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An Introduction to Trigonometry
Slideshow 44, Mathematics Mr. Richard Sasaki, Room 307
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Objectives Learn and recall some components of the right-angled triangle Understand the meaning of sine, cosine and tangent for triangles Understand the graphs π¦= π ππ π₯ , π¦= πππ π₯ , π¦=π‘ππβ‘(π₯)
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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 Angle (Theta) Adjacent
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Special Case Triangles
We saw two basic cases, the triangle and the triangle. 2 2 1 1 30 π 45 π 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
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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 π ππ, πππ and π‘ππ 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 You will need to remember these links. Adjacent
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Sine, Cosine and Tangent
In fact, sine, cosine and tangent are functions on angles which equates to the ratio of the corresponding two edges. πππππ ππ‘π π»π¦πππ‘πππ’π π π ππ π = Hypotenuse π΄πππππππ‘ π»π¦πππ‘πππ’π π Opposite πππ π = π Adjacent πππππ ππ‘π π΄πππππππ‘ π‘ππ π =
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hypotenuse opposite adjacent β As π‘ππ π = πππππ ππ‘π π΄πππππππ‘ , neither are the hypotenuse so there is no restriction between their sizes. 1 2 3 3 3 2 2 2 1 2 2
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Trigonometric Functions
Using the same methods, we can calculate trigonometric values for 60 π . 3 2 πππππ ππ‘π π»π¦πππ‘πππ’π π = 2 π ππ 60 = 60 π 1 30 π π΄πππππππ‘ π»π¦πππ‘πππ’π π = 1 2 3 πππ 60 = πππππ ππ‘π π΄πππππππ‘ = π‘ππ 60 = 3 With a little more imagination, we can do the same for 0 π and 90 π .
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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: β1 β€ π ππ π β€ 1 β1 β€ πππ π β€ 1 ββ < π‘ππ π < β This means the π¦ β axes corresponding to these graphs must obey these rules. What do these graphs look like?
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π¦=π π₯ =sinβ‘(π₯) We know the graph must satisfy β1β€π¦β€1. 2 2
2 2 We saw that sin 30 = , sin 45 = and sin 60 = This isnβt enough data to draw it, but it looks like this: 1 2 3 2 π(π₯) 1 90 180 360 540 720 900 β1 Note: We are using degrees, not radians.
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π¦=π π₯ =cosβ‘(π₯) Again, the graph satisfies β1β€π¦β€1. 3 2 2 2
3 2 2 2 We saw that cos 30 = , cos 45 = and cos 60 = Itβs the opposite, right? Basically itβs like sinβ‘(π₯) but shifted back 90 π : 1 2 π(π₯) 1 90 180 360 540 720 900 β1
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π¦=π π₯ =π‘ππ(π₯) This graph has no boundaries about π¦. 3 3
3 3 What is happening? tan 30 = , tan 45 = and tan 60 = 1 3 The sizes are increasing. There is a cycle howeverβ¦like this: π(π₯) 4 90 180 360 540 720 900 β4 Note: tanβ‘(90) tends to infinity.
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360 sinβ‘(360π+π₯) 360 cosβ‘(360π+π₯) 180 tanβ‘(180π+π₯) π=5 π=70 π=50 π=75 π=7 π=75 1 2 1 2 1 1
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