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

Prove that: sin⁡2B=sin⁡2A+sin⁡2(A−B)−2sin⁡Acos⁡Bsin⁡(A−B){{\sin }^{2}}B={{\sin }^{2}}A+{{\sin }^{2}}\left( A-B \right)-2\sin A\cos B\sin \left( A-B \right).

Explanation

Solution

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
⇒sin⁡2A+sin⁡2(A−B)−2sin⁡Acos⁡Bsin⁡(A−B)\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
⇒sin⁡2A+sin⁡(A−B)[sin⁡(A−B)−2sin⁡Acos⁡B]\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
⇒sin⁡2A+sin⁡(A−B)[sin⁡Acos⁡B−cos⁡Asin⁡B−2sin⁡Acos⁡B] ⇒sin⁡2A+sin⁡(A−B)[−cos⁡Asin⁡B−sin⁡Acos⁡B] ⇒sin⁡2A−sin⁡(A−B)[cos⁡Asin⁡B+sin⁡Acos⁡B] \begin{aligned} & \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{aligned}
We know that the cos⁡Asin⁡B+sin⁡Acos⁡B=sin⁡(A+B)\cos A\sin B+\sin A\cos B=\sin \left( A+B \right), using this we simplify our expression as,
⇒sin⁡2A−sin⁡(A−B)sin⁡(A+B)…(1)\Rightarrow {{\sin }^{2}}A-\sin \left( A-B \right)\sin \left( A+B \right)\ldots \left( 1 \right)
Now, expanding sin⁡(A−B)sin⁡(A+B)\sin \left( A-B \right)\sin \left( A+B \right), we get
sin⁡(A−B)sin⁡(A+B)=(sin⁡Acos⁡B−cos⁡Asin⁡B)⋅(sin⁡Acos⁡B+cos⁡Asin⁡B)\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 (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}, we have:

& \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{aligned}$$ Now, applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. $$\begin{aligned} & \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{aligned}$$ Putting the value of equation (2) in equation (1), we get $\begin{aligned} & \Rightarrow {{\sin }^{2}}A-\left( {{\sin }^{2}}A-{{\sin }^{2}}B \right) \\\ & \Rightarrow {{\sin }^{2}}A-{{\sin }^{2}}A+{{\sin }^{2}}B \\\ & \Rightarrow {{\sin }^{2}}B \\\ \end{aligned}$ 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.