Question

If \[A+B+C=\pi \] then \[{{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}C=2\sin A\cos B\sin C\].

Answer 1

Hint: Let us assume the value of \[{{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}C\] is equal to X. Let us substitute \[C=\pi -(A+B)\] in X. We know that \[\sin \left( \pi -\theta \right)=\sin \theta \]. We know that \[\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A\]. We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]. We know that \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]. We know that \[{{\cos }^{2}}B-{{\sin }^{2}}B=\cos 2B\]. By using these concepts, we can find the value of X.

Complete step by step answer:
From the question, we were given that \[A+B+C=\pi \].
Now we should find the value of \[{{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}C\]. Let us assume the value of \[{{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}C\] is equal to X.
\[\Rightarrow X={{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}\left( \pi -\left( A+B \right) \right)\]
We know that \[\sin \left( \pi -\theta \right)=\sin \theta \].
\[\Rightarrow X={{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}\left( A+B \right)\]
We know that \[\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A\].
\[\Rightarrow X={{\sin }^{2}}A-{{\sin }^{2}}B+{{\left( \sin A\cos B+\sin B\cos A \right)}^{2}}\]
We know that \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
\[\begin{align}
  & \Rightarrow X={{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}B{{\cos }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B+2\operatorname{sinBcosAsinAcosB} \\
 & \Rightarrow X={{\sin }^{2}}A\left( 1+{{\cos }^{2}}B \right)-{{\sin }^{2}}B\left( 1-{{\cos }^{2}}A \right)+2\sin B\cos A\sin A\cos B \\
\end{align}\]
We know that \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \].
\[\begin{align}
  & \Rightarrow X={{\sin }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B-{{\sin }^{2}}B{{\sin }^{2}}A+2\sin B\cos A\sin A\cos B \\
 & \Rightarrow X={{\sin }^{2}}A\left( 1+{{\cos }^{2}}B-{{\sin }^{2}}B \right)+2\sin B\cos A\sin A\cos B \\
\end{align}\]
We know that \[{{\cos }^{2}}B-{{\sin }^{2}}B=\cos 2B\].
\[\Rightarrow X={{\sin }^{2}}A\left( 1+\cos 2B \right)+2\sin B\cos A\sin A\cos B\]
We know that \[1+\cos 2B=2{{\cos }^{2}}B\].
\[\Rightarrow X={{\sin }^{2}}A\left( 2{{\cos }^{2}}B \right)+2\sin B\cos A\sin A\cos B\]
\[\Rightarrow X=2\sin A\cos B\left( \sin A\cos B+\sin B\cos A \right)\]
We know that \[\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A\].
\[\Rightarrow X=2\sin A\cos B\sin \left( A+B \right)\]
We know that \[\sin \left( \pi -\theta \right)=\sin \theta \].
\[\Rightarrow X=2\sin A\cos B\sin \left( \pi -\left( A+B \right) \right)\]
\[\begin{align}
  & \Rightarrow X=2\sin A\cos B\operatorname{sinC} \\
 & \\
\end{align}\]
So, we can say that if \[A+B+C=\pi \] then \[{{\sin }^{2}}A-{{\sin }^{2}}B+{{\sin }^{2}}C=2\sin A\cos B\sin C\].

Note:
 Students may have misconception that \[\sin \left( A+B \right)=\sin A\cos B-\sin B\cos A\]. If this misconception is followed, then we cannot get the correct value of X. So, this misconception should be avoided to get the correct answer and to obtain the proof in a correct manner.