Question

Prove that: \[{{\sin }^{2}}B={{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)-2\sin A\cos B\sin \left( A-B \right)\].

Answer 1

Hint: For solving this question, first we take sin (A -B) common from the terms in which sin (A – B) is present and apply the expansion of sin (A – B) in the right-hand side. After this, by applying algebraic and trigonometry formulas, we easily prove left hand side equal to right hand side.

Complete step-by-step answer:
Some of the useful trigonometric formulas used in solving this problem.
sin (A + B) = sin A cos B + cos A sin B
sin (A − B) = sin A cos B − cos A sin B
According to the problem statement, we consider the right-hand side of the equation for proving equivalence of both sides. First, we expand the right-hand side using the above-mentioned formulas.
Considering the right-hand side of the question, we have
\[\Rightarrow {{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)-2\sin A\cos B\sin \left( A-B \right)\]
Taking sin (A – B) common from the terms, we get
\[\Rightarrow {{\sin }^{2}}A+\sin \left( A-B \right)\left[ \sin \left( A-B \right)-2\sin A\cos B \right]\]
Expand the sin (A – B) inside the square bracket, we get
$\begin{align}
  & \Rightarrow {{\sin }^{2}}A+\sin \left( A-B \right)\left[ \sin A\cos B-\cos A\sin B-2\sin A\cos B \right] \\
 & \Rightarrow {{\sin }^{2}}A+\sin \left( A-B \right)\left[ -\cos A\sin B-\sin A\cos B \right] \\
 & \Rightarrow {{\sin }^{2}}A-\sin \left( A-B \right)\left[ \cos A\sin B+\sin A\cos B \right] \\
\end{align}$
We know that the \[\cos A\sin B+\sin A\cos B=\sin \left( A+B \right)\], using this we simplify our expression as,
$\Rightarrow {{\sin }^{2}}A-\sin \left( A-B \right)\sin \left( A+B \right)\ldots \left( 1 \right)$
Now, expanding $\sin \left( A-B \right)\sin \left( A+B \right)$, we get
$\sin \left( A-B \right)\sin \left( A+B \right)=\left( \sin A\cos B-\cos A\sin B \right)\cdot \left( \sin A\cos B+\cos A\sin B \right)$
By using the algebraic identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, we have:
\[\begin{align}
  & \sin \left( A-B \right)\sin \left( A+B \right)={{\left( \sin A\cos B \right)}^{2}}-{{\left( \cos A\sin B \right)}^{2}} \\
 & ={{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B \\
\end{align}\]
Now, applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
\[\begin{align}
  & \sin \left( A-B \right)\sin \left( A+B \right)={{\sin }^{2}}A\left( 1-{{\sin }^{2}}B \right)-{{\sin }^{2}}B\left( 1-{{\sin }^{2}}A \right) \\
 & ={{\sin }^{2}}A-{{\sin }^{2}}A\cdot {{\sin }^{2}}B-{{\sin }^{2}}B+{{\sin }^{2}}B\cdot {{\sin }^{2}}A \\
 & ={{\sin }^{2}}A-{{\sin }^{2}}B\ldots \left( 2 \right) \\
\end{align}\]
Putting the value of equation (2) in equation (1), we get
$\begin{align}
  & \Rightarrow {{\sin }^{2}}A-\left( {{\sin }^{2}}A-{{\sin }^{2}}B \right) \\
 & \Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}A+{{\sin }^{2}}B \\
 & \Rightarrow {{\sin }^{2}}B \\
\end{align}$
Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric formulas associated with different functions. One important step is the simplification by taking sin (A-B) common, and then applying a suitable identity.