Transcript Trigonometry

Trigonometry
Triangle terminology
hypotenuse
angle
opposite side
adjacent side
Definitions
Pythagoras’ theorem
sine cosine & tangent
a2 + b2 = c2
sin2() + cos2() = 1
Further topics
Exit
Triangle terminology
Opposite side
Contents
D terminology
Adjacent side
Definitions
Hypotenuse
Exit
Opposite side: the side opposite the angle
angle
angle
opposite
Clear
Contents
D terminology
opposite
opposite
opposite
angle
angle
Adjacent side
Definitions
Hypotenuse
Exit
Adjacent side: the side beside the angle
adjacent
adjacent
Opposite side
Contents
D terminology
Clear
Definitions
angle
adjacent
angle
angle
adjacent
angle
Hypotenuse
Exit
Hypotenuse: the longest side
Opposite side
Contents
D terminology
Adjacent side
Definitions
Clear
Exit
Definitions
B
c
a
90o
A
C
b
Tangent
Contents
D terminology
Definitions
Sine
Cosine
Exit
Tangent of angle A
tan(A) =
opposite
adjacent
=
a
b
B
c
a
tan(B)
A
90o
C
b
clear
Contents
D terminology
Definitions
Sine
Cosine
Exit
Tangent of angle B
tan(B) =
opposite
adjacent
=
b
a
B
c
a
tan(A)
90o
A
C
b
clear
Contents
D terminology
Definitions
Sine
Cosine
Exit
Sine of angle A
sin(A) =
opposite
=
hypotenuse
a
c
B
c
a
sin(B)
A
90o
C
b
Tangent
Contents
D terminology
Definitions
Clear
Cosine
Exit
Sine of angle B
sin(B) =
opposite
=
hypotenuse
b
c
B
c
a
sin(A)
90o
A
C
b
Tangent
Contents
D terminology
Definitions
Clear
Cosine
Exit
Cosine of angle A
cos(A) =
adjacent
=
hypotenuse
b
c
B
c
a
cos(B)
A
90o
C
b
Tangent
Contents
D terminology
Definitions
Sine
Clear
Exit
Cosine of angle B
cos(B) =
adjacent
=
hypotenuse
a
c
B
c
a
cos(A)
90o
A
C
b
Tangent
Contents
D terminology
Definitions
Sine
Clear
Exit
Trigonometry: further topics
Sine rule
Cosine rule
Area of triangle
Identity
Contents
Further topics
a
sin(A)
=
b
sin(B)
=
c
sin(C)
b2 + c2 - a2
cos(A) =
2bc
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
Exit
Sine Rule:
sin(A)
a
sin(B)
b
=
A&B
=
sin(C)
c
B&C
B
c
a
A
C
b
Contents
Further topics
back
Clear
Exit
Sine Rule:
sin(A)
a
=
sin(B)
b
=
sin(C)
c
B
sin(A) =
c
a
h
A
b
Contents
Further topics
back
C
Clear
Exit
Sine Rule:
sin(A)
a
=
sin(B)
b
=
sin(C)
c
B
h
sin(A) =
b
c
sin(A)
=>
=
a
a
h
A
b
Contents
Further topics
back
C
Clear
Exit
Sine Rule:
h
sin(A) =
b
sin(A)
a
=
sin(B)
b
=
sin(C)
c
B
sin(B) =
c
sin(A)
h
=>
=
a
ba
a
h
A
b
Contents
Further topics
back
C
Clear
Exit
Sine Rule:
h
sin(A) =
b
sin(A)
h
=>
=
a
ba
sin(A)
a
=
sin(B)
b
=
sin(C)
c
B
h
sin(B) =
a
sin(B)
=>
=
b
c
a
h
A
b
Contents
Further topics
back
C
Clear
Exit
Sine Rule:
h
sin(A) =
b
sin(A)
h
=>
=
a
ba
sin(A)
a
sin(B)
b
=
h
sin(B) =
a
sin(B)
h
=>
=
b
ab
=
sin(C)
c
B
c
a
h
A
=>
b
Contents
Further topics
back
C
Clear
Exit
Sine Rule:
h
sin(A) =
b
Contents
sin(B)
b
=
h
sin(B) =
a
sin(B)
h
=>
=
b
ab
sin(A)
h
=>
=
a
ba
sin(A)
=>
a
sin(A)
a
=
Further topics
sin(B)
b
=
sin(C)
c
B
c
a
h
A
b
back
C
Clear
Exit
Sine Rule:
sin(A)
a
=
sin(B)
b
=
sin(C)
c
B
sin(B) =
c
a
h
A
Contents
Further topics
C
b
back
Clear
Exit
Sine Rule:
sin(A)
a
=
sin(B)
b
=
sin(C)
c
B
h
sin(B) =
c
c
sin(B)
=>
=
b
h
A
Contents
Further topics
a
C
b
back
Clear
Exit
Sine Rule:
h
sin(B) =
c
sin(A)
a
=
sin(C)
c
B
c
a
h
A
Further topics
sin(B)
b
sin(C) =
sin(B)
h
=>
=
b
cb
Contents
=
C
b
back
Clear
Exit
Sine Rule:
h
sin(B) =
c
sin(B)
h
=>
=
b
cb
sin(A)
a
h
sin(C) =
b
sin(C)
=>
=
c
A
Contents
=
Further topics
sin(B)
b
=
sin(C)
c
B
c
a
h
C
b
back
Clear
Exit
Sine Rule:
h
sin(B) =
c
sin(B)
h
=>
=
b
cb
=
h
sin(C) =
b
sin(C)
h
=>
=
c
bc
=>
Contents
sin(A)
a
A
Further topics
sin(B)
b
=
sin(C)
c
B
c
a
h
C
b
back
Clear
Exit
Sine Rule:
h
sin(B) =
c
Contents
=
h
sin(C) =
b
sin(C)
h
=>
=
c
bc
sin(B)
h
=>
=
b
cb
sin(B)
=>
b
sin(A)
a
=
Further topics
sin(C)
c
A
sin(B)
b
=
sin(C)
c
B
c
a
h
C
b
back
Clear
Exit
sin(P+Q) :
= sin(P) cos(Q) + sin(Q) cos(P)
Show proof
P
Q
Contents
Further topics
back
Clear
Exit
sin(P+Q) :
= sin(P) cos(Q) + sin(Q) cos(P)
B
=> sin(P+Q) =
Sine rule
p
q
P
Q
b
Contents
Further topics
back
Clear
Exit
sin(P+Q) :
=> sin(P+Q) =
= sin(P) cos(Q) + sin(Q) cos(P)
B
sin(B)
(p+q)
b
p
c
=> sin(P+Q) =
Substitute sin(B) = cos(P)
h
q
P
Q
b
;sin(B)=
Contents
Further topics
h
= cos(P)
c
back
Clear
Exit
sin(P+Q) :
=> sin(P+Q) =
=> sin(P+Q) =
=> sin(P+Q) =
= sin(P) cos(Q) + sin(Q) cos(P)
B
sin(B)
(p+q)
b
(p+q)
p
c
cos(P)
b
Expand bracket
h
q
P
Q
b
;sin(B)=
Contents
Further topics
h
= cos(P)
c
back
Clear
Exit
sin(P+Q) :
=> sin(P+Q) =
=> sin(P+Q) =
=> sin(P+Q) =
= sin(P) cos(Q) + sin(Q) cos(P)
B
sin(B)
(p+q)
b
(p+q)
p
c
cos(P)
b
p
q
cos(P) +
b
b
cos(P)
h
q
P
Q
b
=> sin(P+Q) =
Contents
Substitute:
cos(P) = h/c ; q/b =sin(Q)
Further topics
;cos(P)=
h
c
q
=sin(Q)
b
back
Clear
Exit
sin(P+Q) :
=> sin(P+Q) =
=> sin(P+Q) =
=> sin(P+Q) =
= sin(P) cos(Q) + sin(Q) cos(P)
B
sin(B)
(p+q)
b
(p+q)
p
c
cos(P)
b
p
q
cos(P) +
b
b
h
q
P
Q
cos(P)
b
=> sin(P+Q) =
p
h
x
b
c
=> sin(P+Q) =
Substitute:
p/ = sin(P) ; h/ = cos(Q)
c
b
Contents
Further topics
+ sin(Q) cos(P)
;
p
h
= sin(P) , = cos(Q)
c
b
back
Clear
Exit
sin(P+Q) :
=> sin(P+Q) =
=> sin(P+Q) =
=> sin(P+Q) =
= sin(P) cos(Q) + sin(Q) cos(P)
B
sin(B)
(p+q)
b
(p+q)
p
c
cos(P)
b
p
q
cos(P) +
b
b
h
q
P
Q
cos(P)
b
=> sin(P+Q) =
p
h
x
b
c
+ sin(Q) cos(P)
=> sin(P+Q) = sin(P)cos(Q) +sin(Q)cos(P)
Contents
Further topics
;
p
h
= sin(P) , = cos(Q)
c
b
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
sin(A) =
c
a
A
b
Contents
Further topics
C
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
=
sin(A) =
c
c
a
x
A
A
b
Contents
Further topics
C
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
A
A
b
Contents
Further topics
C
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
Contents
b
c
=
Further topics
A
b
C
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
b
x
=
=> b2 = cx
c
b
A
b
C
=>
Contents
Further topics
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
Contents
b
x
=
=> b2 = cx
c
b
A
b
C
=> a2 + b2 = cy +cx
Note:
=>
y+x=c
=> c(y + x ) = c2
Further topics
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
b
x
=
=> b2 = cx
c
b
=> a2 + b2 = cy +cx
=>
a2
+
b2
=
c2
A
b
C
Note:
y+x=c
=> c(y + x ) = c2
=>
Contents
Further topics
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
b
x
=
=> b2 = cx
c
b
A
b
C
=> a2 + b2 = cy +cx = c(y + x ) = c2
=> a2 + b2 = c2
a2 b2
c2
=> 2 + 2 = 2
c c
c
=>
Contents
Further topics
back
Clear
Exit
a2 + b2 = c2
sin2(A) +cos2(A) = 1
Pythagoras’ theorem:
B
y
a
y
sin(A) =
=
=> a2 = cy
c
a
c
a
x
cos(A) =
b
x
=
=> b2 = cx
c
b
A
b
C
=> a2 + b2 = cy +cx = c(y + x ) = c2
=> a2 + b2 = c2
a2 b2
c2
=> 2 + 2 = 2
c c
c
=> sin2(A) + cos2(A) =
Contents
Further topics
1
back
Clear
Exit
Cosine Rule:
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
B
Show proof
c
a
A
C
b
Contents
Further topics
back
Clear
Exit
Cosine Rule:
a2
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
B
= Pythagoras’ Theorem
c
A
a
h
x
b-x
b
Contents
Further topics
back
Clear
C
Exit
Cosine Rule:
a2
=
(b-x)2
+
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
B
h2
=> a2 = Expand bracket
c
A
a
h
x
b-x
b
Contents
Further topics
back
Clear
C
Exit
Cosine Rule:
a2
+
h2
=> a2 = b2 - 2bx + x2
+ h2
=> a2
=
=
(b-x)2
Substitute
h2
=
c2
-
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
B
c
x2
A
a
h
x
b-x
b
Note:
C
h2 = c2 -
x2
Contents
Further topics
back
Clear
Exit
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
Cosine Rule:
a2
+
h2
=> a2 = b2 - 2bx + x2
+ h2
=> a2
=
=
=> a2 =
(b-x)2
b2 -
2bx +
x2
+
c2
B
-
c
x2
a
h
Simplify
A
x
b-x
b
Note:
C
h2 = c2 -
x2
Contents
Further topics
back
Clear
Exit
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
Cosine Rule:
a2
+
h2
=> a2 = b2 - 2bx + x2
+ h2
=> a2
=
=
(b-x)2
b2 -
2bx +
x2
=> a2 = b2 - 2bx
+
c2
B
-
c
x2
a
h
+ c2
=> a2 = Substitute x = c cos(A)
A
x
b-x
b
C
Note: x = c cos(A)
Contents
Further topics
back
Clear
Exit
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
Cosine Rule:
a2
+
h2
=> a2 = b2 - 2bx + x2
+ h2
=> a2
=
=
(b-x)2
b2 -
2bx +
x2
=> a2 = b2 - 2bx
+
c2
Contents
-
c
x2
a
h
+ c2
=> a2 = b2 - 2bc cos(A) + c2
=> a2 =
B
A
x
b-x
b
C
Note: x = c cos(A)
Rearrange
Further topics
back
Clear
Exit
b2 + c2 - a2
cos(A) =
2bc
a2 = b2 + c2 - 2bc cos(A)
Cosine Rule:
a2
+
h2
=> a2 = b2 - 2bx + x2
+ h2
=> a2
=
=
(b-x)2
b2 -
2bx +
x2
=> a2 = b2 - 2bx
+
c2
B
-
c
x2
a
h
+ c2
=> a2 = b2 - 2bc cos(A) + c2
A
x
b-x
b
C
=> a2 = b2 + c2 - 2bc cos(A)
Contents
Further topics
back
Clear
Exit
Area of triangle:
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
B
c
A
a
b
C
Next
Contents
Further topics
back
Clear
Exit
Area of triangle:
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
B
c
A
height
a
b
C
base
Area of rectangle
Area of triangle ABC
= base x height
=
ready
Next
Contents
Further topics
back
Clear
Exit
Area of triangle:
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
B
c
A
height
a
b
C
base
Area of rectangle
Area of triangle ABC
Area of triangle
= base x height
= ½ of LEFT rect. + ½ of RIGHT rect.
= ½ base x height
ready
Next
Contents
Further topics
back
Clear
Exit
Area of triangle:
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
B
c
A
height
a
b
C
base
Area of triangle
= ½ base x height
ready
Next
Contents
Further topics
back
Clear
Exit
Area of triangle:
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
B
c
A
height
a
b
C
base
Area of triangle
= ½ base x height
Area of triangle
= ½ base x c x height
c
ready
Next
Contents
Further topics
back
Clear
Exit
Area of triangle:
Area = ½ a b sin(C)= ½ b c sin(A) = ½ c a sin(B)
B
c
A
height
a
C
b
base
Area of triangle
= ½ base x height
Area of triangle
= ½ base x c x height
Area of triangle
= ½
c
b
c
Sin(A)
complete
Contents
Further topics
back
Clear
Exit