Question

cos4AcosA-cos6Acos9A=sin10Asin5A

Answer 1

2/2(cos4AcosA-cos6Acos9A)
1/2(2cos4acosa-2cos6acos9a)
1/2(cos5a-cos3a+cos3a-cos15a)
1/2(cos5a-cos15a)
2/2(sin10asin5a)
sin10asin5a
is proven

Answer 2

hello users .......

we have to prove that :
cos4AcosA-cos6Acos9A=sin10Asin5A

solution :-
formula used :-
cos a cos b = 1/2 [ cos (a+b) + cos (a - b) ]
&
cos C - cos D = 2 sin (C + D)/2 sin (D - C)/2

Here,
taking LHS;
cos4AcosA-cos6Acos9A

= 1/2 [ cos ( 4A + A ) + cos ( 4A - A ) ] - 1/2 [ cos ( 6A+ 9 A ) + cos ( 6A - 9 A ) ]

= 1/2 [ cos ( 5A ) + cos ( 3A ) ] - 1/2 [ cos ( 15 A ) + cos ( -3A ) ]

= 1/2 cos ( 5A ) + 1/2 cos ( 3A ) - 1/2 cos ( 15 A ) - 1/2 cos ( -3A )

= 1/2 cos ( 5A ) + 1/2 cos ( 3A ) - 1/2 cos ( 15 A ) - 1/2 cos ( 3A ) ....

(because cos (-x) = cos x )

=> 1/2 cos 5A - 1/2 cos 15A

=> 1/2 [ cos 5A - cos 15A ]

=> 1/2 × 2 × sin { ( 5A +15 A)/2} sin { (15A - 5A)/2 }

=> sin 10A sin 5A = RHS

Hence;
Proved ......


⭐✡ hope it helps ✡⭐