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AS Core 2 Trigonometry

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Hypotenuse = x. Adjacent = x Opposite = x 3. 2 2. sin 60o = 3 cos 60o = 1 tan 60o = 3 ... Hypotenuse = x. Opposite = x Adjacent = x 3. 2 2. sin 30o = 1 cos 30o ... – PowerPoint PPT presentation

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Title: AS Core 2 Trigonometry


1
AS Core 2Trigonometry
Created by Ali V Ali
2
Measuring Angles In Degrees or Radians
?
360o
The angle, ?, can be measured in degrees. This
represents the turn required to move from one
line to the other in the direction shown.
This turn is measured in degrees. Degrees are a
unit measuring turning where 360o is a full turn.
3
Radians is another measure for angles. This time
you represent the angle as the distance point A
moves around the circumference of an imaginary
circle.
A
If we imagine a circle of radius 1 unit, then a
full turn would be a full circle and the point A
moves would be the same as the circumference of
the circle
  • 360o 2? radians (or 2? c)
  • 1o 2? c
  • 360o
  • 1 c 360o
  • 2?

4
Uses of radians
L
Here we have a sector draw with angle ?. This
sector has an arc length of L and an area of A.
r
Area,A
?
r
In degrees
In radians
Length of arc, L L (2? r) ?
360o Area of sector, A A (?r 2) ?
360o
L (2? r) ? r ? 2?
A (?r2) ? ½ r2 ? 2?
5
Hypotenuse
Opposite
Adjacent
sine ? opposite hypotenuse cosine ?
adjacent hypotenuse tangent ? opposite
adjacent
6
Some Standard Solutions
x
0o
45o
x
Hypotenuse Adjacent Opposite 0 sin 0o 0
cos 0o 1 tan 0o 0
Adjacent Opposite x Hypotenuse x?2 sin 45o
1/?2 cos 45o 1/?2 tan 45o 1
60o
30o
x
x
60o
60o
90o
x
For 60o Hypotenuse x Adjacent x Opposite x
?3 2
2 sin 60o ?3 cos 60o 1 tan 60o ?3
2 2
For 30o Hypotenuse x Opposite x Adjacent x
?3 2
2 sin 30o 1 cos 30o ?3 tan 30o
1 2 2
?3
Hypotenuse Opposite Adjacent 0 sin 90o 1
cos 90o 0 tan 90o undefined
7
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8
90o
90o
Opposite Adjacent
Opposite Adjacent -
0o
180o
180o
0o
270o
270o
90o
90o
Opposite - Adjacent -
Opposite - Adjacent
0o
180o
180o
0o
270o
270o
9


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Images from BBC AS Guru
10
CAST Diagram
11
Solve sin x 0.5 for the range 0 x 360o
? x arcsin 0.5 30o
but sin is positive in two quadrants so
x 30o or (180 30)150o
12
Solve sin x 0.5 for the range 0 x 360o
? x arcsin 0.5 30o
but sin is positive in two quadrants so
x 30o or (180 30)150o
13
The End
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