Question

prove that cos A+cosB+cosC+cos(A+B+C)=4cos(A+B)/2 ×cos(B+C)/2×cos(C+A)/2​

Answer 1

Answer:

cos a + cos b + cos c + cos (a+b+c)

=2 cos ((a+b)/2) cos ((a-b)/2) + 2 cos ((c+a+b+c)/2) cos ((c-a-b+c)/2)

= 2cos((a+b)/2)cos((a-b)/2) + 2cos((a+b+2c)/2)cos((-a-b)/2)

= 2cos((a+b)/2)cos((a-b)/2) + 2cos((a+b+2c)/2)cos((-(a+b))/2)

Remember cos(-x) = cos x and cos a + cos b is 2cos((a+b)/2)

= 2cos((a+b)/2)cos((a-b)/2) + 2cos((a+b+2c)/2)cos((a+b)/2)

2cos((a+b)/2) is taken common

= 2 cos ((a+b)/2) { cos ((a-b)/2) + cos ((a+b+2c)/2) }

= 2 cos ((a+b)/2) { 2cos((a-b+a+b+2c)/2*2) cos(a-b-a-b-2c)/2*2) }

= 2 cos ((a+b)/2) { 2 cos ((2a+2c)/4) cos((-2b-2c)/4) }

= 4 cos ((a+b)/2) cos ((a+c)/2) cos ((b+c)/2)

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