a cos A + b cos B + c cos C = 2b sin A sin C

Question:

a cos + b cos B + c cos C = 2sin sin C

Solution:

By sine rule, we know that

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ (say)

$\Rightarrow a=k \sin A, b=k \sin B, c=k \sin C$

Now,

LHS $=a \cos A+b \cos B+c \cos C$

$=k \sin A \cos A+k \sin B \cos B+k \sin C \cos C$

$=\frac{k}{2}(2 \sin A \cos A+2 \sin B \cos B+2 \sin C \cos C)$

$=\frac{k}{2}(\sin 2 A+\sin 2 B+2 \sin C \cos C)$

$=\frac{k}{2}\left(2 \sin \frac{2 A+2 B}{2} \cos \frac{2 A-2 B}{2}+2 \sin C \cos C\right)$

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

$=\frac{k}{2}(2 \sin (\pi-\mathrm{C}) \cos (A-B)+2 \sin C \cos C)$   $(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

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

$=\frac{k}{2} \times 2 \sin C(\cos (A-B)+\cos C)$

$=k \sin C\left(2 \cos \left(\frac{A-B+C}{2}\right) \cos \left(\frac{A-B-C}{2}\right)\right)$

$=k \sin C\left(2 \cos \left(\frac{\pi-B-B}{2}\right) \cos \left(\frac{B+C-A}{2}\right)\right)$ $(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

$=k \sin C\left(2 \cos \left(\frac{\pi-2 B}{2}\right) \cos \left(\frac{\pi-2 A}{2}\right)\right)$  $(\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi)$

$=k \sin C\left(2 \cos \left(\frac{\pi}{2}-B\right) \cos \left(\frac{\pi}{2}-A\right)\right)$

$=2 k \sin C(\sin B \sin A)$

$=2(k \sin B) \sin A \sin C$

$=2 b \sin A \sin C$

$=\mathrm{RHS}$

$\therefore$ LHS $=$ RHS

Hence, a cos + b cos B + c cos C = 2sin sin C.

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