Question

cos(Q+R-P) + cos(R+P-Q) + cos(P+Q-R) = 1 + 4cosP. cosQ. cosR when P=+Q+R = 180* where *=degrees.

Answer 1

$\textbf{Given:}$

$P+Q+R=180^{\circ}$

$\textbf{To prove:}$

$cos(Q+R-P)+cos(R+P-Q)+cos(P+Q-R)=1+4\,cos\,P\,cos\,Q\,cos\,R$

$\text{Consider,}$

$cos(Q+R-P)+cos(R+P-Q)+cos(P+Q-R)$

$\text{Using the identity,}$

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

$=2\,cos(\frac{Q+R-P+R+P-Q}{2})\,cos(\frac{Q+R-P-R-P+Q}{2})+cos(P+Q-R)$

$=2\,cos\,R\,cos(\frac{2Q-2P}{2})+cos(P+Q+R-2R)$

$=2\,cos\,R\,cos(Q-P)+cos(180^{\circ}-2R)$

$=2\,cos\,R\,cos(-(P-Q))-cos2R$

$=2\,cos\,R\,cos(P-Q)-[2cos^2R-1]$

$=2\,cos\,R\,cos(P-Q)-2cos^2R+1$

$=1+2\,cos\,R[cos(P-Q)-cosR]$

$=1+2\,cos\,R[cos(P-Q)-cos(180^{\circ}-(P+Q))]$

$=1+2\,cos\,R[cos(P-Q)+cos(P+Q)]$

$\text{Using the identity,}$

$\boxed{\bf\,cos(A+B)+cos(A-B)=2\,cos\,A\;cosB}$

$=1+2\,cos\,R[2\,cos\,P\,cos\,Q]$

$=1+4\,cos\,P\,cos\,Q\,cos\,R$

$\therefore\boxed{\bf\,cos(Q+R-P)+cos(R+P-Q)+cos(P+Q-R)=1+4\,cos\,P\,cos\,Q\,cos\,R}$

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