Question

यदि A+B+C = 1800 तो सिद्ध कीजिए: CosA + Cos B + cos C = 1+4sinA/2 sinB/2 sinC/2

Answer 1

$\textbf{Given:}$

$A+B+C=180^{\circ}$

$\text{consider,}$

$cosA+cosB+cosC$

$\text{Using, }\;\;\boxed{\bf\;cosC+cosD=2\;cos(\frac{C+D}{2})\;cos(\frac{C-D}{2})}$

$=2\;cos(\frac{A+B}{2})\;cos(\frac{A-B}{2})+1-2\;sin^2\frac{C}{2}$

$=2\;cos(90^{\circ}-\frac{C}{2})\;cos(\frac{A-B}{2})+1-2\;sin^2\frac{C}{2}$

$\text{using, }\;\;\boxed{\bf\;cos(90^{\circ}-A)=sinA}$

$=1+2\;sin\frac{C}{2}\;cos(\frac{A-B}{2})-2\;sin^2\frac{C}{2}$

$=1+2\;sin\frac{C}{2}[cos(\frac{A-B}{2})-sin\frac{C}{2}]$

$=1+2\;sin\frac{C}{2}[cos(\frac{A-B}{2})-sin(90^{\circ}-\frac{A+B}{2})]$

$\text{using, }\;\;\boxed{\bf\;sin(90^{\circ}-A)=cosA}$

$=1+2\;sin\frac{C}{2}[cos(\frac{A-B}{2})-cos(\frac{A+B}{2})]$

$=1+2\;sin\frac{C}{2}[cos(\frac{A}{2}-\frac{B}{2})-cos(\frac{A}{2}-\frac{B}{2})]$

$\text{Using, }\;\;\boxed{\bf\;cos(A-B)-cos(A+B)=2\;sinA\;sinB}$

$=1+2\;sin\frac{C}{2}[2\;sin\frac{A}{2}\;sin\frac{B}{2}]$

$=1+4\;sin\frac{A}{2}\;sin\frac{B}{2}\;sin\frac{C}{2}$

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If A+B+C=180, sinA-sinB+sinC=

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