QUESTION

# If A + B + C + D = 2π, then prove the trigonometric equation:sin A – sin B + sin C – sin D = $- 4{\text{ cos}}\dfrac{{{\text{A + B}}}}{2}{\text{ sin}}\dfrac{{{\text{A + C}}}}{2}{\text{ cos}}\dfrac{{{\text{A + D}}}}{2}$A. TrueB. False

Hint: In order to solve the equation we start by solving the LHS and make it equal to the RHS. We use the identity of sinA ± sinB and use the given, A + B + C + D = 2π to simplify the equation further.

Given Data, A + B + C + D = 2π
$\Rightarrow \dfrac{{\text{A}}}{2} + \dfrac{{\text{B}}}{2} + \dfrac{{\text{C}}}{2} + \dfrac{{\text{D}}}{2} = \pi \\ \Rightarrow \dfrac{{{\text{A + B}}}}{2} = \pi - \dfrac{{{\text{C + D}}}}{2} \\$

Now, LHS = sin A – sin B + sin C – sin D
We know the formula for sinx – siny =$2\cos \left( {\dfrac{{{\text{x + y}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{x - y}}}}{2}} \right)$, using this in the equation, we get
${\text{ = 2cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{A - B}}}}{2}} \right) + {\text{2cos}}\left( {\dfrac{{{\text{C + D}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right)$
From given, A+B = 2π-(C+D), also we know cos (π-θ) = -cos θ
${\text{ = 2cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{A - B}}}}{2}} \right) + {\text{2cos}}\left( {\dfrac{{2\pi {\text{ - }}\left( {{\text{A + B}}} \right)}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right)$
${\text{ = 2cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{A - B}}}}{2}} \right){\text{ - 2cos}}\left( {\dfrac{{\left( {{\text{A + B}}} \right)}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right)$
${\text{ = 2cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right)\left[ {{\text{sin}}\left( {\dfrac{{{\text{A - B}}}}{2}} \right){\text{ - sin}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right)} \right]$
Now, ${\text{sin}}\left( {\dfrac{{{\text{A - B}}}}{2}} \right){\text{ - sin}}\left( {\dfrac{{{\text{C - D}}}}{2}} \right)$is again in the form of sin x – sin y =$2\cos \left( {\dfrac{{{\text{x + y}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{x - y}}}}{2}} \right)$, applying the formula we get
${\text{ = 2cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right)\left[ {{\text{2cos}}\left( {\dfrac{{{\text{A - B + C - D}}}}{4}} \right) \times {\text{sin}}\left( {\dfrac{{{\text{A - B - C + D}}}}{4}} \right)} \right] \\ {\text{ = 4cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{cos}}\left( {\dfrac{{{\text{A - B + C - D}}}}{4}} \right) \times {\text{sin}}\left( {\dfrac{{{\text{A - B - C + D}}}}{4}} \right) \\$
Using A + B + C + D = 2π, we convert the terms inside the function as,
${\text{ = 4cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{cos}}\left( {\dfrac{{{\text{A + C - 2}}\pi {\text{ + A + C}}}}{4}} \right) \times {\text{sin}}\left( {\dfrac{{{\text{A + D - 2}}\pi {\text{ + A + D}}}}{4}} \right) \\ = {\text{4cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{cos}}\left( {\dfrac{{{\text{A + C}}}}{2} - \dfrac{\pi }{2}} \right) \times {\text{sin}}\left( {\dfrac{{{\text{A + D}}}}{2} - \dfrac{\pi }{2}} \right) \\ = {\text{4cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{cos}}\left( -(\dfrac{\pi }{2}-{\dfrac{{{\text{A + C}}}}{2})} \right) \times {\text{sin}}\left( -( \dfrac{\pi }{2}-{\dfrac{{{\text{A + D}}}}{2})} \right) \\$
Now we know sin (-x) = -sin x ,cos (-x) = cos x ,cos (90°-x) = sin x and sin (90°-x) = cos x, then equation becomes
$= - {\text{4cos}}\left( {\dfrac{{{\text{A + B}}}}{2}} \right){\text{sin}}\left( {\dfrac{{{\text{A + C}}}}{2}} \right){\text{cos}}\left( {\dfrac{{{\text{A + D}}}}{2}} \right)$
= RHS, hence proved.
Hence the equation sin A – sin B + sin C – sin D = $- 4{\text{ cos}}\dfrac{{{\text{A + B}}}}{2}{\text{ sin}}\dfrac{{{\text{A + C}}}}{2}{\text{ cos}}\dfrac{{{\text{A + D}}}}{2}$ holds True, Option A is the correct answer.

Note: In order to solve this type of questions the key is to express the given terms in the form of a difference between π and the remaining angle. Also, we express the sine function in terms cosine function and vice versa to solve the equation. Adequate knowledge in trigonometric formulae and identities of sine and cosine functions is necessary.