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Question: How do you prove \[\sin (x + y) \cdot \sin (x - y) = {(\sin x)^2} - {(\sin y)^2}\]?...

How do you prove sin⁡(x+y)⋅sin⁡(x−y)=(sin⁡x)2−(sin⁡y)2\sin (x + y) \cdot \sin (x - y) = {(\sin x)^2} - {(\sin y)^2}?

Explanation

Solution

According to the question, we will first take our LHS and RHS. Then we will try to solve either RHS or LHS and solving one of them will give us the either part. To solve this problem we use the basic trigonometric formulas like sin(a+b)sin(a+b) and cos (a+b)(a+b). Here, for this question, solving the LHS part will be much easier than solving the RHS part.

Complete step-by-step solution:
We will try to separate the question in the form of LHS and RHS. So, here our LHS is sin⁡(x+y)⋅sin⁡(x−y)\sin (x + y) \cdot \sin (x - y). Now, our RHS is (sin⁡x)2−(sin⁡y)2{(\sin x)^2} - {(\sin y)^2}. Now, we will try to prove both LHS and RHS equal.
First, we will try to solve our LHS.
LHS=sin⁡(x+y)⋅sin⁡(x−y) = \sin (x + y) \cdot \sin (x - y)
Here, we will apply the two basic trigonometric formulas:
sin⁡(a+b)=(sin⁡acos⁡b+cos⁡asin⁡b)\sin (a + b) = (\sin a\cos b + \cos a\sin b) and cos⁡(a+b)=(sin⁡acos⁡b−cos⁡asin⁡b)\cos (a + b) = (\sin a\cos b - \cos a\sin b)
When we apply these formulas in LHS, then we will get:
LHS=(sin⁡acos⁡b+cos⁡asin⁡b)(sin⁡acos⁡b−cos⁡asin⁡b) = (\sin a\cos b + \cos a\sin b)(\sin a\cos b - \cos a\sin b)
Now, we will try to open the bracket, and multiply the terms. After that, we will get:
=(sin⁡2xcos⁡2y−cos⁡2xsin⁡2y)= ({\sin ^2}x{\cos ^2}y - {\cos ^2}x{\sin ^2}y)
Now, we will apply the famous trigonometric formula:
sin⁡2a+cos⁡2a=1{\sin ^2}a + {\cos ^2}a = 1
Here, we will modify or rearrange the formula, and we will get:
⇒cos⁡2a=1−sin⁡2a\Rightarrow {\cos ^2}a = 1 - {\sin ^2}a
Now, we will apply this formula in our equation, and we will get:
LHS =sin⁡2x(1−sin⁡2y)−(1−sin⁡2x)sin⁡2y = {\sin ^2}x(1 - {\sin ^2}y) - (1 - {\sin ^2}x){\sin ^2}y
When we open the brackets, and multiply the terms, we get:
=sin⁡2x−sin⁡2xsin⁡2y−sin⁡2y+sin⁡2ysin⁡2x= {\sin ^2}x - {\sin ^2}x{\sin ^2}y - {\sin ^2}y + {\sin ^2}y{\sin ^2}x
Here we can see that some terms can get canceled out. So, we will cancel out −sin⁡2xsin⁡2y - {\sin ^2}x{\sin ^2}yand sin⁡2ysin⁡2x{\sin ^2}y{\sin ^2}xfrom LHS, and we will get:
=sin⁡2x−sin⁡2y= {\sin ^2}x - {\sin ^2}y
Here, RHS=(sin⁡x)2−(sin⁡y)2 = {(\sin x)^2} - {(\sin y)^2}
Therefore, our LHS=RHS (proved)

Note: To solve this problem we have used the basic trigonometric formulas as the question contains the basic terms as sinxsinx and cosxcosx. The above method was easy and the question gets solved quickly. But there is another method to solve the question. We can also start solving the RHS part first, and then we can get the LHS part. We can expand the RHS part. But expanding is difficult. So, we should try solving the LHS part first.