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

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

Explanation

Solution

Here, we are given sin⁡2x−sin⁡2y=sin⁡(x+y)sin⁡(x−y){\sin ^2}x - {\sin ^2}y = \sin (x + y)\sin (x - y) and we will solve the RHS part and compare it with the LHS part. To solve this, we will use the trigonometric ratios formulas: sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin (x + y) = \sin x\cos y + \cos x\sin y, sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin (x - y) = \sin x\cos y - \cos x\sin y and sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 . Also we will use the formula (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} too. After applying all these to solve the RHS part, we will get the LHS part and so will prove both are equal.

Complete step by step answer:
Given that,
sin⁡2x−sin⁡2y=sin⁡(x+y)sin⁡(x−y){\sin ^2}x - {\sin ^2}y = \sin (x + y)\sin (x - y)
We will solve the RHS part and compare it with LHS as below:
RHS=sin⁡(x+y)sin⁡(x−y)RHS= \sin (x + y)\sin (x - y)
We know that,
sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\sin (x + y) = \sin x\cos y + \cos x\sin y and sin⁡(x−y)=sin⁡xcos⁡y−cos⁡xsin⁡y\sin (x - y) = \sin x\cos y - \cos x\sin y
Apply this and we will get,
RHS=(sin⁡xcos⁡y+cos⁡xsin⁡y)(sin⁡xcos⁡y−cos⁡xsin⁡y)RHS = \left( {\sin x\cos y + \cos x\sin y} \right)\left( {\sin x\cos y - \cos x\sin y} \right)

Let a=sin⁡xcos⁡ya = \sin x\cos y and b=cos⁡xsin⁡yb = \cos x\sin y
Substituting the value, we will get,
RHS=(a+b)(a−b)RHS = (a + b)(a - b)
We know the formula (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} and so applying this, we will get,
RHS=a2−b2RHS = {a^2} - {b^2}
Again substitute the values of a and b, we will get,
RHS=(sin⁡xcos⁡y)2−(cos⁡xsin⁡y)2RHS = {\left( {\sin x\cos y} \right)^2} - {\left( {\cos x\sin y} \right)^2}
RHS=sin⁡2xcos⁡2y−cos⁡2xsin⁡2yRHS = {\sin ^2}x{\cos ^2}y - {\cos ^2}x{\sin ^2}y
We will use the identity of trigonometric ratio as:
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1
⇒cos⁡2x=1−sin⁡2x\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x

So, applying this, we will get,
RHS=sin⁡2x(1−sin⁡2y)−(1−sin⁡2x)sin⁡2yRHS= {\sin ^2}x\left( {1 - {{\sin }^2}y} \right) - \left( {1 - {{\sin }^2}x} \right){\sin ^2}y
Expanding this, we will get,
RHS=(sin⁡2x−sin⁡2xsin⁡2y)−(sin⁡2y−sin⁡2xsin⁡2y)RHS = \left( {{{\sin }^2}x - {{\sin }^2}x{{\sin }^2}y} \right) - \left( {{{\sin }^2}y - {{\sin }^2}x{{\sin }^2}y} \right)
Removing the brackets, we will get,
RHS=sin⁡2x−sin⁡2xsin⁡2y−sin⁡2y+sin⁡2xsin⁡2yRHS = {\sin ^2}x - {\sin ^2}x{\sin ^2}y - {\sin ^2}y + {\sin ^2}x{\sin ^2}y
Rearranging this, we will get,
RHS=sin⁡2x−sin⁡2y−sin⁡2xsin⁡2y+sin⁡2xsin⁡2yRHS = {\sin ^2}x - {\sin ^2}y - {\sin ^2}x{\sin ^2}y + {\sin ^2}x{\sin ^2}y
Cancelling the terms, we will get,
RHS=sin⁡2x−sin⁡2y=LHSRHS = {\sin ^2}x - {\sin ^2}y =LHS
Thus, RHS = LHS.

Hence, sin⁡2x−sin⁡2y=sin⁡(x+y)sin⁡(x−y){\sin ^2}x - {\sin ^2}y = \sin (x + y)\sin (x - y) is proved.

Note: Trigonometry is the relationship between the sides and angles of a right-angled triangle. Trigonometry is one of those divisions in mathematics that helps in finding the angles and missing sides of a triangle with the help of trigonometric ratios. The angles are either measured in radians or degrees. The trigonometric ratios of a triangle are also called the trigonometric functions. Sine, cosine, and tangent are 3 important trigonometric functions and are abbreviated as sin, cos and tan.