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Question Number 86540 by jagoll last updated on 29/Mar/20
prove that cos ((A/2))+cos ((B/2))+cos ((C/2)) = 4 cos (((π+A)/4))cos (((π+B)/4))cos (((π−C)/4)) where A+B+C = π
$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{B}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\:=\: \\ $$$$\mathrm{4}\:\mathrm{cos}\:\left(\frac{\pi+\mathrm{A}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi+\mathrm{B}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{where}\:\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\:\pi \\ $$
Commented by jagoll last updated on 29/Mar/20
RHS 2 { 2cos (((π+A)/4))cos (((π+B)/4))} cos (((π−C)/4)) 2 {cos (((2π+A+B)/4))+cos (((A−B)/4))} cos (((π−C)/4)) 2 {−sin (((A+B)/4)) + cos (((A−B)/4))}sin (((π+C)/4)) 2{cos (((A−B)/4))sin (((π+C)/4))−sin (((A+B)/4))sin (((π+C)/4))}
$$\mathrm{RHS} \\ $$$$\mathrm{2}\:\left\{\:\mathrm{2cos}\:\left(\frac{\pi+\mathrm{A}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\pi+\mathrm{B}}{\mathrm{4}}\right)\right\}\:\mathrm{cos}\:\left(\frac{\pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{2}\:\left\{\mathrm{cos}\:\left(\frac{\mathrm{2}\pi+\mathrm{A}+\mathrm{B}}{\mathrm{4}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{4}}\right)\right\}\:\mathrm{cos}\:\left(\frac{\pi−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{2}\:\left\{−\mathrm{sin}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{4}}\right)\:+\:\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{4}}\right)\right\}\mathrm{sin}\:\left(\frac{\pi+\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{2}\left\{\mathrm{cos}\:\left(\frac{\mathrm{A}−\mathrm{B}}{\mathrm{4}}\right)\mathrm{sin}\:\left(\frac{\pi+\mathrm{C}}{\mathrm{4}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{A}+\mathrm{B}}{\mathrm{4}}\right)\mathrm{sin}\:\left(\frac{\pi+\mathrm{C}}{\mathrm{4}}\right)\right\} \\ $$
Commented by john santu last updated on 29/Mar/20
A+B+C = π⇒B−C = π−(A+C) LHS A = π−(B+C) (A/2) = (π/2)−(((B+C)/2)) cos ((A/2)) = sin (((B+C)/2)) ⇒cos ((B/2))+cos ((C/2)) = 2cos (((B+C)/4))cos (((B−C)/4)) sin (((B+C)/2)) = 2sin (((B+C)/4))cos (((B+C)/4)) ⇒2cos (((B+C)/4)) {sin (((B+C)/4))+cos (((B−C)/4))} 2 cos (((π−A)/4)) {sin (((π−A)/4))+cos (((π−(A+C))/4))}
$$\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\:\pi\Rightarrow\mathrm{B}−\mathrm{C}\:=\:\pi−\left(\mathrm{A}+\mathrm{C}\right) \\ $$$$\mathrm{LHS} \\ $$$$\mathrm{A}\:=\:\pi−\left(\mathrm{B}+\mathrm{C}\right) \\ $$$$\frac{\mathrm{A}}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{2}}−\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{2}}\right) \\ $$$$\mathrm{cos}\:\left(\frac{\mathrm{A}}{\mathrm{2}}\right)\:=\:\mathrm{sin}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{cos}\:\left(\frac{\mathrm{B}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{C}}{\mathrm{2}}\right)\:=\:\mathrm{2cos}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\mathrm{B}−\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\mathrm{sin}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{2}}\right)\:=\:\mathrm{2sin}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{4}}\right)\mathrm{cos}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\mathrm{2cos}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{4}}\right)\:\left\{\mathrm{sin}\:\left(\frac{\mathrm{B}+\mathrm{C}}{\mathrm{4}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{B}−\mathrm{C}}{\mathrm{4}}\right)\right\} \\ $$$$\mathrm{2}\:\mathrm{cos}\:\left(\frac{\pi−\mathrm{A}}{\mathrm{4}}\right)\:\left\{\mathrm{sin}\:\left(\frac{\pi−\mathrm{A}}{\mathrm{4}}\right)+\mathrm{cos}\:\left(\frac{\pi−\left(\mathrm{A}+\mathrm{C}\right)}{\mathrm{4}}\right)\right\} \\ $$$$ \\ $$
Commented by jagoll last updated on 29/Mar/20
sorry sir A+B+C = π
$$\mathrm{sorry}\:\mathrm{sir}\:\mathrm{A}+\mathrm{B}+\mathrm{C}\:=\:\pi \\ $$

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