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29 October, 00:08

Prove that sin^2a+sin^2b-sin^2c=2sinasinbsinc

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  1. 29 October, 00:35
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    = > (i) B + C = 180 - A

    = > (ii) A = 180 - (B + C) = B + C

    Now,

    Sin^2A + sin^2B - sin^2C

    = > sin^2A + sin (B + C) sin (B - C)

    = > sin^2A + sin (180 - A) sin (B - C)

    = > sin^2A + sinA sin (B - C)

    = > sinA (sinA + sin (B - C))

    = > sinA (sin (B + C) + sin (B - C)) [ from (ii) ]

    We know that sin (A + B) + sin (A - B) = 2 sinAcosB

    = > sinA (2sinB cosc)

    = > 2sinAsinBcosC.
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