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Identities
• The set of real numbers for which an equation is defined is
called the domain of the equation. If an equation is true for
all values in its domain it is called an identity.
• Example 1. The equation
x 2 1
 x 1
x 1
has domain equal to all real numbers except 1.
• Example 2. The equation 1  tan 2 x  sec 2 x has domain equal to
all real numbers except those of the form 2  n . Since it is
true for all numbers in its domain, it is an identity.
• Is the equation in Example 1 an identity?
Conditional equations
• If an equation is only true for some values in its domain, it
is called a conditional equation.
• Example. The equation 2x +1 = 3 has domain equal to all
real numbers, but it is only true for x = 1. Therefore, it is a
conditional equation.
• Problem. Which equation is conditional?
(a) cos 2 x  sin 2 x  1
(b) cos 2 x  sin 2 x  1
Fundamental identities
• Reciprocal identities
1
sin u 
csc u 
tan u 
1
sec u 
cot u 
1
tan u
cos u
sin u
1
cot u
sec u
csc u
1
1
cos u 
• Quotient identities
tan u 
sin u
cot u 
cos u
cos u
sin u
• Pythagorean identities
2
2
sin u  cos u  1
2
2
1  tan u  sec u
2
2
1  cot u  csc u
Fundamental identities, continued
• Cofunction identities
 u)  cos u
cos ( 2

 u)  cot u
cot( 2

 u)  csc u
csc( 2

sin( 2
tan ( 2
sec( 2

 u)  sin u

 u )  tan u

 u)  sec u
• Even/Odd identities
sin (  u)   sin u
cos (  u)  cos u
tan( u)   tan u
csc (  u)   csc u
sec (  u)  sec u
cot (  u)   cot u
Two possible ways to describe a right triangle
4  x2
x
2 sec
2 tan

arctan(x/2 )
2
2
x is independen t variable ,
0x
 is independen t variable ,
0    2
x  2  tan  ,
4  x 2  4(1  tan 2  )  4  sec 2  2  sec 
Using fundamental identities--an example
• Use fundamental identities to verify the trig identity
2  sin t  3  sin( t)  4  cos( π2  t)  9  sin t.
• We simplify the left-side of the identity.
2  sin t  3  sin(  t)  4  cos( π2  t)
 2  sin t  3  sin t  4  cos( π2  t) since sine is odd
 2  sin t  3  sin t  4  sin t using a cofunction identity
 9  sin t using algebra.
Guidelines for verifying trigonometric identities
1.
Work with one side of the equation at a time.
2.
Look to factor an expression, add fractions, square a
binomial, or create a monomial denominator.
3.
Look to use the fundamental identities. Note which
functions are in the final expression you want. Sines and
cosines pair up well, as do secants and tangents, and
cosecants and cotangents.
4.
If the preceding guidelines do not help, then try converting
all terms to sines and cosines.
5.
Always try something.
Practice problems--verify the identity
1. tan x  cot x  1
2. sec x  cos x  1
3. (cot 2 x)(sec 2 x  1)  1
4. (1  sin x)(1  sin x)  cos 2 x
cot x
5.
 csc x  sin x
sec x
1
6. tan(sin x) 
x
1 x2
Solving trigonometric equations
• If you are given a trigonometric equation, your task is to
manipulate it using algebra and the fundamental
trigonometric identities until you can write an equation of
the form
trig function  number.
• Example. Solve 2∙cos x = 1. Clearly, this is equivalent to
cos x = 1/2. To solve for x, note that there are two
solutions in [0, 2), namely /3 and 5/3. Also, because
cos x has a period of 2, there are infinitely many other
solutions, which can be written as
π
5π
x   2nπ and x 
 2nπ
3
3
where n is an integer.
• Since the latter solution gives all possible solutions of the
equation, it is called the general solution.
Graphical approaches to solving cos x = 1/2
• The graph indicates how an infinite number of solutions
can occur.
5π
π
π
y


x
1
2

x
3
x
3


3
 2π

y  cos x
x
5π
3
 4π x  π3  2π x  5π3  2π
• Also, the unit circle shows infinitely many solutions occur.


cos π  2nπ  1
3
2


cos 5π  2nπ  1
3
2
Solving trigonometric equations--an example
• Solve (tan 3x)(tan x) = tan 3x. This equation can be
rewritten as (tan 3x)(tan x – 1) = 0. Next, set the factors to
zero. We have
(i) tan 3x  0,
(ii) tan x  1.
• To solve (i) for 3x in the interval [0, ), we have 3x = 0. In
general, we have 3x = n so that x = n/3, n an integer.
• To solve (ii) for x in the interval [0, ), we have x = /4.
In general, we have x = /4 + n, n an integer.
Solving trigonometric equations--another example
• Solve csc x + cot x = 1. First, let's convert to sines and
cosines. We obtain
1
cos x

 1  1  cos x  sin x.
sin x sin x
• Since there is no obvious way to solve 1 + cos x = sin x
directly, we will try squaring both sides in the hope that the
Pythagorean identity will result in a simplification. Of course,
we know that squaring may introduce extraneous solutions.
We have
(1  cos x) 2  sin 2 x  1  2cos x  cos 2 x  1  cos 2 x.
• After canceling 1, the latter equation becomes
2cos x  2cos 2 x  0  cos x  (1  cos x)  0.
• cos x = 0 yields /2 + 2n, n an integer, cos x = –1 yields
 + 2n, n an integer, which is extraneous since csc x and cot x
are undefined at x = .
Solving trigonometric equations--using inverse functions
• Solve sec 2 x  2 tan x  4
1  tan 2 x  2 tan x  4  0
tan 2 x  2 tan x  3  0
(tan x  3)(tan x  1)  0
x  arctan 3 and x  arctan( 1)
are two solutions in ( 2 , 2 ).
• The general solution is:
x  arctan( 3)  nπ and x  ( 4 )  nπ
Guidelines for solving trigonometric equations
1. Try to isolate the trigonometric function on one side of the
equation.
2.
Look to use standard techniques such as collecting like terms
and factoring (or use the quadratic formula).
3.
Look to use the fundamental identities.
4.
To solve equations that contain forms such as sin kx or cos kx,
first solve for kx and then divide by k.
5.
If you can't get a solution using exact values, use inverse
trigonometric functions to solve.
Practice problems--solving trigonometric equations
1.
3  csc x  2  0
2. 3  sec 2 x  4  0
3. 3  sin x  1  sin x
4. sin 2 x  3  cos 2 x
5. (sin 3x)(sin x  1)  0
6. 2  cos 2 x  cos x  1  0
7. tan 3x  1  0
8. cos 2 x  sin 2 x  1
9. tan 2 x  2 tan x  0
Application of sum and difference formulas
• Sum and Difference Formulas
sin(u  v)  sin u  cos v  cos u  sin v
sin(u  v)  sin u  cos v  cos u  sin v
cos(u  v)  cos u  cos v  sin u  sin v
cos(u  v)  cos u  cos v  sin u  sin v
tan(u  v) 
tan(u  v) 
tan u  tan v
1  tan u  tan v
tan u  tan v
1  tan u  tan v
Application of sum and difference formulas--two examples
• Find the exact value of cos 12π .
cos 12π  cos( π3  π4 )  cos π3  cos π4  sin π3  sin
 12 

2
2

3
2

2
2
2 6
4
• Verify that 2 sin( x  π4 )  sin x  cos x.
2 sin( x 
π
4
)  2 (sin x  cos π4  cos x  sin π4 )
 2 (sin x 
1
2
 sin x  cos x
 cos x 
1
2
)
π
4
Double-angle and power-reducing formulas
• Double-Angle Formulas
sin(2u)  2sin u  cos u
2
2
cos(2u)  cos u  sin u
2
 2  cos u  1
2
 1  2  sin u
tan(2u) 
2  tan u
2
1  tan u
• Power-Reducing Formulas (double-angle formulas restated)
2
sin u 
1  cos 2u
2
cos u 
2
2
tan u 
1  cos 2u
2
1  cos 2u
1  cos 2u
Using double-angle formulas to solve an equation
• Solve cos 2x + cos x = 0. First use the double-angle formula
for cos 2x.
(2  cos 2 x  1)  cos(x)  0
2  cos 2 x  cos(x)  1  0
quadratic with unknown cos x
(cos x  1)(cos x  12 )  0
factor the quadratic, divide by 2
cos x   1  x  π  2nπ
π
5π
1
cos x  2  x   2nπ , x 
 2nπ
3
3
n an integer
Using the power-reducing formulas
• Rewrite sin4x in terms of first powers of the cosines of
multiple angles.
sin 4 x  (sin 2 x) 2
algebra
 1  cos 2x 

power - reduction

2


1
 1  2cos 2x  cos 2 2x
expand
4
1
 1  cos 4x  
 1  2cos 2x  
power - reduction
 
4
2


2



1
3  4 cos 2x  cos 4x 
8
algebra
Other formulas not covered
• Half-Angle formulas, Product-to-Sum Formulas, and Sumto-Product Formulas are interesting, but will not be
covered in this course.
More practice problems--verify using all available identities
csc x
1. csc 2x 
2  cos x
2. 1  cos 10x  2  cos 2 5x
3. cos 4 x  sin 4 x  cos 2x
4. (sin x  cos x) 2  1  sin 2x
5. cos 3x  4  cos 3 x  3  cos x
6. cos π3  x   cos π3  x   cos x
More practice problems--solve for x using all available identities
1. sin 2x  cos x  0
2. cos x  (sin x)(ta n x)  2
3. sin(x  2 )  sin( x  3π2 )  1
4. tan( x  π)  2  sin(x  π)  0
5. sin(x  π2 )  cos 2 x  0
6. cos(x  π)  cos x  1  0
7. sin( x 2)  cos x  0
Solution of number 7 from previous slide
• Let u = x/2, then sin( x 2)  cos x  0 becomes sin(u)  cos 2u  0.
• Using a double angle formula, we have sin(u)  (1  2 sin 2 u)  0.
• This becomes 2 sin 2 u  sin u  1  0, which factors as
(2 sin u  1)(sin u  1)  0.
• We have two cases (i) sin u = −1/2, (ii) sin u = 1.
• (i) yields u = 7π/6, u = 11π/6, (ii) yields u = π/2.
• Thus x = 7π/3, x = 11π/3, x = π , which are all the solutions in
[0, 4π).
S u man ddiffe re n ceform u las
Double- an gleform ulas
sin(u  v)  sin u  cos v  cos u  sin v
sin(2u)  2sin u  cos u
sin(u  v)  sin u  cos v  cos u  sin v
2
cos(u  v)  cos u  cos v  sin u  sin v
2
 2  cos u  1
cos(u  v)  cos u  cos v  sin u  sin v
tan(u  v) 
tan(u  v) 
2
 1  2  sin u
tan u  tan v
1  tan u  tan v
tan(2u) 
tan u  tan v
2  tan u
2
1  tan u
1  tan u  tan v
Powe r - re ducingform ulas
2
sin u 
2
cos(2u)  cos u  sin u
1  cos 2u
2
cos u 
2
2
tan u 
1  cos 2u
2
1  cos 2u
1  cos 2u