prove that sin2A+sin2B+sin2C=4sinAsinBsinc

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Prove. Sin2A+Sin2B+Sin2C=4SinASinBSinc Prove. Sin2A+Sin2B+Sin2C=4SinASinBSinc
If A + B + C = 180, Prove that sin2A + sin2B + sin2C = 4 sinA sinB sinC If A + B + C = 180, Prove that sin2A + sin2B + sin2C = 4 sinA sinB sinC
If A+B+C=pi then prove that : sin2A+sin2B+sin2C=4sinAsinBsinC If A+B+C=pi then prove that : sin2A+sin2B+sin2C=4sinAsinBsinC
Proof related to triangles.**sin2A+sin2B+sin2c=4sinAsinBsinC Proof related to triangles.**sin2A+sin2B+sin2c=4sinAsinBsinC
If A+B+C = 180° then prove the following identity:: Sin 2A + Sin2B+ + Sin2C =4SinASinBSinC If A+B+C = 180° then prove the following identity:: Sin 2A + Sin2B+ + Sin2C =4SinASinBSinC
If `A+B+C =pi ,` prove that `sin 2A+sin 2B+sin 2C=4 sin Asin B sin C.` If `A+B+C =pi ,` prove that `sin 2A+sin 2B+sin 2C=4 sin Asin B sin C.`
Theorem 1 :If `A+B+C=pi` then prove that `sin2A+sin2B+sin2C=4sinAsinBsinC` Theorem 1 :If `A+B+C=pi` then prove that `sin2A+sin2B+sin2C=4sinAsinBsinC`
If A+B+C=180 then show that sin2A+sin2B+sin2C=4sinAsinBsinC If A+B+C=180 then show that sin2A+sin2B+sin2C=4sinAsinBsinC