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| 1. |
If A,B,C are angles of a triangle then prove that cosA+cos(B-C)=2sinB sinC |
| Answer» We know that A + B +C = 180° (sum of 3 angles of a triangle)So A = 180-(B+C)Also cos(180-X) = -cos X .and\xa0cos(X +/- Y) =cosXcosY -/+ sinXsinY.Now cosA + cos(B-C)= cos{180-(B+C)}+cos(B-C)=--cos(B+C)+cos(B-C)=-(cosBcosC-sinBsinC)+cosBcosC+sinB sinC=-cosBcosC+sinB sinc+cosBcosC+sinB sinC=2sinBsinC.\xa0\xa0\xa0\xa0 | |