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If A+B+C=pi, then prove the following. sin2A+sin2B-sin2C=4cosA cdot cosB cdot sinC |
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Answer» Solution :L.H.S. =sin2A+sin2B-SIN2C `2sinfrac{2A+2B}{2}cosfrac{2A-2B}{2}-sin2C` `=2SIN(A+B)COS(A-B)-2sinCcosC` `=2sinCcos(A-B)-2sinCcosC` `[becauseA+B=pi-C` or, sin(A+B)=sin(pi-C)=sinC]` `=2sinC[cos(A-B)-cosC]` `=2sinC[cos(A-B)-cos(A+B)]` `=2sinC cdot 2cosA cdot cosB` `=4cosA cosBsinC=R.H.S` |
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