enough, then the value of x is: $$1/{cosα}$$

Some records report that at least one estimate gave this ratio as 400, resulting in the the distance to the Sun as 400 times that of the distance to the Moon. This means that the angle they measured would have been 89.86°.

The actual value is: (Distance to Sun)/(Distance to Moon) = ${1.5 × 10^8}/{385.5 × 10^3} = 390$ km

This is a true angle of 89.85°. That's an accuracy of one part in ten thousand for naked eye measurement - not bad at all!

Tycho Brahe used a similar technique, an astrolabe, to obtain the plethora of data which was of such great use to Kepler and Galileo.

Content © Andrew Bone. All rights reserved. Created : December 30, 2014 Last updated :March 5, 2016

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