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An Introduction to Trigonometry

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1 An Introduction to Trigonometry
Slideshow 44, Mathematics Mr. Richard Sasaki, Room 307

2 Objectives Learn and recall some components of the right-angled triangle Understand the meaning of sine, cosine and tangent for triangles Understand the graphs 𝑦= 𝑠𝑖𝑛 π‘₯ , 𝑦= π‘π‘œπ‘  π‘₯ , 𝑦=π‘‘π‘Žπ‘›β‘(π‘₯)

3 The Right-Angled Triangle
Trigonometry is like Pythagoras but includes angles. When we have a specified angle, the vocabulary is different. Opposite Hypotenuse Simple trigonometry involves 2 edges and an angle. If one thing is missing, how do we find it? πœƒ Angle (Theta) Adjacent

4 Special Case Triangles
We saw two basic cases, the triangle and the triangle. 2 2 1 1 30 π‘œ 45 π‘œ 3 1 We need to think about the relationship between the edge lengths and the angles. Any ideas? The relationship lies with the three main trigonometric functions, , and sine cosine tangent

5 Sine, Cosine and Tangent
Sine, cosine and tangent are the relationships between edge lengths and angles. In calculation, sine, cosine and tangent are shown as 𝑠𝑖𝑛, π‘π‘œπ‘  and π‘‘π‘Žπ‘› respectively. We still usually refer to them by their full names though. Each refer to two of the edges. Cosine Tangent Sine S C T Hypotenuse Opposite O H A H O A πœƒ You will need to remember these links. Adjacent

6 Sine, Cosine and Tangent
In fact, sine, cosine and tangent are functions on angles which equates to the ratio of the corresponding two edges. π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ 𝑠𝑖𝑛 πœƒ = Hypotenuse π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ Opposite π‘π‘œπ‘  πœƒ = πœƒ Adjacent π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘Žπ‘› πœƒ =

7 hypotenuse opposite adjacent ∞ As π‘‘π‘Žπ‘› πœƒ = π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ , neither are the hypotenuse so there is no restriction between their sizes. 1 2 3 3 3 2 2 2 1 2 2

8 Trigonometric Functions
Using the same methods, we can calculate trigonometric values for 60 π‘œ . 3 2 π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = 2 𝑠𝑖𝑛 60 = 60 π‘œ 1 30 π‘œ π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ = 1 2 3 π‘π‘œπ‘  60 = π‘‚π‘π‘π‘œπ‘ π‘–π‘‘π‘’ π΄π‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ = π‘‘π‘Žπ‘› 60 = 3 With a little more imagination, we can do the same for 0 π‘œ and 90 π‘œ .

9 Boundaries We do not have time to explore further but after looking at many values inserted in the trigonometric functions, we would have the following boundaries: βˆ’1 ≀ 𝑠𝑖𝑛 πœƒ ≀ 1 βˆ’1 ≀ π‘π‘œπ‘  πœƒ ≀ 1 βˆ’βˆž < π‘‘π‘Žπ‘› πœƒ < ∞ This means the 𝑦 – axes corresponding to these graphs must obey these rules. What do these graphs look like?

10 𝑦=𝑓 π‘₯ =sin⁑(π‘₯) We know the graph must satisfy βˆ’1≀𝑦≀1. 2 2
2 2 We saw that sin 30 = , sin 45 = and sin 60 = This isn’t enough data to draw it, but it looks like this: 1 2 3 2 𝑓(π‘₯) 1 90 180 360 540 720 900 βˆ’1 Note: We are using degrees, not radians.

11 𝑦=𝑓 π‘₯ =cos⁑(π‘₯) Again, the graph satisfies βˆ’1≀𝑦≀1. 3 2 2 2
3 2 2 2 We saw that cos 30 = , cos 45 = and cos 60 = It’s the opposite, right? Basically it’s like sin⁑(π‘₯) but shifted back 90 π‘œ : 1 2 𝑓(π‘₯) 1 90 180 360 540 720 900 βˆ’1

12 𝑦=𝑓 π‘₯ =π‘‘π‘Žπ‘›(π‘₯) This graph has no boundaries about 𝑦. 3 3
3 3 What is happening? tan 30 = , tan 45 = and tan 60 = 1 3 The sizes are increasing. There is a cycle however…like this: 𝑓(π‘₯) 4 90 180 360 540 720 900 βˆ’4 Note: tan⁑(90) tends to infinity.

13 360 sin⁑(360π‘˜+π‘₯) 360 cos⁑(360π‘˜+π‘₯) 180 tan⁑(180π‘˜+π‘₯) πœƒ=5 πœƒ=70 πœƒ=50 πœƒ=75 πœƒ=7 πœƒ=75 1 2 1 2 1 1


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