Introduction to Trigonometry 11

Some History\textbf{Some History}

In the 18001800s, when India was still a British colony, a British surveyor stationed not far away from the foothills of the Himalayas sighted an extremely high peak in the distance. When a mathematician calculated its height based on measurements taken from a survey, he realised that it was the world's highest mountain. It was then named Mount Everest, in honour of Sir George Everest, the Surveyor-General of India.

How was the mountain's height measured?

\bullet Firstly, a point PP on the ground away from the mountain was chosen.

\bullet Next, the distance dd from PP to the base of the mountain was measured as precisely as possible (this was extremely difficult to do due to hilly terrain).

\bullet The angle of elevation θ\theta of the summit from the point PP was then measured.

\bullet Using trigonometry, the mountain's height was finally determined.

What, then, is trigonometry? Basically, it is the study of the relationship between angles and lengths.

1. The Trigonometric Ratios\textbf{1. The Trigonometric Ratios}

a. Defining the trigonometric ratios at acute angles

Let θ\theta be an acute angle, i.e. 0<θ<900^{\circ} < \theta < 90^{\circ}. Draw a right-angled triangle ABC\triangle ABC with A=θ\angle A=\theta and B=90\angle B=90^{\circ}.

The sine, cosine and tangent ratios at the angle θ\theta are then defined as follows:

sinθ=Opposite sideHypothenuse=BCAC;\sin \theta = \frac{\text{Opposite side}}{\text{Hypothenuse}}=\frac{BC}{AC};

cosθ=Adjacent sideHypotenuse=ABAC;\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{AB}{AC};

tanθ=Opposite sideAdjacent side=BCAB.\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{BC}{AB}.

To remember these definitions more easily, memorise the mnemonic 'toa-cah-soh' or 'soh-cah-toa'.

b. Exercises

11. Use the concept of similar triangles to show that the definitions of the three basic trigonometric ratios at an acute angle do not depend on the size of the right-angled triangle drawn.

22. Show that, for acute θ\theta, cos(90θ)=sinθ\cos(90^{\circ}-\theta)=\sin \theta.

33. Construct ABC\triangle ABC such that B=90,AB=BC=1,AC=2\angle B=90^{\circ}, AB=BC=1, AC=\sqrt{2}. Construct a second DEF\triangle DEF such that E=90,DE=3,EF=1\angle E=90^{\circ}, DE=\sqrt{3},EF=1 and DF=2DF=2. Use the triangles to determine the following values:

sin30,sin45,sin60,\sin 30^{\circ}, \sin 45^{\circ}, \sin 60^{\circ},

cos30,cos45,cos60,\cos 30^{\circ}, \cos 45^{\circ}, \cos 60^{\circ},

tan30,tan45,tan60.\tan 30^{\circ}, \tan 45^{\circ}, \tan 60^{\circ}.

44. Construct ABC\triangle ABC such that B=90\angle B=90^{\circ} and A=θ\angle A=\theta. If tanθ=43\tan \theta = \frac{4}{3}, find the values of sinθ\sin \theta and cosθ\cos \theta.

55. Show that, for acute angle θ\theta, sin2θ+cos2θ=1\sin^{2}\theta +\cos^{2}\theta = 1.

Details and Assumptions:\textbf{Details and Assumptions:}

sin2θ=(sinθ)2,cos2θ=(cosθ)2\sin^{2}\theta = (\sin\theta)^{2}, \cos^{2}\theta=(\cos\theta)^{2}.

In the next note, Introduction to Trigonometry 22 (coming soon!), we will define the three basic trigonometric ratios at other angles.

Note by Victor Loh
6 years, 11 months ago

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@Victor Loh Explanatory note well done.

Mardokay Mosazghi - 6 years, 11 months ago

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Krishna Ar - 6 years, 11 months ago

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Prove that 1÷seca-1+1÷seca+1=2coseca×cota

Neelesh Sahu - 6 years, 10 months ago

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