Question

cos3A-cos5A+cos7A-cos9A=4snAcos2Asin6A

Answer 1

cos9A + cos3A + cos7A + cos5A

{ 2 cos ( 9A + 3A / 2 ) cos ( 9A - 3A / 2 ) } + { 2 cos ( 7A+ 5A / 2) cos ( 7A - 5A / 2 ) }

( 2 cos6A cos3A ) + ( 2 cos6A cosA )

2 cos6A ( cos3A + cosA )

2 cos6A { 2 cos ( 3A + A / 2) cos ( 3A - A / 2 ) }

2 cos6A ( 2 cos2A cosA )

4 cosA cos2A cos6A = RHS.....

HOPE U GETV IT..!!!!


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Answer 2

hello users .......

we have to prove that .
Cos3A-cos5A+cos7A-cos9A = 4sinAcos2Asin6A

solution:-
we know that :
cos C - cos D = 2 sin ( c+d)/2 sin (d-c)/2
and 
sin C + sin D = 2 sin (c+d)/2 cos(c-d)/2

here,
(Cos3A-cos5A)+(cos7A-cos9A)

={ 2× sin (3A + 5A )/2× sin (5A - 3A)/2 } + { 2×sin (7A + 9A )/2× sin (9A - 7A) /2 }

= {2× sin 8A/2×sin 2A/2 }+ {2× sin 16A/2 × sin 2A/2}

= 2×sin 4A × sin A + 2× sin 8A × sin A 

= 2 sin A { sin 4A + sin 8A }

= 2 sin A × { 2 × sin (4A+ 8A)/2 ×cos (4A- 8A)/2 } 

= 2 sin A × 2 × sin 12A /2 × cos (-4A/2)

= 4 sin A ×sin 6A ×cos 2A = RHS 

here, 
A property used cos (-x°)  = cos x°

❂❂ hope it helps ❂❂