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# Introduction to Trigonometry $$1$$

$$\textbf{Some History}$$

In the $$1800$$s, 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 $$P$$ on the ground away from the mountain was chosen.

$$\bullet$$ Next, the distance $$d$$ from $$P$$ 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 $$\theta$$ of the summit from the point $$P$$ 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.

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

a. Defining the trigonometric ratios at acute angles

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

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

$\sin \theta = \frac{\text{Opposite side}}{\text{Hypothenuse}}=\frac{BC}{AC};$

$\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{AB}{AC};$

$\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{BC}{AB}.$

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

b. Exercises

$$1$$. 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.

$$2$$. Show that, for acute $$\theta$$, $$\cos(90^{\circ}-\theta)=\sin \theta$$.

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

$\sin 30^{\circ}, \sin 45^{\circ}, \sin 60^{\circ},$

$\cos 30^{\circ}, \cos 45^{\circ}, \cos 60^{\circ},$

$\tan 30^{\circ}, \tan 45^{\circ}, \tan 60^{\circ}.$

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

$$5$$. Show that, for acute angle $$\theta$$, $$\sin^{2}\theta +\cos^{2}\theta = 1$$.

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

$$\sin^{2}\theta = (\sin\theta)^{2}, \cos^{2}\theta=(\cos\theta)^{2}$$.

In the next note, Introduction to Trigonometry $$2$$ (coming soon!), we will define the three basic trigonometric ratios at other angles.

Note by Victor Loh
3 years, 5 months ago

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@Victor Loh Explanatory note well done.

- 3 years, 5 months ago

Superb!

- 3 years, 5 months ago

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

- 3 years, 4 months ago