sin (A + 3B)+sin (3A + B) 4 -sin 2A +sin 2B(a) 2 cos (A + B) (b) 2 sin

1.	sin (A + 3B)+sin (3A + B) 4 -sin 2A +sin 2B(a) 2 cos (A + B) (b) 2 sin (A-B)(c) 2 sin (A + B) (d) 2 cos (A-B)
Answer» Answer:a)2cos (A+B) Explanation :{sin(A +3B) + sin(3A +B)}/{sin2A +sin2B} use formula,sinA + sinB = 2sin(A+B)/2.cos(A-B)/2 sin2A = 2sinA.cosA cos(-a) = cosa now, {2sin(A + 3B + 3A + B)/2.cos(A +3B-3A-B)/2 }/{2sin(A + B).cos(A -B)} => sin(2A +2B).cos(B - A)/sin(A + B).cos(A -B) =>2sin(A +B).cos(A + B).cos(B -A)/sin(A + B).cos(A - B) => 2cos( A + B) option a is the right answer

sin (A + 3B)+sin (3A + B) 4 -sin 2A +sin 2B(a) 2 cos (A + B) (b) 2 sin (A-B)(c) 2 sin (A + B) (d) 2 cos (A-B)

Answer»

Answer:a)2cos (A+B)

Explanation :{sin(A +3B) + sin(3A +B)}/{sin2A +sin2B}

use formula,sinA + sinB = 2sin(A+B)/2.cos(A-B)/2 sin2A = 2sinA.cosA cos(-a) = cosa

now,

{2sin(A + 3B + 3A + B)/2.cos(A +3B-3A-B)/2 }/{2sin(A + B).cos(A -B)}

=> sin(2A +2B).cos(B - A)/sin(A + B).cos(A -B)

=>2sin(A +B).cos(A + B).cos(B -A)/sin(A + B).cos(A - B)

=> 2cos( A + B)

option a is the right answer