Question

If A+B+C=2S then prove that cos (S-A) + cos (S-B) + cos (S-C) + cos S = 4 cos A/2 cos B/2 cos C/2

Answer 1

Given: A+B+C=2S

To Prove: cos (S-A) + cos (S-B) + cos (S-C) + cos S = 4 cos A/2 cos B/2 cos C/2

Proof:

Consider the Left hand side of the equation

cos (S-A) + cos (S-B) + cos (S-C) + cos S

=[ cos S + cos (S-A)] + [cos (S-B) + cos (S-C)]

By using the identity:

$\cos A+\cos B = 2\cos (\frac{A+B}{2})\cos (\frac{A-B}{2})$

Now,

=[ cos S + cos (S-A)] + [cos (S-B) + cos (S-C)]

= $[2\cos(\frac{S+S-A}{2})\cos (\frac{S-S+A}{2})]+[2\cos(\frac{S-B+S-C}{2})\cos (\frac{S-B-S+C}{2})]$

=$[2\cos(\frac{2S-A}{2})\cos (\frac{A}{2})]+[2\cos(\frac{2S-B-C}{2})\cos (\frac{C-B}{2})]$

=$[2\cos(\frac{2S-A}{2})\cos (\frac{A}{2})]+[2\cos(\frac{A}{2})\cos (\frac{C-B}{2})]$

Taking $2\cos \frac{A}{2}$ common

= $2\cos \frac{A}{2}[\cos(\frac{2S-A}{2})+\cos (\frac{C-B}{2})]$

= $2\cos \frac{A}{2}[\cos(\frac{B+C}{2})+\cos (\frac{C-B}{2})]$

= $2\cos \frac{A}{2}[2 \cos (\frac{B+C+C-B}{2 \times2})\cos (\frac{B+C-C+B}{2 \times2})]$

= $2\cos \frac{A}{2}[2 \cos (\frac{2C}{4})\cos (\frac{2B}{4})]$

= $4\cos \frac{A}{2} \cos (\frac{B}{2})\cos (\frac{C}{2})$

= RHS

Hence, proved.

Answer 2

Answer:

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