Question

2sin[(C+D)/2].cos[(C­-D)/2] is equal to

A) sinC ­- sinD B) cosC + cosD C) cosC -­ cosD D) sinC + sinD

Answer 1

Step-by-step explanation:

Step-by-step explanation:

w/k,sin(A+B)+sin(A-B)=2sinAcosB

let,A+B=C ......(i)

A-B=D......(ii)

(i)+(ii)2A=C+D

A=C+D/2

(i)-(ii)2B=C-D

B=C-D/2

so,sin(A+B)+sin(A-B)=2sinAcosB

sinC+sinD=2sin(C+D/2)cos(C-D/2)

Answer 2

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B) cosC + cosD