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Question: Find the value of \({\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B)\)....

Find the value of sin⁡2B−sin⁡2A−sin⁡2(A−B)+2sin⁡Acos⁡Bsin⁡(A−B){\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B).

Explanation

Solution

We will try to simplify the given expression by using the formula: sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B and (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab. Doing all this and simplifying, we will get the desired value.

Complete step-by-step answer:
We have with us the expression in the question which is: sin⁡2B−sin⁡2A−sin⁡2(A−B)+2sin⁡Acos⁡Bsin⁡(A−B){\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B).
Let us put in the value of sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin (A - B) = \sin A\cos B - \cos A\sin B.
We will now get:-
⇒sin⁡2B−sin⁡2A−(sin⁡Acos⁡B−cos⁡Asin⁡B)2+2sin⁡Acos⁡B(sin⁡Acos⁡B−cos⁡Asin⁡B)\Rightarrow {\sin ^2}B - {\sin ^2}A - {(\sin A\cos B - \cos A\sin B)^2} + 2\sin A\cos B(\sin A\cos B - \cos A\sin B)
Now, we will apply (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab on (sin⁡Acos⁡B−cos⁡Asin⁡B)2{(\sin A\cos B - \cos A\sin B)^2}by taking a=sin⁡Acos⁡Ba = \sin A\cos B and b=cos⁡Asin⁡Bb = \cos A\sin B to get (sin⁡Acos⁡B−cos⁡Asin⁡B)2=(sin⁡Acos⁡B)2+(cos⁡Asin⁡B)2−2.sin⁡Acos⁡B.cos⁡Asin⁡B{(\sin A\cos B - \cos A\sin B)^2} = {(\sin A\cos B)^2} + {(\cos A\sin B)^2} - 2.\sin A\cos B.\cos A\sin B
So, we get:

⇒sin⁡2B−sin⁡2A−(sin⁡Acos⁡B)2−(cos⁡Asin⁡B)2+2.sin⁡Acos⁡B.cos⁡Asin⁡B+ 2sin⁡Acos⁡B(sin⁡Acos⁡B−cos⁡Asin⁡B)  \Rightarrow {\sin ^2}B - {\sin ^2}A - {(\sin A\cos B)^2} - {(\cos A\sin B)^2} + 2.\sin A\cos B.\cos A\sin B + \\\ 2\sin A\cos B(\sin A\cos B - \cos A\sin B) \\\

Simplifying the expression further to get the following expression:

⇒sin⁡2B−sin⁡2A−sin⁡2Acos⁡2B−cos⁡2Asin⁡2B+2sin⁡Asin⁡Bcos⁡Acos⁡B+ 2sin⁡2Acos⁡2B−2sin⁡Asin⁡Bcos⁡Acos⁡B  \Rightarrow {\sin ^2}B - {\sin ^2}A - {\sin ^2}A{\cos ^2}B - {\cos ^2}A{\sin ^2}B + 2\sin A\sin B\cos A\cos B + \\\ 2{\sin ^2}A{\cos ^2}B - 2\sin A\sin B\cos A\cos B \\\

Simplifying the expression further by clubbing the like terms to get the following expression:
⇒sin⁡2B−sin⁡2A+sin⁡2Acos⁡2B−cos⁡2Asin⁡2B\Rightarrow {\sin ^2}B - {\sin ^2}A + {\sin ^2}A{\cos ^2}B - {\cos ^2}A{\sin ^2}B
We can rewrite it as:-
⇒sin⁡2B(1−cos⁡2A)−sin⁡2A(1−cos⁡2B)\Rightarrow {\sin ^2}B(1 - {\cos ^2}A) - {\sin ^2}A(1 - {\cos ^2}B)
Now, we will apply the formula: sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 in here to get: sin⁡2B=1−cos⁡2B{\sin ^2}{\rm B} = 1 - {\cos ^2}{\rm B} and sin⁡2A=1−cos⁡2A{\sin ^2}{\rm A} = 1 - {\cos ^2}{\rm A}.
⇒sin⁡2Bsin⁡2A−sin⁡2Asin⁡2B\Rightarrow {\sin ^2}B {\sin ^2}A - {\sin ^2}A {\sin ^2}B
⇒0\Rightarrow 0
Hence, the value of the expression sin⁡2B−sin⁡2A−sin⁡2(A−B)+2sin⁡Acos⁡Bsin⁡(A−B){\sin ^2}B - {\sin ^2}A - {\sin ^2}(A - B) + 2\sin A\cos B\sin (A - B) is 0.
Hence, the answer is 0.

Note: The student may get confused and get worried on seeing so many terms for once and think about approaching some other way but this is a very easy way.
There is a general answer to such a question since, we are given to find a particular value of such an expression with no mentioned value of A and B. So, there must be some calculations involved to cut out most of the terms. But, this does not mean that students shall write the answer on a guess basis.
Fun Fact about Trigonometry:-
The etymology of trigonometry comes from the Greek words "trigonon" (triangle) and "metron" (measure). The person usually associated with inventing trigonometry was a Greek mathematician named Hipparchus. Hipparchus was originally an accomplished astronomer, who observed and applied trigonometric principles to study the zodiac.
Trigonometry is associated with music and architecture.
Engineers use Trigonometry to figure out the angles of the sound waves and how to design rooms.