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```MATH 201 - Week 11
Ferenc Balogh
Concordia University
2008 Winter
Based on the textbook
J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson
Overview
Addition and Subtraction Formulas - Section 7.2
Formulas
Applications
A sin x + B cos x
Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3
Squares of Trigonometric Functions
Half-Angle Formulas
Product-Sum Formulas
sin(s + t) = sin s cos t + cos s sin t
cos(s + t) = cos s cos t − sin s sin t
tan(s + t) =
tan s + tan t
1 − tan s tan t
Subtraction Formulas
sin(s − t) = sin s cos t − cos s sin t
cos(s − t) = cos s cos t + sin s sin t
tan(s − t) =
tan s − tan t
1 + tan s tan t
Example. Prove the addition formula for the tangent using the
addition formulas for the sine and cosine functions.
Solution.
tan(s + t) =
=
=
=
sin(s + t)
cos(s + t)
sin s cos t + cos s sin t
cos s cos t − sin s sin t
cos s cos t tan s + tan t
cos s cos t 1 − tan s tan t
tan s + tan t
.
1 − tan s tan t
π
Example. Find the exact value of sin 5π
12 and tan 12 .
Solution. Since
sin
5π
12
5π
12
=
=
=
=
Similarly, using
π
12
=
tan
π
3
3π
12
+
2π
12
=
π
4
+
π
6
we have
π
+
4 6 π
π π
π
sin
cos
+ cos
sin
6
4
6
√4
√
1 1
1 3
3+1
√ .
√
+√
=
2 2
22
2 2
sin
−
π
4
π
we get
π
=
12
=
tan π3 − tan π4
1 + tan π3 tan π4
√
√
3−1
3−1
√
√ .
=
1+ 3·1
1+ 3
Example. Find the exact value of
sin 20◦ cos 70◦ + cos 20◦ sin 70◦ .
Solution.
sin 20◦ cos 70◦ + cos 20◦ sin 70◦ = sin(20◦ + 70◦ )
= sin 90◦ = 1.
Example. Prove the cofunction identity
π
sin
− x = cos x.
2
Solution.
LHS
= sin
π
−x
2
π
π
= sin cos x − cos sin x
2
2
= 1 · cos x − 0 · sin x
= cos x = RHS.
Example. Prove the identity
sin(x + y ) − sin(x − y ) = 2 cos x sin y .
Solution.
LHS
= sin(x + y ) − sin(x − y )
= (sin x cos y + cos x sin y ) − (sin x cos y − cos x sin y )
= sin x cos y + cos x sin y − sin x cos y + cos x sin y
= 2 cos x sin y = RHS.
Example. Write the expression
√
3
1
sin x + cos x
2
2
in terms of sine only.
Solution. Notice that
√
3
π
,
cos =
6
2
sin
π
1
= .
6
2
Therefore
√
3
1
sin x + cos x
2
2
π
π
= cos sin x + sin cos x
6
6
π
= sin x +
.
6
In general, an expression of the form
A sin x + B cos x,
where A and B are real numbers and not both of them are zero, can be
expressed in terms of sine only.
To apply the trick above, we have to find an angle φ such that
cos φ = √
A
+ B2
sin φ = √
A2
B
.
+ B2
A2
Then
A sin x + B cos x
A
B
√
sin x + √
cos x
2
2
2
A +B
A + B2
=
p
A2
=
p
A2 + B 2 (cos φ sin x + sin φ cos x)
=
p
A2 + B 2 sin(x + φ).
+
B2
A sin x + B cos x
A sin x + B cos x = k sin(x + φ),
where
k=
and
cos φ = √
p
A2 + B 2
A
,
+ B2
A2
sin φ = √
B
.
+ B2
A2
Example. Write the expression
12 sin x + 5 cos x
in terms of sine only. What are the period, amplitude and phase shift?
Solution. In this example, A = 12, B = 5 and hence
p
p
√
√
k = A2 + B 2 = 122 + 52 = 144 + 25 = 169 = 13.
12 sin x + 5 cos x
12
5
= 13
sin x +
cos x
13
13
= 13 sin(x + φ),
where
12
5
,
sin φ =
.
13
13
So the amplitude is 13, the period is 2π and the phase shift is given by
cos φ =
φ = arccos
12
≈ 22.62◦ .
13
Double-Angle Formulas
sin 2x
= 2 sin x cos x
cos 2x
= cos2 x − sin2 x
= 1 − 2 sin2 x
= 2 cos2 x − 1
tan 2x
=
2 tan x
1 − tan2 x
Proofs.
sin 2x
= sin(x + x)
= sin x cos x + cos x sin x
= 2 sin x cos x.
cos 2x
= cos(x + x)
= cos x cos x − sin x sin x
= cos2 x − sin2 x.
tan 2x
= tan(x + x)
tan x + tan x
=
1 − tan x tan x
2 tan x
=
.
1 − tan2 x
Example. Find sin 2x, cos 2x and tan 2x if
sin x = −
3
5
and x belongs to the third quadrant.
Solution. We have to find the value of cos x first:
r
r
p
9
16
4
2
cos x = ± 1 − sin x = ± 1 −
=±
=± .
25
25
5
The sign must be negative because x belongs to the third quadrant.
Therefore tan x = 43 and
3
4
24
· −
=
sin 2x = 2 sin x cos x = 2 · −
5
5
25
cos 2x
= cos2 x − sin2 x =
tan 2x
=
16
9
7
−
=
25 25
25
2 · 34
2 tan x
=
9 =
1 − tan2 x
1 − 16
6
4
7
16
=
6 16
24
·
=
.
4 7
7
Squares of Trigonometric Functions
sin2 x
=
1 − cos 2x
2
cos2 x
=
1 + cos 2x
2
tan2 x
=
1 − cos 2x
.
1 + cos 2x
These are used to lower the powers of the trigonometric functions
in simplifying certain expressions.
Example. Lower the powers of the trigonometric functions in the
following expression:
sin4 x cos2 x.
Solution.
sin4 x cos2 x
=
sin2 x
=
=
=
=
2
cos2 x
1 − cos 2x
2
2
1 + cos 2x
2
1 − 2 cos 2x + cos2 2x 1 + cos 2x
4
2
1 + cos 4x
1
1 − 2 cos 2x +
(1 + cos 2x)
8
2
1
(3 − 4 cos 2x + cos 4x) (1 + cos 2x)
16
=
1
3 − 4 cos 2x + cos 4x + 3 cos 2x − 4 cos2 2x + cos 4x cos 2x
16
=
1
(3 − cos 2x + cos 4x − 2(1 + cos 4x) + cos 4x cos 2x)
16
=
1
(1 − cos 2x − cos 4x + cos 4x cos 2x) .
16
Squares of Trigonometric Functions
u
sin
2
u
cos
2
tan
u
2
r
= ±
r
= ±
1 − cos u
2
1 + cos u
2
=
1 − cos u
sin u
=
sin u
.
1 + cos u
The signs are determined by the quadrant in which the terminal
point of u2 lies.
Example. Find the exact value of sin 9π
8 .
Solution.
s
sin
The angle
9π
8
9π
8
=
±
1 − cos 9π
4
.
2
is in the third quadrant so its sine is negative and
cos
π
9π
π
1
= cos
+ 2π = cos = √ .
4
4
4
2
Hence
sin
9π
8
s
1 − cos 9π
4
2
s
1−
= −
= −
2
√1
2
s√
=−
2−1
√
2 2
Product-to-Sum Formulas
sin u cos v
=
cos u sin v
=
cos u cos v
=
sin u sin v
=
1
sin(u + v ) + sin(u − v )
2
1
sin(u + v ) − sin(u − v )
2
1
cos(u + v ) + cos(u − v )
2
1
cos(u + v ) − cos(u − v ) .
2
Example. Prove the product-to-sum formula
cos u sin v =
1
sin(u + v ) − sin(u − v ) .
2
Solution. Using the addition formulas we get
RHS
=
=
=
=
=
1
sin(u + v ) − sin(u − v )
2
1
(sin u cos v + cos u sin v ) − (sin u cos v − cos u sin v )
2
1
sin u cos v + cos u sin v − sin u cos v + cos u sin v
2
1
2 cos u sin v
2
cos u sin v = LHS.
Example. Write sin 2x cos 3x as a sum of trigonometric functions.
Solution. Using the first product-to-sum formula with u = 2x,
v = 3x we have
1
sin 2x cos 3x =
sin(2x + 3x) + sin(2x − 3x)
2
1
=
sin 5x + sin(−x)
2
1
sin 5x − sin x .
=
2
Product-to-Sum Formulas
sin u + sin v
= 2 sin
u+v
u−v
cos
2
2
sin u − sin v
= 2 cos
u+v
u−v
sin
2
2
cos u + cos v
= 2 cos
u−v
u+v
cos
2
2
cos u − cos v
= −2 sin
u+v
u−v
sin
.
2
2
Example. Write sin 2x − sin 7x as a product of trigonometric
functions.
Solution. Using the second sum-to-product formula with u = 2x,
v = 7x we have
sin 2x − sin 7x
= 2 cos
2x − 7x
2x + 7x
sin
2
2
= 2 cos
9x
−5x
sin
2
2
= −2 cos
5x
9x
sin .
2
2
```
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