In ∆ ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.

Open in App

Solution

Suppose $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ ...(1)

Consider the LHS of the given equation.

$LHS = a \cos A + b \cos B + c \cos C$
$= k (sinAcos A + \sin B \cos B + \sin C \cos C) . . . (from (1)) = \frac{k}{2} (2 sinAcos A + 2 \sin B \cos B + 2 \sin C \cos C) (Multiplying and dividing by 2) = \frac{k}{2} (\sin 2 A + \sin 2 B + \sin 2 C) = \frac{k}{2} {2 \sin (A + B) \cos (A - B) + \sin 2 C} = \frac{k}{2} {2 \sin (π - C) \cos (A - B) + 2 \sin C \cos C} (∵ A + B + C = π) = k {\sin C \cos (A - B) + \sin C \cos C} = k \sin C {\cos (A - B) + \cos C} = 2 k \sin C \cos (\frac{A - B + C}{2}) \cos (\frac{A - B - C}{2}) = 2 k \sin C \cos (\frac{π}{2} - \frac{2 B}{2}) \cos (\frac{π}{2} - \frac{2 A}{2}) = 2 k \sin C \sin B \sin A = 2 a \sin B \sin C (∵ \frac{a}{\sin A} = k) = RHS$

Hence proved.

Suggest Corrections

Similar questions

△ A B C

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

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

In any $Δ A B C,$ prove that

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

a cos A + b cos B + c cos C = 4 R sin A sin B sin C

△ A B C

a cos A + b cos B + c cos C = \frac{8 Δ^{2}}{a b c}

Join BYJU'S Learning Program