Prove that
cos (a+b) + sin (a-b) = 2sin(45+a).cos(45+b)

Step-by-step explanation:

cos (A+B) + sin (A-B) = 2 sin (45°+A) cos (45° + B)

= 2 (sin45°cosA + cos45°sinA)(cos45°cosB - sin45°sinB)

But sin45=cos45 =(sqrt2)/2

= 2 ((sqrt2)/2 *cosA + (sqrt2)/2 *sinA)((sqrt2)/2 *cosB -(sqrt2)/2 *sinB)

= 2 ((sqrt2)/2 *(cosA + sinA))*((sqrt2)/2 *(cosB - sinB))

= 2*(sqrt2)/2 * (sqrt2)/2 * (cosA + sinA)*(cosB - sinB)

= (cosA + sinA)*(cosB - sinB)

= cosAcosB +sinAcosB -cosAsinB - sinAsinB

Regrouping:

= (cosAcosB- sinAsinB) + (sinAcosB -cosAsinB)

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

the answer for this question is 38