Question

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

Answer 1

Hint: Here, we are given \[{\sin ^2}x - {\sin ^2}y = \sin (x + y)\sin (x - y)\] and we will solve the RHS part and compare it with the LHS part. To solve this, we will use the trigonometric ratios formulas: \[\sin (x + y) = \sin x\cos y + \cos x\sin y\], \[\sin (x - y) = \sin x\cos y - \cos x\sin y\] and \[{\sin ^2}x + {\cos ^2}x = 1\] . Also we will use the formula \[(a + b)(a - b) = {a^2} - {b^2}\] 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,
\[{\sin ^2}x - {\sin ^2}y = \sin (x + y)\sin (x - y)\]
We will solve the RHS part and compare it with LHS as below:
\[ RHS= \sin (x + y)\sin (x - y)\]
We know that,
\[\sin (x + y) = \sin x\cos y + \cos x\sin y\] and \[\sin (x - y) = \sin x\cos y - \cos x\sin y\]
Apply this and we will get,
\[RHS = \left( {\sin x\cos y + \cos x\sin y} \right)\left( {\sin x\cos y - \cos x\sin y} \right)\]

Let \[a = \sin x\cos y\] and \[b = \cos x\sin y\]
Substituting the value, we will get,
\[RHS = (a + b)(a - b)\]
We know the formula \[(a + b)(a - b) = {a^2} - {b^2}\] and so applying this, we will get,
\[RHS = {a^2} - {b^2}\]
Again substitute the values of a and b, we will get,
\[RHS = {\left( {\sin x\cos y} \right)^2} - {\left( {\cos x\sin y} \right)^2}\]
\[RHS = {\sin ^2}x{\cos ^2}y - {\cos ^2}x{\sin ^2}y\]
We will use the identity of trigonometric ratio as:
\[{\sin ^2}x + {\cos ^2}x = 1\]
\[ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x\]

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

Hence, \[{\sin ^2}x - {\sin ^2}y = \sin (x + y)\sin (x - y)\] 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.