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If A, B and C are interior angles of a triangle ABC, then show that `sin((B+C)/2)=cosA/2`. |
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Answer» In `deltaABC` `A+B+C=180^(@)` `rArrB+C=180^(@)-A` `rArr(B+C)/(2)=90^(@)-(A)/(2)` `rArr"sin" (B+C)/(2)=sin(90^(@)-(A)/(2))` `rArr"sin"(B+C)/(2)="cos"(A)/(2)` |
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