For triangle ABC, show that: sin(A+B)/2=cosC/2
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Step-by-step explanation:
In Triangle
A+,B+C=180
divide by2
A/2+B/2+C/2=90
A/2+B/2=90-C/2
sin(A+B/2)=sin(90-c/2)
sin (A+B)/2=cos C/2
Now since
A+B+C=180
(A+B+C)/2=90°
(A+B)/2=90°−C/2
sin(A+B)/2=sin(90°−C/2). . . Taking sines on both sides
sin(A+B)/2=cos(C/2)
Proof done
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