How would you rewrite the expressions using the power-reducing formulas in Trigonometry?

Question

In terms of the first power of the cosine.

1. cos^4(x)
2. sin^2(x)cos^2(x)

On the book, the answer to the first is:
(1/8)(3 + 4cos2x + cos4x)
and the second:
(1/8)(1 - cos4x)

Answer 1

 = cos⁴(x)

 = [ cos²x]²

 {Note : cos²x = ( 1 + cos 2x) / 2 }

 = [ ( 1/2 )( 1 + cos 2x)]²

{Note : (AB)² = A²B²}

=   ( 1/2 )²( 1 + cos 2x)²                                                    

{ Note :  (AB)² =A^{2 +} B^{2 +}  2AB }

 = ( 1/4 ) ( 1 + 2cos2x + cos²2x)

{Note : cos²x = ( 1 + cos 2x) / 2  , cos⁴x  = ( 1/2 )( 1 + cos 4x ) }

 = ( 1/4 ) [ 1 + 2cos2x +( 1/2 )( 1 + cos 4x) ]

Simplfy

 = ( 1/4 ) [ 1 + 2cos2x +( 1/2 ) + ( 1/2 )cos 4x ]

Simplfy

 = ( 1/4 ) +( 1/4 ) 2cos2x +( 1/4 )( 1/2 ) + ( 1/4 )( 1/2 )cos 4x ]

= ( 1/4 ) +( 1/4 ) 2cos2x + ( 1/8 ) + ( 1/8 ) cos 4x

 = ( 1/4 ) + ( 1/8 ) +( 1/2 ) cos2x+ ( 1/8 ) cos 4x

 = ( 3/8 ) + ( 1/2 ) cos2x+ ( 1/8 ) cos 4x                          LCM in 4 , 8 is 8

 Take out common factors.

 = ( 1/8 )[ 3 + 4cos 2x + cos 4x ]
 Ther fore

cos⁴(x) = ( 1/8 )[ 3 + 4cos 2x + cos 4x ]

Answer 2

 = cos⁴(x)

 = [ cos²x]²

 {Note : cos²x = ( 1 + cos 2x) / 2 }

 = [ ( 1/2 )( 1 + cos 2x)]²

{Note : (AB)² = A²B²}

=   ( 1/2 )²( 1 + cos 2x)²                                                    

{ Note :  (AB)² =A^{2 +} B^{2 +}  2AB }

 = ( 1/4 ) ( 1 + 2cos2x + cos²2x)

{Note : cos²x = ( 1 + cos 2x) / 2  , cos⁴x  = ( 1/2 )( 1 + cos 4x ) }

 = ( 1/4 ) [ 1 + 2cos2x +( 1/2 )( 1 + cos 4x) ]

Simplfy

 = ( 1/4 ) [ 1 + 2cos2x +( 1/2 ) + ( 1/2 )cos 4x ]

Simplfy

 = ( 1/4 ) +( 1/4 ) 2cos2x +( 1/4 )( 1/2 ) + ( 1/4 )( 1/2 )cos 4x ]

= ( 1/4 ) +( 1/4 ) 2cos2x + ( 1/8 ) + ( 1/8 ) cos 4x

 = ( 1/4 ) + ( 1/8 ) +( 1/2 ) cos2x+ ( 1/8 ) cos 4x

 = ( 3/8 ) + ( 1/2 ) cos2x+ ( 1/8 ) cos 4x                          LCM in 4 , 8 is 8

 Take out common factors.

 = ( 1/8 )[ 3 + 4cos 2x + cos 4x ]
 Ther fore

cos⁴(x) = ( 1/8 )[ 3 + 4cos 2x + cos 4x ]

Answer 3

= sin²(x)cos²(x)

= [sin(x)cos(x)]²

Divide , Multiply by 1/4

( 1/4 )4[sin(x)cos(x)]²

Simplfy

( 1/4 )[2sin(x)cos(x)]²

[Note : 2sinAcosA = sin2A]

( 1/4 )[sin2(x)]²

[Note : sin²A = ( 1- cos2A) / 2  ,  sin²2A = ( 1- cos4A) / 2 ]

( 1/4 )[( 1- cos4x) / 2 ]

Simplfy

( 1/8 )( 1 - cos4x )

There fore

sin²(x)cos²(x) = ( 1/8 )( 1 - cos4x )

Answer 4

1.

cos⁴(x) =  [ cos²x]²

Recall double angle trigonometric identities: cos²x = ( 1 + cos 2x) / 2 }

= [(1/2)( 1 + cos 2x)]²

= ( 1/4) ( 1 + cos 2x)²                                                    

= ( 1/4 ) ( 1 + 2cos2x + cos²2x)

= ( 1/4 ) [ 1 + 2cos2x +( 1/2 )( 1 + cos 4x) ]

= ( 1/4 ) [ 1 + 2cos2x +( 1/2 ) + ( 1/2 )cos 4x ]

= ( 1/4 ) +( 1/4 ) 2cos2x +( 1/4 )( 1/2 ) + ( 1/4 )( 1/2 )cos 4x ]

= ( 1/4 ) +( 1/4 ) 2cos2x + ( 1/8 ) + ( 1/8 ) cos 4x

= ( 1/4 ) + ( 1/8 ) +( 1/2 ) cos2x+ ( 1/8 ) cos 4x

Write the expression with common denominator i.e., 8

= ( 2/8 ) + ( 1/8 ) +( 4/8 ) cos2x+ ( 1/8 ) cos 4x

= ( 3/8 ) + ( 4/8 ) cos2x+ ( 1/8 ) cos 4x

Take out common factors.

= ( 1/8 )[ 3 + 4cos 2x + cos 4x ]

cos⁴(x) = ( 1/8 )[ 3 + 4cos 2x + cos 4x ]