Question

sin 2A+sin 2B+sin 2(A-B)=4 sin A.cos B. cos(A-B) prove it.​

Answer 1

Proof:

L.H.S. = sin2A + sin2B + sin{2(A - B)}

= (sin2A + sin2B) + sin{2(A- B)}

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

= 2 sin(A + B) cos(A - B) + 2 sin(A - B) cos(A - B)

= 2 cos(A - B) {sin(A + B) + sin(A - B)}

= 2 cos(A - B) * 2 sin{(A + B + A - B)/2} cos{(A + B - A + B)/2}

= 2 cos(A - B) * 2 sinA cosB

= 4 sinA cosB cos(A - B) = R.H.S.

Hence, proved.

Rules:

• sinC + sinD = 2 sin{(C + D)/2} cos{(C - D)/2}
• sinC - sinD = 2 cos{(C + D)/2} sin{(C - D)/2}
• cosC + cosD = 2 cos{(C + D)/2} cos{(C - D)/2}
• cosC - cosD = 2 sin{(C + D)/2} sin{(D - C)/2}

Answer 2

Answer:

We have to prove the given expression  ;

sin 2A+sin 2B+sin 2(A-B)=4 sin A.cos B. cos(A-B)

At, first we will take only LHS;

i.e ;

$Sin 2A + Sin2B + Sin2(A-B)$

Applying some trigonometric identity in above expression;

$2 \times [ sin (2A+2B)/2 \times cos (2A-2B)/2] + sin 2(A-B)$

$2 \times [ sin (A+B) \times cos (A-B)] + sin 2(A-B)$

$2 \times sin (A+B) \times cos (A-B) + 2 \times sin (A-B)cos (A-B)$

$2 \times cos (A-B) [ sin (A+B) + sin (A-B)]$

$2 \times cos (A-B) \times  2 sin (A+B+A-B)/2 \times cos (A+B-A+B)/2$

=  $4 \times cos (A-B) \times sin A \times cos B$ = R.H.S

∴ It is being proved that

=  sin 2A+sin 2B+sin 2(A-B)=4 sin A.cos B. cos(A-B)