# Introduction to Database Systems

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```					                  Double-Angle and Half-Angle
Formulas

Dr .Hayk Melikyan
Departmen of Mathematics and CS
melikyan@nccu.edu
H.Melikyan/1200                                        1
Double-Angle Identities

sin2 = 2 sin cos

cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1
2tan
tan 2 =      1  tan 2

H.Melikyan/1200                                       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

H.Melikyan/1200                                    3
Power-Reducing Formulas

1  cos 2
sin  
2

2
1  cos 2
cos  
2

2
1  cos 2
tan  
2

1  cos 2

H.Melikyan/1200                        4
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  cos 2 x 
2         1  2 cos 2 x              

                           
 
2      
             4                               4             

                             
 2  4 cos 2 x  1  cos 2 x  3  3 cos 2 x 
                               
              8                         8

H.Melikyan/1200                                                                 5
Half-Angle Identities

x           1 – cos x
sin 2 = ±           2

x           1 + cos x
cos 2 = ±           2

x           1 – cos x     sin x    1 – cos x
tan 2 = ±       1 + cos x = 1 + cos x = sin x

x
where the sign is determined by the quadrant in which 2 lies.

H.Melikyan/1200                                                 6
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  cos225       2 
            
2            2
2 2    2 2
      
4      2

H.Melikyan/1200                                                                     7
Half-Angle Formulas for:

1  cos

tan 
2    sin 
    sin 
tan 
2 1  cos

H.Melikyan/1200             8
Example

   Verify the following identity:

(sin   cos )  1  sin 2
2

Solution

(sin   cos ) 2
 sin 2   2 sin  cos  cos2 
1  cos 2 1  cos 2
                          2 sin  cos
2            2
2
  2 sin  cos  1  sin 2
2
H.Melikyan/1200                                                  9
Product-to-Sum and Sum-to-Product Formulas

Product-to-Sum Formulas

sin  sin   cos(   )  cos(   )
1
             
2
cos cos   cos(   )  cos(   )
1
              
2
sin  cos   sin(   )  sin(   )
1
2
cos sin   sin(   )  sin(   )
1
2
H.Melikyan/1200                                10
Example
   Express the following product as a sum or difference:

cos3x cos2x
Solution

cos cos   cos(   )  cos(   )
1
             
2
cos3x cos 2 x

 cos(3x  2 x)  cos(3x  2 x)
1
2
 cos(x)  cos(5 x)
1
2

H.Melikyan/1200                                              11
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.
a.            sin  sin  = 1/2 [cos( - ) - cos( + )]

sin 8x sin 3x  1/2[cos (8x  3x)  cos(8x  3x)]  1/2(cos 5x  cos 11x)

sin  cos  = 1/2[sin( + ) + sin( - )]
b.
sin 4x cos x  1/2[sin (4x  x)  sin(4x  x)]  1/2(sin 5x  sin 3x)

H.Melikyan/1200                                                                       12
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

H.Melikyan/1200                              13
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 cos3 x
2        2

H.Melikyan/1200                                                   14
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

H.Melikyan/1200                                               15
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
        2          2  tan x  y cot x  y
x y      x y          2        2
cos        sin
2          2
H.Melikyan/1200                                             16
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.

H.Melikyan/1200                                                         17
Trigonometric Equations

y

1                           y = cos x
y = 0.5
x
–4            –2                        2          4

–1

cos x = 0.5 has infinitely many solutions for –  < x < 

y

1                      y = cos x
0.5
x
2
–1                          cos x = 0.5 has two solutions for 0 < x < 2

H.Melikyan/1200                                                                              18
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                This is the given equation.

3 sin x  5 sin x  2  5 sin x  5 sin x – 1      Subtract 5 sin x from both sides.

2 sin x  2  1                           Simplify.

2 sin x  1                        Add 2 to both sides.

sin x  -1/2                     Divide both sides by 2 and solve for sin x.

H.Melikyan/1200                                                                                      19
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          Factor. Notice that 2t2 + t – 1 factors as (t – 1)(2t + 1).

2 cos x  1 0      or    cos x  1 0    Set each factor equal to 0.

2 cos x  1 cos x 1 cos x  1/2     Solve for cos x.

x   x  2 x  
The solutions in the interval [0, 2) are /3, , and 5/3.
H.Melikyan/1200                                                                                                 20
Example
   Solve the following equation:

7 cos  9  2 cos
Solution:     7 cos  9  2 cos
9 cos  9
cos  1
   ,3 ,5
    2n
H.Melikyan/1200                             21
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
H.Melikyan/1200                                     22
Example
Solve the equation on the interval [0,2)

cos x  2 cos x  3  0
2
Solution:
cos2 x  2 cos x  3  0
(cos x  3)(cosx  1)  0
cos x  3  0 cos x  1  0
cos x  3 cos x  1
no solution x  0
x0
H.Melikyan/1200                                  23
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
H.Melikyan/1200                                   24

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