Question

cos2a+cos2b+cos2c=1+4sinasinbsinc​

Answer 1

Answer:

cos2a + cos2b + cos2c

= cos2a + [ cos(2b) + cos(2c) ]

= cos2a + 2*cos(b + c)*cos(b - c)

= (2cos²a - 1) + 2cos(180 - a)*cos(b - c)

= 2cos²(a) - 1 - 2cos(a)*cos(b - c)

= - 1 + 2cos(a) * [cos(a) - cos(b - c)]

= - 1 + 2cos(a) [ cos {180 - (b + c) } - cos(b - c) ]

= - 1 + 2cos(a) [ - cos (b + c) - cos(b - c) ]

= - 1 - 2cos(a) [ cos (b + c) - cos(b - c) ]

= - 1 - 2cos(a) [ 2*cos(b)*cos(c) ]

= - 1 - 4 cosa cosb cosc

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