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Trigonometric Ratios of Multiples of an Angle

In any Δ ABC,...

Question

In any $Δ A B C,$ prove that

$a c o s A + b c o s B + c c o s C = 2 a s i n B s i n C .$

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Solution

By the sine rule, we have

$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = k (s a y)$

$\Rightarrow a = k s i n A, b = k s i n B, a n d c = k s i n C .$

$∴ L H S = a c o s A + b c o s B + c c o s C$

= $k s i n A c o s A + k s i n B + k s i n C c o s C$

= $\frac{1}{2} (s i n 2 A + s i n 2 B + s i n 2 C)$

= $\frac{1}{2} k [2 s i n (A + B) c o s (A - B) + 2 s i n C c o s c]$

= $k s i n (A + B) c o s (A - B) + s i n C c o s C$

= $k s i n (π + C) c o s (A - B) + s i n C c o s C$

= $k s i n C [c o s (A - B) + c o s C]$

= $k s i n C [c o s (A - B) + c o s π - (A - B)]$

= $k s x i n C [c o s (A - B) - c o s (A + B)]$

= $K s i n c \times 2 s i n A s i n A$

= $2 (k s i n A) s i n B s i n C$

= $2 a s i n B s i n C = R H S .$

Hence, $a c o s A + b c o s B + c c o s C = 2 a s i n B s i n C .$

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