Question

Prove that
\[sin(A + B + C) = \sin A\cos B\cos C + \cos AsinB\cos C + \sin C\cos A\cos B - \sin A\sin B\sin C\]

Answer 1

Hint: Here we will assume \[A + B = x\] and \[C = y\] , we will then use the following identities \[\operatorname{Sin} (x + y) = \sin x\cos y + \cos x\sin y\]
\[\cos (x + y) = \cos x\cos y - \sin x\sin y\]
to get the desired answer.

Complete step-by-step answer:
Considering the Left hand side we get:-
\[LHS = \sin (A + B + C)\]
Let \[A + B = x\]………………(1)
 \[C = y\]…………………….(2)
Hence we get:-
\[LHS = \sin (x + y)\]
Now applying the following identity
\[\operatorname{sin} (x + y) = \sin x\cos y + \cos x\sin y\]
We get:-
\[LHS = \sin x\cos y + \cos x\sin y\]
Now putting back the values of x and y from equation 1 and equation 2 we get:-
\[LHS = \sin \left( {A + B} \right)\cos C + \cos \left( {A + B} \right)\sin C\]
Again applying the identities
\[\operatorname{sin} (x + y) = \sin x\cos y + \cos x\sin y\]
\[\cos (x + y) = \cos x\cos y - \sin x\sin y\]
 We get:-
\[LHS = \left[ {\sin A\cos B + \cos A\sin B} \right]\cos C + \left[ {\cos A\cos B - \sin A\sin B} \right]\sin C\]
Multiplying \[\cos C\] and \[\sin C\] into the brackets we get:-
\[
  LHS = \sin A\cos B\cos C + \cos AsinB\cos C + \sin C\cos A\cos B - \sin A\sin B\sin C \\
  {\text{ }} = RHS \\
 \]
Therefore, L.H.S=R.H.S
Hence proved

Note: When the question contains A+B+C then just let A+B =x and C =y and apply the standard identities and solve them further. And substitute the term as mentioned in identities. The students should apply the correct identities in order to get the desired answer.