Prove 4 c o s <mrow class="MJX-TeXAtom-ORD">

Banguizb

Banguizb

Answered question

2022-07-09

Prove
4 c o s 2 ( 2 x ) 4 c o s 2 ( x ) + 3 s i n 2 ( x ) 4 c o s 2 ( 5 π 2 x ) s i n 2 2 ( x π ) = 8 c o s ( 2 x ) + 1 2 ( c o s ( 2 x ) 1 )

Answer & Explanation

thatuglygirlyu

thatuglygirlyu

Beginner2022-07-10Added 14 answers

In the numerator,
4 cos 2 ( 2 x ) 7 cos 2 ( x ) + 3
As suggested in the comments, use,
cos 2 ( x ) = 1 + cos ( 2 x ) 2
Thus
4 cos 2 ( 2 x ) 7 cos 2 ( x ) + 3 = 4 cos 2 ( 2 x ) 7 2 7 2 cos ( 2 x ) + 3 = 1 2 ( 8 cos 2 ( 2 x ) 7 cos ( 2 x ) 1 ) = 1 2 ( 8 cos ( 2 x ) + 1 ) ( cos ( 2 x ) 1 )
Dividing this with the expression for the denominator you have got gives the desired result.

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