cosA +cosB – cosC = -1 + 4cosA/2 cosB/2 sinC/2

Question

cosA +cosB – cosC = -1 + 4cosA/2 cosB/2 sinC/2

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Amara 1 month 2021-08-12T09:52:09+00:00 1 Answer 0 views 0

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    2021-08-12T09:53:47+00:00

    Step-by-step explanation:

    We have to prove that :

    cosa+cosb−cosc=4cos(a/2)cos(b/2)sin(c/2)−1

    Here, we make an assumption that a+b+c=π

    i.e, the sum of the angles a, b and c is 180°

    L.H.S=(cosa+cosb)−cosc

    =2cosa+b2cosa−b2−cosc

    =2cos(π−c2)cos(a−b2)−cosc

    =2cos(π2−c2)cos(a−b2)−cosc

    =2sinc2cos(a−b2)−(1−2sin2c2)

    =2sinc2(cosa−b2+sinc2)−1

    =2sinc2(cosa−b2+sinπ−(a+b)2)−1

    =2sinc2(cosa−b2+sin[π2−(a+b)2])−1

    =2sinc2(cosa−b2+cos[(a+b)2])−1

    =2sinc2(cos2a4+cos[(−2b)4])−1

    =2sinc2(cosa2cos[−b2])−1

    =4cos(a2)(cosb2)sin(c2)−1

    =R.H.S

    Happy math!

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