Question

Prove that: Cos(A + B) + sin(A - B) = 2sin(45° + A)cos(45° + B)​

Answer 1

Answer

It will be very easy if we take the R.H.S

R.H.S

2 sin(45° + A)cos(45° + B)​

= Sin (45 + A +45 + B) + Sin ( 45 + A - 45 - B )

= Sin ( 90 +A + B ) + Sin ( A - B )

= Cos (A+B) + Sin(A+B)  [L.H.S]

L.H.S = R.H.S (Verified)

                                                                                                           

Answer 2

Answer:

Step-by-step explanation:

LHS

= cosAcosB - sinAsinB + sinAcosB - cosAsinB

RHS

= 2(sin45°cosA + cos45°sinA)(cos45cosB - sin45sinB)

= 2(√2/2cosA + √2/2sinA)(√2/2cosB - √2/2sinB)

= 2( (1/2)cosAcosB - (1/2)cosAsinB + (1/2)sinAcosB - (1/2)sinAsinB)

= cosAcosB - cosAsinB + sinAcosB - sinAsinB

= LHS