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An Insight into Trigonometric Identities and Trigonometric Equations
Elementary Trigonometric Results
Recall definition of sine, cosine and tangent of an angle.
Let P( x, y ) be a point on a circle with radius r = x 2 + y 2 in the Cartesian Plane
uuuv uuuv
As usual, ∠XOP is measured in the direction from OX to OP , taking anticlockwise
as positive.
Let θ = ∠XOP
x
sin θ y
y
Then sin θ = , cos θ = , tan θ =
=
cos θ x
r
r
Relations between trigonometric functions of angles of opposite signs
Observe the following diagram where points A and A’, B and B’ are reflection of
each other in the x-axis (think of A' to A as well as A to A' )
B(c,d)
A(a,b)
A’(a,-b)
B’(c,-d)
We can summarize the change of coordinates corresponding to change of sign of
angle (namely a reflection) as this :
⎛x⎞ ⎛ x ⎞
⎜ ⎟ → ⎜ ⎟ as θ → −θ ,for all values of θ , positive or negative
⎝ y⎠ ⎝−y⎠
We thus have the result
sin(−θ ) = − sin θ , cos(−θ ) = cos θ and tan(−θ ) = − tan θ ………..
(1)
Relations between trigonometric functions of angles by a difference of one right
angle
The following diagram shows how coordinates are changed as a result of adding one
right angle to a given one. The various cases are:
1
∠XOP is changed to ∠XOQ , ∠XOQ is changed to ∠XOR ,………
Y
Q(-b,a)
P(a,b)
X
O
R(-a,-b)
S(b,-a)
We can summarize the change of coordinates as a result of the addition of one right
angle as this:
⎛ x ⎞ ⎛−y⎞
π
⎜ ⎟ → ⎜ ⎟ as θ → θ + , where θ is in any quadrant
2
⎝ y⎠ ⎝ x ⎠
We thus have
x
y
If cos θ = and sin θ = ,
r
r
π
π
−y
x
then sin(θ + ) = = cos θ , cos(θ + ) =
= − sin θ
2
r
2
r
So,
π
π
π
sin(θ + ) = cos θ , cos(θ + ) = − sin θ and tan(θ + ) = − tan θ
2
2
2
for all values of θ ………………………………………………………… (2)
(The reader should not just take ∠XOP as an example of θ , instead, think about
∠XOQ, ∠XOR and ∠XOS as well )
With results (1) and (2), which are valid for all values of θ , we deduce all results
related to complimentary and supplementary angles.
For example,
cos(1800 − θ ) = cos(900 + 900 − θ ) = - sin(900 − θ ) = - cos( −θ ) = - cosθ
which is true for all values of θ
Formulae for addition and subtraction of angles.
2
Y
C
B
p
q
O
A
X
The above figure shows two angles p and q of any size, given by
∠AOC and ∠AOB respectively, where A, B and C are points on a circle, centre O,
radius r . (It need not be p > q as depicted)
Let OA be lying on the OX axis and so the angles are measured using OX as initial
direction.
We find C(r cos p, r sin p ) , B(r cos q, r sin q )
By using the cosine rule, we have, for all possible positions of B and C,
2(OB)(OC) cos( p − q ) = (OB) 2 + (OC) 2 − (BC) 2
So,
2r 2 cos( p − q ) = r 2 + r 2 − [(r cos p − r cos q ) 2 + (r sin p − r sin q ) 2 ]
= 2r 2 − r 2 [cos 2 p − 2 cos p cos q + cos 2 q] − r 2 [sin 2 p − 2sin p sin q + sin 2 q ]
= 2r 2 − r 2 [1 + 1 − 2 cos p cos q − 2sin p sin q ]
= 2r 2 [cos p cos q + sin p sin q]
So, cos( p − q) = cos p cos q + sin p sin q , for all values of p and q ……………… (3)
From results (1) , (2) and (3) we can establish all formulae related to additions and
subtractions of angles,
For example
cos( p + q) = cos( p − −q) = cos p cos − q + sin p sin − q = cos p cos q − sin p sin q
Trigonometric Equations
Let us investigate the three simple equations
sin θ = a, where -1 ≤ a ≤ 1
cos θ = b, where -1 ≤ b ≤ 1
tan θ = c, for all values of c
3
First note that all three equations have two solutions in any interval of length 2π , for
example, we can find two solutions in each of (0.5π , 2.5π ), (-0.9π , 1.1π ) etc
So, to begin with, it is usually convenient to obtain two in the range of
[−π , π ]
Next, we note that if α is a solution, say to sin θ = a, where -1 ≤ a ≤ 1 , then
sin α = a, sin(α + 2π ) = a, sin(α − 2π ) = a, sin(α + 2nπ ) = a, where n is any integer
So, if α , β are two solutions to sin θ = a, where -1 ≤ a ≤ 1 , in the interval [−π , π ] ,
then the complete solution set is θ = α + 2nπ , β + 2nπ , though of course, it works out
to be θ = nπ + (−1) n (α ) , which is equivalent to θ = nπ + (−1) n ( β )
(More discussions are left as an exercise)
Exercise (without answers attached)
Recall these results:
sin(−θ ) = − sin θ , cos(−θ ) = cos θ and tan(−θ ) = − tan θ ,
for all values of θ
………………………….
π
π
(1)
π
sin(θ + ) = cos θ , cos(θ + ) = − sin θ and tan(θ + ) = − tan θ
2
2
2
………………………
for all values of θ
cos( p − q ) = cos p cos q + sin p sin q , for all values of p and q ………………
(2)
(3)
1.
Testify the rule :
⎛ x ⎞ ⎛−y⎞
0
⎜ ⎟ → ⎜ ⎟ as θ → θ + 90 , where P ( x, y ) is the point on a circle of radius
y
x
⎝ ⎠ ⎝ ⎠
r , θ = ∠XOP , for the following cases
2.
By using Results (1) and (2) above, express the following in terms of
sin θ , cosθ and tanθ only
(i)
(iv)
3.
cos(1800 − θ ) (iii)
cos(1800 + θ ) (vi)
tan(1800 − θ )
tan(1800 + θ )
By using Results (1) and (2) above, express the following in terms of
sin θ , cosθ , tanθ and cotθ only
(i)
(iv)
4.
sin(1800 − θ ) (ii)
sin(1800 + θ ) (v)
sin(2700 − θ ) (ii)
sin(2700 + θ ) (v)
cos(2700 − θ ) (iii)
cos(2700 + θ ) (vi)
tan(2700 − θ )
tan(2700 + θ )
By using Results (1), (2) and (3) above, obtain the formulae
sin( p + q) = sin p cos q + cos p sin q and
sin( p − q ) = sin p cos q − cos p sin q
4
5.
α , β are two different solutions of the equation
sin θ = a, where a ≤ 1
State a relation between α and β if
(i)
α and β are in the same quadrant
(ii)
α and β are in different quadrant
(iii)
there is no information about the quadrants of α and β .
6.
α , β are two different solutions of the equation
cos θ = a, where a ≤ 1
State a relation between α and β
7.
α , β are two different solutions of the equation
tan θ = a
What is the values of α − β ?
8
In the solution for the equation cos(2 x − 700 ) = 0.5 , values of x required for
are in the interval (−1500 , 5700 )
How many possible values of x are there?
5
Exercise (with answers attached)
Recall these results:
sin(−θ ) = − sin θ , cos(−θ ) = cos θ and tan(−θ ) = − tan θ ,
………………………….
for all values of θ
π
π
(1)
π
sin(θ + ) = cos θ , cos(θ + ) = − sin θ and tan(θ + ) = − tan θ
2
2
2
for all values of θ
………………………
cos( p − q) = cos p cos q + sin p sin q , for all values of p and q ………………
(2)
(3)
1. Testify the rule :
⎛ x ⎞ ⎛−y⎞
0
⎜ ⎟ → ⎜ ⎟ as θ → θ + 90 , where P ( x, y ) is the point on a circle of radius
y
x
⎝ ⎠ ⎝ ⎠
r , θ = ∠XOP , for the following cases
(i)
(ii)
(iii)
θ = 2400
θ = −2400
θ = 5000
2. By using Results (1) and (2) above, express the following in terms of
sin θ , cosθ and tanθ only
(i)
sin(1800 − θ ) (ii)
cos(1800 − θ ) (iii)
tan(1800 − θ )
(iv)
sin(1800 + θ ) (v)
cos(1800 + θ ) (vi)
tan(1800 + θ )
Answers
sin θ
− tan θ
(i)
(ii)
− cos(θ )
(iii)
(iv)
− sin θ
(v)
− cos θ
(vi)
tan θ
3. By using Results (1) and (2) above, express the following in terms of
sin θ , cosθ , tanθ and cotθ only
(i)
(iv)
Answers:
(i)
(iv)
sin(2700 − θ ) (ii)
sin(2700 + θ ) (v)
cos(2700 − θ ) (iii)
cos(2700 + θ ) (vi)
tan(2700 − θ )
tan(2700 + θ )
− cos θ
− cos θ
− sin θ
sin θ
cot θ
− cot θ
(ii)
(v)
(iii)
(vi)
4. By using Results (1), (2) and (3) above, obtain the formulae
sin( p + q) = sin p cos q + cos p sin q and
sin( p − q) = sin p cos q − cos p sin q
Solution
π
π
π
π
π
sin( p + q ) = − cos( + p + q) = −[cos( + p ) cos q − sin( + p ) sin q ]
2
2
2
= sin p cos q + cos p sin q
π
sin( p − q ) = − cos( + p − q) = −[cos( + p ) cos − q − sin( + p ) sin − q
2
2
2
6
= sin p cos q − cos p sin q
5.
α , β are two different solutions of the equation
sin θ = a, where a ≤ 1
State a relation between α and β if
α and β are in the same quadrant
(i)
α and β are in different quadrant
(ii)
(iv)
Answers:
(i)
(ii)
(iii)
6.
β = 2nπ + α , where n ∈
β = 2nπ + π − α , where n ∈ π − α
β = π − α , 2π + α , 3π − α ,.... -π − α , -2π + α , -3π − α ,......
So, β = nπ + (−1) n α , where n ∈
α , β are two different solutions of the equation
cos θ = a, where a ≤ 1
State a relation between α and β
Answer
7.
there is no information about the quadrants of α and β .
β = 2nπ ± α , where n ∈
α , β are two different solutions of the equation
tan θ = a
What is the values of α − β ?
Answer
α − β = nπ , for n ∈
In the solution for the equation cos(2 x − 700 ) = 0.5 , values of x required for
are in the interval (−1500 , 5700 )
How many possible values of x are there?
Answer
The interval length for x is 5700 − −1500 = 7200 = 2 × 3600
So, the interval length for (2 x − 700 ) is 4 × 3600
So, there are 8 possible values for (2 x − 700 ) , and therefore also 8 possible
values for x .
8
7