Question

If A, B , C are angles of a triangle, then \[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right)\] is equal to
(A) $4\sin \left( A \right)\cos \left( B \right)\cos \left( C \right)$
(B) $4\cos \left( A \right)$
(C) $4\sin \left( A \right)\cos \left( A \right)$
(D) $4\cos \left( A \right)\cos \left( B \right)\sin \left( C \right)$

Answer 1

Hint: As we know, the sum of all angles of a triangle is 180 degrees. So, we use this to solve the question. Also, we uses the direct result of \[\sin \left( X \right) + \sin \left( Y \right) = 2\sin \left( {\dfrac{{X + Y}}{2}} \right)\cos \left( {\dfrac{{X - Y}}{2}} \right)\] and double angle formula of sin, that is, \[\sin \left( {2X} \right) = 2\sin \left( X \right)\cos \left( X \right)\] and also the result of property \[\cos \left( {X - Y} \right) + \cos \left( {X - Y} \right) = 2\cos \left( X \right)\cos \left( Y \right)\].

Complete step-by-step answer:
In this question, we are given that A, B, C are angles of a triangle. We know that the sum of three angles of a triangle is \[\pi \].
This implies that,
\[A + B + C = \pi \]
Multiplying both sides by 2, we get,
\[2A + 2B + 2C = 2\pi \]
\[2C = 2\pi - \left( {2A + 2B} \right)\]_ _ _ _ _ _ _ _ _ _ _(i)
We have to find the value of \[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right)\]. We solve this equation by first applying \[\sin (X) + \sin \left( Y \right)\] formula, that is, \[\sin \left( X \right) + \sin \left( Y \right) = 2\sin \left( {\dfrac{{X + Y}}{2}} \right)\cos \left( {\dfrac{{X - Y}}{2}} \right)\] in \[\sin \left( {2A} \right) + \sin \left( {2B} \right)\], so we get,
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) - \sin \left( {2C} \right)\]
Substituting value of 2C from equation (i)
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) - \sin \left( {2\pi - \left( {2A + 2B} \right)} \right)\]
We know that, \[\sin \left( {2\pi - \theta } \right) = - \sin \left( \theta \right)\], so in the equation we will change \[\sin \left( {2\pi - \left( {2A + 2B} \right)} \right) = - \sin \left( {2A + 2B} \right)\]
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) - \left( { - \sin \left( {2A + 2B} \right)} \right)\]
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right) + \;\sin \left( {2A + 2B} \right)\]
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + \;\sin \left( {2\left( {A + B} \right)} \right)\]
We know that, \[\sin \left( {2X} \right) = 2\sin \left( X \right)\cos \left( X \right)\]
On applying this formula, we get,
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {A + B} \right)\cos \left( {A - B} \right) + \;2\sin \left( {A + B} \right)\cos \left( {A + B} \right)\]
Taking \[2\sin (A + B)\] as common,
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {A + B} \right)\left[ {\cos \left( {A - B} \right) + \;\cos \left( {A + B} \right)} \right]\]
Now, on directly using the following result, we get,
\[\cos \left( {X - Y} \right) + \cos \left( {X + Y} \right) = 2\cos \left( X \right)\cos \left( Y \right)\]
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 2\sin \left( {A + B} \right)\left[ {2\cos \left( A \right)\;\cos \left( B \right)} \right]\]_ _ _ _ _ _ _ _ _ _ _ _(ii)
Know, we know that, sum of three angles of a triangle is 180 degrees, that is,
\[A + B + C = \pi \]
This implies that,
\[A + B = \pi - C\]
Substituting value of (A+B) in equation (ii), we get,
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 4\sin \left( {\pi - C} \right)\cos \left( A \right)\cos \left( B \right)\]
We also know that, \[\sin \left( {\pi - \theta } \right) = \sin \left( \theta \right)\], so we get,
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 4\sin \left( C \right)\cos \left( A \right)\cos \left( B \right)\]
That can also be written as,
\[\sin \left( {2A} \right) + \sin \left( {2B} \right) - \sin \left( {2C} \right) = 4\cos \left( A \right)\cos \left( B \right)\sin \left( C \right)\]
So, the correct answer is “Option D”.

Note: The main thing to keep in mind while doing the questions is that you would have to remember trigonometric formulas by heart. You can directly use some of the results of the identities. You should be careful while checking the sign of trigonometric functions, like sine and cosecant functions are positive in 1st and 2nd quadrants, tangent and cotangent functions are positive in 1st and 3rd quadrants, while cosine and secant functions are positive in 1st and 4th quadrants. Take care while doing the calculations.