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Double-Angle and Half-Angle
Formulas
Dr .Hayk Melikyan
Departmen of Mathematics and CS
[email protected]
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Double-Angle Identities
sin2 = 2 sin cos
cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1
tan 2 =
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2tan
1  tan 2
2
Three Forms of the Double-Angle Formula for cos2
cos 2  cos   sin 
2
2
cos 2  2 cos   1
2
cos 2  1  2 sin 
2
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Power-Reducing Formulas
1  cos 2
sin  
2
1  cos 2
2
cos  
2
1  cos 2
2
tan  
1  cos 2
2
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Example
Write an equivalent expression for sin4x that does not contain
powers of trigonometric functions greater than 1.
Solution
 1  cos 2 x  1  cos 2 x 
sin 4 x  sin 2 x sin 2 x  


2
2



1  cos 2 x 

1

2
cos
2
x

2


 1  2 cos 2 x  cos 2 x 
2


  
4
4


 



 2  4 cos 2 x  1  cos 2 x  3  3 cos 2 x 


8
8


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Half-Angle Identities
x
sin 2 = ±
1 – cos x
2
x
cos 2 = ±
1 + cos x
2
x
tan 2 = ±
1 – cos x
sin x
1 – cos x
=
=
1 + cos x 1 + cos x
sin x
x
where the sign is determined by the quadrant in which 2 lies.
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Text Example
Find the exact value of cos 112.5°.
Solution
Because 112.5°  225°/2, we use the halfangle formula for cos /2
with   225°. What sign should we use when we apply the formula? Because
112.5° lies in quadrant II, where only the sine and cosecant are positive, cos
112.5° < 0. Thus, we use the  sign in the halfangle formula.
225
cos112.5  cos
2
 2 

1 


1  cos 225
 2 


2
2

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2 2
2 2

4
2
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Half-Angle Formulas for:

1  cos 
tan 
2
sin 

sin 
tan 
2 1  cos 
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Verifying a Trigonometric Identity
Verify the identity:
x
x
tan  cot  2csc x
2
2
sin x
sin x

 2csc x
1  cos x 1  cos x
sin x(1  cos x)
sin x(1  cos x)

 2csc x
(1  cos x)(1  cos x) (1  cos x)(1  cos x)
2sin x
 2csc x
2
1  cos x
2sin x
 2csc x
2
sin x
1
2
 2csc x
sin x
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The Half-Angle Formulas for Tangent
1  cos 
 
tan   
Quadrant I or III
1  cos 
2
1  cos 
 
tan    
Quadrant II or IV
1  cos 
2
   1  cos 
tan   
in any quadrant
sin 
2
sin 
 
tan   
in any quadrant
 2  1  cos 
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Example

Verify the following identity:
(sin   cos ) 2  1  sin 2
Solution
(sin   cos  ) 2
 sin 2   2 sin  cos   cos 2 
1  cos 2 1  cos 2


 2 sin  cos 
2
2
2
  2 sin  cos   1  sin 2
2
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Product-to-Sum and Sum-to-Product Formulas
Product-to-Sum Formulas
1
sin  sin   cos(   )  cos(   )
2
1
cos  cos   cos(   )  cos(   )
2
1
sin  cos   sin(    )  sin(    )
2
1
cos  sin   sin(    )  sin(    )
2
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Example

Express the following product as a sum or difference:
cos 3x cos 2x
Solution
1
cos  cos   cos(   )  cos(   )
2
cos 3x cos 2 x
1
 cos(3x  2 x)  cos(3x  2 x)
2
1
 cos( x)  cos(5 x)
2
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Text Example
Express each of the following products as a sum or difference.
a. sin 8x sin 3x
b. sin 4x cos x
Solution
The product-to-sum formula that we are using is shown in each
of the voice balloons.
sin  sin  = 1/2 [cos( - ) - cos( + )]
a.
sin 8x sin 3x  1/2[cos (8x  3x)  cos(8x  3x)]  1/2(cos 5x  cos 11x)
b.
sin  cos  = 1/2[sin( + ) + sin( - )]
sin 4x cos x  1/2[sin (4x  x)  sin(4x  x)]  1/2(sin 5x  sin 3x)
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Evaluating the Product of a Trigonometric Expression
Determine the exact value of the expression
sin  cos  
1
sin(   )  sin(   )
2
 3    
sin   cos  
 8  8
1   3  
 3   
 sin     sin    
2  8 8 
 8 8 
1 

 sin  sin 
2 2
4
1
1 
 1 

2
2
2 1
1  2  1
 
 
2 2
2 2 
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Sum-to-Product Formulas
sin   sin   2 sin

cos

2
2


sin   sin   2 sin
cos
2
2

 
cos   cos   2 cos
cos
2
2

 
cos   cos   2 sin
sin
2
2
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Example

Express the difference as a product:
sin 4x  sin 2x
Solution
sin   sin   2 sin

cos

2
2
4x  2x
4x  2x
sin 4 x  sin 2 x  2 sin
cos
2
2
2x
6x
 2 sin
cos
 2 sin x cos 3x
2
2
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Example

Express the sum as a product:
sin x  sin 4x
Solution
sin   sin   2 sin

cos

2
2
x  4x
x  4x
sin x  sin 4 x  2 sin
cos
2
2
5x
 3x
 2 sin
cos
2
2
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Example

Verify the following identity:
sin x  sin y
x y
x y
 tan
cot
sin x  sin y
2
2
Solution
x y
x y
2 sin
cos
sin x  sin y
2
2

sin x  sin y 2 sin x  y cos x  y
2
2
x y
x y
sin
cos
x y
x y
2
2

 tan
cot
x y
x y
2
2
cos
sin
2
2
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Example
Express the following as a product and if possible find
the exact value. cos750  cos 150
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Example
Verify that the following is an identity:
sin 3x  sin x
tan x 
cos 3x  cos x
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Equations Involving a Single Trigonometric
Function
•
To solve an equation containing a single trigonometric function:
• Isolate the function on one side of the equation.
sinx = a (-1 ≤ a ≤ 1 )
cosx = a
(-1 ≤ a ≤ 1 )
tan x = a ( for any real a )
• Solve for the variable.
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Trigonometric Equations
y
y = cos x
1
y = 0.5
–4 
–2 
2
4
x
–1
cos x = 0.5 has infinitely many solutions for –  < x < 
y
y = cos x
1
0.5
2
–1
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x
cos x = 0.5 has two solutions for 0 < x < 2
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Text Example
Solve the equation: 3 sin x  2  5 sin x  1.
Solution
The equation contains a single trigonometric function, sin x.
Step 1 Isolate the function on one side of the equation. We can solve for
sin x by collecting all terms with sin x on the left side, and all the constant
terms on the right side.
3 sin x  2  5 sin x  1
3 sin x  5 sin x  2  5 sin x  5 sin x – 1
2 sin x  2  1
2 sin x  1
sin x  -1/2
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This is the given equation.
Subtract 5 sin x from both sides.
Simplify.
Add 2 to both sides.
Divide both sides by 2 and solve for sin x.
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Text Example
Solve the equation:
2 cos2 x  cos x  1  0,
0  x < 2.
Solution
The given equation is in quadratic form 2t2  t  1  0 with t 
cos x. Let us attempt to solve the equation using factoring.
2 cos2 x  cos x  1  0
This is the given equation.
(2 cos x  1)(cos x  1)  0
2 cos x  1 0
or
Factor. Notice that 2t2 + t – 1 factors as (t – 1)(2t + 1).
cos x  1 0
2 cos x  1 cos x 1 cos x  1/2
Set each factor equal to 0.
Solve for cos x.
x   x  2 x  
The solutions in the interval [0, 2) are /3, , and 5/3.
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Example

Solve the following equation:
7 cos  9  2 cos
Solution:
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7 cos  9  2 cos
9 cos  9
cos  1
   ,3 ,5
    2n
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Example

Solve the equation on the interval [0,2)

3
tan 
2
3
Solution:

3
tan 
2
3

7
 and
2 6
6

7
  and
3
3
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
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Example
Solve the equation on the interval [0,2)
cos x  2 cos x  3  0
2
Solution:
cos 2 x  2 cos x  3  0
(cos x  3)(cos x  1)  0
cos x  3  0 cos x  1  0
cos x  3 cos x  1
no solution x  0
x0
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Example

Solve the equation on the interval [0,2)
sin 2x  sin x
Solution:
sin 2 x  sin x
2 sin x cos x  sin x
2 cos x  1
1
cos x 
2
 5
x ,
3
3
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