Download Name: Period: Date: Trig Identities and their proofs Sometimes

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Date: ________________
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Trig Identities and their proofs
Sometimes, instead of memorizing, it is much easier to remember mathematical
information if you understand where it comes from. This packet provides proofs of all of
the trigonometric identities; you do not need to know these, but they might help you
remember the identities (and, it’s interesting to see these anyway, so enjoy!)
1)
cos( A  B)  cos A cos B  sin A sin B
(Cosine of a difference)
Proof: The diagram below is a unit circle with center O. Two central angles, A and B,
are labeled, as are points P and Q, which are on the circle. Note that the radius of
the unit circle is 1.
Since points P and Q are on the unit
circle, their coordinates can be
represented using cosine and sine.
The angle formed by segment OP and
OQ is the angle A-B
We will now use the distance formula to
find the distance from point P to point Q
(in other words, find the length of segment PQ):
d  ( x2  x1 ) 2  ( y 2  y1 ) 2
Distance formula
d 2  ( x2  x1 ) 2  ( y 2  y1 ) 2
Square both sides (makes it easier to
work with)
PQ 2  (cos A  cos B) 2  (sin A  sin B) 2
Substituting in the coordinates of P
and Q
PQ 2  cos 2 A  cos 2 B  2 cos A cos B  sin 2 A  sin 2 B  2 sin A sin B
Squaring the terms and expanding
PQ 2  cos 2 A  sin 2 A  cos 2 B  sin 2 B  2 cos A cos B  2 sin A sin B
Commutative property of addition
PQ 2  1  1  2(cos A cos B  sin A sin B)
Pythagorean Identity. And factoring
out a (-2)
Let’s examine the triangle that can be
created by drawing line segment PQ
(diagram to the right). This triangle is a
non-right triangle, and we can use the law
of cosines to get an expression for the
length of segment PQ:
a 2  b 2  c 2  2bc cos A
(The law of cosines)
PQ 2  OP 2  OQ 2  2(OP)(OQ) cos( A  B)
(Substituting in our values)
PQ 2  1  1  2 cos( A  B)
OP and OQ have lengths of 1
We now have two expressions for PQ². Let’s set them equal
1  1  2(cos A cos B  sin A sin B)  1  1  2 cos( A  B)
Subtract 2 from both sides, then divide both sides by -2:
cos A cos B  sin A sin B  cos( A  B)
2)
cos( B)  cos B
Proof: cos(0  B)  cos 0 cos B  sin 0 sin B
cos( B)  (1) cos B  (0) sin B
cos( B)  cos B
3)
sin( B)   sin B
 
 

  
Proof: sin    B     cos  B     
2 
 2 
2 

(Negative Angle Identity for Cosine)
Cosine of a difference
Calculation
Simplify
(Negative Angle Identity for Sine)
Cofunctions of complementary
angles are equal
 
 
sin( B)  cos B cos     sin B sin   
 2
 2
sin( B)  0  (sin B)(1)
sin( B)   sin B
Cosine of a difference
Calculation
Simplify
4)
cos( A  B)  cos A cos B  sin A sin B
Proof: cos( A  B)  cos A cos B  sin A sin B
cos( A  (B))  cos A cos(B)  sin A sin(B)
cos( A  B)  cos A cos(B)  sin A sin(B)
cos( A  B)  cos A cos B  sin A sin(B)
cos( A  B)  cos A cos B  sin A( sin B)
cos( A  B)  cos A cos B  sin A sin B
5)
sin( A  B)  sin A cos B  sin B cos A
(Cosine of a sum)
(Identity from #1)
(Let A be the first angle, and
let –B be the second angle)
(Substitution)
(Identity from #2)
(Identity from #3)
(Negative times positive
equals negative)
(Sine of a sum)
Proof:


Start with cos   A  B 
2





cos   A  B   cos   ( A  B) 
2

2



* cos   A  B   sin( A  B)
2

(Distribute the minus sign)
Cofunctions of comp. angles
are equal.


Start with cos   A  B 
2

 




cos   A  B   cos    A   B 
2


 2







cos   A  B   cos   A  cos B  sin   A  sin B
2

2

2



* cos   A  B  = sin A cos B  cos Asin B
2

Associative Prop. of Addition
(Identity from #1)
Cofunctions of
complementary angles are
congruent
Notice the two starred statements above. They are


both cos   A  B  , so we can set them equal to
2

each other:
sin( A  B)  sin A cos B  cos Asin B
6)
sin( A  B)  sin A cos B  sin B cos A
(Sine of a difference)
Proof:
( A  B)   A    B  
sin( A  B)  sin  A  ( B) 
(Subtraction is the same as
adding a negative)
(Take the sine of both sides)
sin( A  B)  sin Acos( B)  cos Asin( B)
sin( A  B)  sin A cos B  cos Asin( B)
(Use Identity #5)
(Use Identity #2)
sin( A  B)  sin A cos B  cos A   sin( B) 
(Use Identity #3)
sin( A  B)  sin A cos B  cos Asin( B)
(Negative times positive is
negative)
7)
(Sine of Double Angle)
sin (2A) = 2 sin A cos A
Proof:
2A = A + A
sin (2A) = sin (A+A)
sin (2A) = sin A cos A + cos A sin A
sin (2A) = sin A cos A + sin A cos A
sin (2A) = 2 sin A cos A
8)
cos (2A) = cos²A – sin²A
(Doubling something is the
same as adding it to itself)
(Take sine of both sides)
(Identity #5)
(Multiplication is
commutative, ie 5 times 3 is
the same as 3 times 5)
(If I add two of the same
thing together, it’s the same
as multiplying by 2)
(Cosine of Double Angle, #1)
Proof:
2A = A + A
cos (2A) = cos (A+A)
cos (2A) = cos A cos A – sin A sin A
cos (2A) = cos²A – sin²A
(Doubling is the same as
adding to itself)
(Take cosine of both sides)
(Use Identity #4)
(Squaring is the same as
multiplying something by
itself)
9)
cos(2A) = 2cos² -1
(Cosine of a Double Angle, #2)
Proof:
cos (2A) = cos²A – sin²A
cos (2A) = cos²A – (1-cos²A)
cos (2A) = cos² A – 1 + cos²A
cos (2A) = 2cos²A – 1
10)
cos(2A) = 1 – 2sin²A
(Use Identity #8)
(Pythagorean Identity:
sin²A+cos²A=1. So,
sin²A = 1 – cos²A.)
(Distribute the minus sign)
(Combine like terms)
(Cosine of a Double Angle, #3)
Proof:
cos (2A) = cos²A – sin²A
cos (2A) = (1 - sin²A – sin²A)
cos (2A) = 1 – 2sin²A
(Use Identity #8)
(Pythagorean Identity:
sin²A+cos²A=1. So,
cos²A = 1 – sin²A.)
(Combine like terms)
11)
sin
1
1  cos B
B
2
2
(Sine of half angle)
Proof:
cos (2A) = 1 – 2sin²A
1
Let A = B
2
1
1
cos (2( B )) = 1 – 2sin²( B )
2
2
1
cos (B) = 1 – 2sin²( B )
2
1
cos(B) – 1 = - 2sin²( B )
2
cos B  1
1

 sin ²( B)
2
2
 cos B  1
1
 sin ²( B)
2
2
1  cos B
1
 sin ²( B)
2
2
(Identity #10)
1
(Substitute B for every A)
2
1
(Subtitute B for every A)
2
(2 times ½ is equal to 1)
(Subtract 1 from both sides)
(Divide both sides by negative 2)
(Distribute the minus sign in the
numerator)
(Move the 1 in the numerator in front
of the cos B)

1  cos B
1
 sin( B)
2
2
Take the square root of both sides
12) cos
1
1  cos B
B
2
2
(Cosine of half angle)
Proof:
cos (2A) = 2cos²A – 1
1
Let A = B
2
1
1
cos(2( B ) = 2 cos²( B )-1
2
2
1
cos(B) = 2 cos²( B )-1
2
1
cos(B) + 1 = 2 cos²( B )
2
cos B  1
1
 cos ²( B)
2
2
1  cos B
1
 cos ²( B)
2
2
(Identity #9)
1
(Substitute B for every A)
2
1
(Substitute B for every A)
2
(2 times ½ equals 1)
(Add 1 to both sides)
(Divide both sides by 2)
(Addition is commutative (in other
words, x+y is the same thing as y+x)

1  cos B
1
 cos( B)
2
2
(Take the square root of both sides)