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

sin(45+A)=sin45cosA+cos45sinA=(cosA+sinA)/√2; similarly cos(45+B)=(cosB-sinB)/√2.

Let A = 2sin(45+A)cos(45+B) = (cosA+sinA)(cosB-sinB) = cosAcosB-cosAsinB+sinAcosB-sinAsinB.

cos(A+B)=cosAcosB-sinAsinB; sin(A-B)=sinAcosB-cosAsinB.

So A=(cosAcosB-sinAsinB)+(sinAcosB-cosAsinB)=cos(A+B)+sin(A-B) QED.

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To prove cos(A+B)+sin(A-B)=2sin(45°+A).cos(45°+B) LHS: cos(A+B)+sin(A-B) =(cosAcosB-sinAsinB)+(sinAcosB-cosAsinB) =cosAcosB-sinAsinB+sinAcosB-cosAsinB Now take RHS =2sin(45°+A).cos(45°+B) Using Identity of sin(A+B) and cos(A+B) =2( (sin45°.cosA + cos45°.cosA).(cos45°.cosA - sin45°sinA) ) Now put value of cos 45°=1/√2 and sin 45°=1/√2 By putting values we get =2( (1/√2×cos A+1/√2×sinA).(1/√2cosB-1/√2sinB) ) Now take LCM. 2 [{(cosA+sinA)÷√2 }.{(cosB-sinB)÷√2}]. Now multiple the terms inside the bracket =2[(cosAcosB+sinAcosB-sinAcosB-sinAsinB)÷(√2×√2)] = 2{(cosAcosB-sinAsinB+sinAcosB-cosAsinB)÷(2)} Now 2 get cancelled with the 2 in the denominator =cosAcosB-sinAsinB+sinAcosB-cosAsinB Now LHS=RHS Hp
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