Question

If cos (A+B+C) = cos A cos B cos C, then prove that
8sin (B+C) sin (C+A) sin (A+B) + sin 2A sin 2B sin 2C = 0

Answer 1

Hint: In order to prove that 8sin (B+C) sin (C+A) sin (A+B) + sin 2A sin 2B sin 2C = 0, we first expand the given condition cos (A+B+C) = cos A cos B cos C by using the formula of Cos (A+B) and substituting it in the equation to obtain the values of sin (B+C), sin (C+A), sin (A+B).

Complete step-by-step answer:
Given data,
cos (A+B+C) = cos A cos B cos C

We know the formula of a trigonometric term of the form, cos (x + y) = cos x cos y – sin x sin y.
Now let us apply this formula to the given term, cos (A+B+C) = cos A cos B cos C

The LHS of this equation is of the form cos (x + y), where x = A + B and y = C.
⟹ cos (A+B+C) = cos A cos B cos C
⟹ cos (A) cos (B+C) – sin (A) sin (B+C) = cos A cos B cos C
⟹ cos (A) [cos (B+C) – cos B cos C] = sin A sin (B+C) -- (1)

From the formula of cos (x + y) = cos x cos y – sin x sin y, we get
cos (x + y) - cos x cos y = – sin x sin y
Similarly, cos (B + C) - cos B cos C = – sin B sin C

Let us substitute the above in equation (1), we get
⟹ cos (A) [– sin B sin C] = sin A sin (B+C)
$\therefore {\text{sin }}\left( {{\text{B + C}}} \right) = \dfrac{{{\text{ - cos A sin B sin C}}}}{{{\text{sin A}}}}$



Similarly we get,
$\therefore {\text{sin }}\left( {{\text{C + A}}} \right) = \dfrac{{{\text{ - cos B sin C sin A}}}}{{{\text{sin B}}}}$
$\therefore {\text{sin }}\left( {{\text{A + B}}} \right) = \dfrac{{{\text{ - cos C sin A sin B}}}}{{{\text{sin C}}}}$

Now we find the value of 8sin (B+C) sin (C+A) sin (A+B) + sin 2A sin 2B sin 2C and should prove it is equal to 0.
⟹8sin (B+C) sin (C+A) sin (A+B) + sin 2A sin 2B sin 2C
We know the formula of sin 2A = 2 sin A cos A, similarly sin 2B = 2 sin B cos B and sin 2C = 2 sin C cos C.

Let us substitute all the obtained values in the above equation, we get
$8\left( {\dfrac{{{\text{ - cos A sin B sin C}}}}{{{\text{sin A}}}} \times \dfrac{{{\text{ - cos B sin C sin A}}}}{{{\text{sin B}}}} \times \dfrac{{{\text{ - cos C sin A sin B}}}}{{{\text{sin C}}}}} \right) + 2{\text{ sinA cos A}}{\text{. 2 sin B cos B}}{\text{. 2 sin C cos C}}$
$ \Rightarrow - 8{\text{ cos A cos B sin C cos C sin A sin B + 8 sin A sin B sin C cos A cos B cos C}}$
⟹0
∴8sin (B+C) sin (C+A) sin (A+B) + sin 2A sin 2B sin 2C = 0
Hence proved.

Note – In order to solve this type of problems the key is to know the formulae of trigonometric terms of the form cos (x + y) and sin 2x and other relevant formulae of this type. While solving these problems the approach followed can make a big difference especially in trigonometric problems. The approach followed can be improved by solving more problems of this type.
Here in the question the given term is not in the form cos (x + y) but we consider x = A and y = B + C and go about solving the problem.