Back to: MATHEMATICS SS3

**Welcome to class! **

In today’s class, we will be talking about the application of surds to trigonometric ratios. Enjoy the class!

**Application of surds to trigonometric ratios**

Apart from solving quadratics, surds also arise in trigonometry.

The angles 30°, 45°, 60° have the following trigonometric ratios.

Triangle *ABC *is equilateral. *AD* is the line interval from *A* to the midpoint of *BD*. Triangles *ABD* and *ACD* are congruent (SSS).

$AnglesABD=60\xb0,BAD=30\xb0andPythagoras\u2019theoremgivesthatAD=\surd 3\phantom{\rule{0ex}{0ex}}$

Hence,

sin 30° = cos 60° = 1/2 , | sin 60° = cos 30° =
$\surd 3/2$ , |
||

tan 30° =
$1/\surd 3$ and |
tan 60° =
$\surd 3$ . |

$U\mathrm{sin}gasquaredividedbyadiagonalweformtwoisoscelesright\u2013angledtrianglesandcansee\mathrm{sin}45\xb0=\mathrm{cos}45\xb0=1/\surd 2and\mathrm{tan}45\xb0=1.$

**The trigonometric functions of ***y* = cos θ and *y* = cos θ.

*y*= cos θ and

*y*= cos θ.

$U\mathrm{sin}gthevalues\mathrm{sin}30\xb0=1/2=0.5and\mathrm{sin}60\xb0=\surd 3/2\approx 0.87,wecandrawupthefollowingtableofvaluesandthenplotthem.$

θ° | 0 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | 330 | 360 | |||||||||||||

sin θ | 0 | 0.5 | 0.87 | 1 | 0.87 | 0.5 | 0 | −0.5 | −0.87 | −1 | −0.87 | −0.5 | 0 |

More points can be used to show that the shape of the graph is as shown below.

Electrical engineers and physicists call this a **sine wave**.

The basic formulae here all reply on the SINE graph as follows:

- The
**sine**of an angle is defined by the**vertical height of a point**as it rotates around a unit circle (that is, its radius is 1) measured from a horizontal line through the centre of the circle. So it cannot be bigger than 1 or less than -1. - the
**cosine**of an angle is defined by the**horizontal distance of a point**as it rotates around the unit circle measured from a vertical line through the centre of the circle. It too must be in the range -1 to 1.

**The surd form of trigonometric ratios:**

The equilateral triangle above has sides each 2 units long and all angles at 60^{o}. It has been halved into 2 right – angled triangles of base 1 unit long. Using Pythagoras:

(ab)^{2} = (ac)^{2} + (bc)^{2}

=> (2)^{2} = (ac)^{2} + (1)^{2}

=> (ac)^{2} = 4 – 1 = 3

and (ac) = root 3.

Using the definitions of trigonometric ratios, summarised in SOHCAHTOA:

sin 30^{o} = bc / ab = 1 / 2

sin 60^{o} = ac / ab = root 3 / 2

cos 30^{o} = ac / ab = root 3 / 2

cos 60^{o} = bc / ab = 1 /2

tan 30^{o} = bc / ac = 1 / root 3

tan 60 = ac / bc = root 3 / 1 = root 3

The isosceles above has equal sides of 1 unit each and the 2 complementary angles each of 45^{o}. Using Pythagoras’ Theorem to find side (ac):

(ac)^{2} = (ab)^{2} + (bc)^{2}

(ac)^{2} = (1)^{2} + (1)^{2} = 2

ac = root 2

Trigonometrical ratio: |
Surd form: |
Approximation |

sin 30^{o} |
1 / 2 | 0.5 |

sin 45^{o} |
1 / root 2 | 0.7071 |

sin 60^{o} |
root 3 / 2 | 0.866 |

cos 30^{o} |
root 3 / 2 | 0.866 |

cos 45 | 1 / root 2 | 0.7071 |

cos 60^{o} |
1 / 2 | 0.5 |

tan 30^{o} |
1 / root 3 | 0.5774 |

tan 45^{o} |
1 | |

tan 60^{o} |
root 3 | 1.7321 |

**For any pair of complementary angles these 2 rules apply:**

sin x = cos (90 – x)

and cos x = sin (90 – x)

sin 30^{o} = cos 60^{o} |

sin 60^{o} = cos 30^{o} |

sin 45^{o} = cos 45^{o} |

sin 20^{o} = cos 70^{o} |

cos 41^{o} = sin 49^{o} |

cos 72^{o} = sin 18^{o} |

sin 36^{o} = cos 54^{o} |

When the angle is 30^{o} on a right-angled triangle, then the side opposite the 30^{o} angle is half the hypotenuse. This is also true for similar triangles. Sides H, A and O can be any value and O = 1 / 2 H. Therefore, sin 30^{o} = 0.5.

In our next class, we will be talking about** Matrices and Determinant.** We hope you enjoyed the class.

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David OladapoThis is of great help.

Thank you

But exercises to test one’s intellect would have been better.