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

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

Explanation

Solution

We use the trigonometric identities that can expand both the values on the left hand side of the equation. Then use distributive property to multiply the values inside the brackets together.

  • Distributive Property: For any four numbers ‘a’, ‘b’, ‘c’ and ‘d’ we can write (a+b)(c+d)=a(c+d)+b(c+d)(a + b)(c + d) = a(c + d) + b(c + d)
  • Trigonometric identities:
    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

Complete step by step solution:
We have to prove sin⁡(x+y)sin⁡(x−y)=cos⁡2y−cos⁡2x\sin (x + y)\sin (x - y) = {\cos ^2}y - {\cos ^2}x
We know the trigonometric identities 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
Substitute the values from identities in left hand side of the equation
⇒sin⁡(x+y)sin⁡(x−y)=(sin⁡xcos⁡y+cos⁡xsin⁡y)(sin⁡xcos⁡y−cos⁡xsin⁡y)\Rightarrow \sin (x + y)\sin (x - y) = \left( {\sin x\cos y + \cos x\sin y} \right)\left( {\sin x\cos y - \cos x\sin y} \right)
Now use distributive property to multiply the terms in the brackets.
⇒sin⁡(x+y)sin⁡(x−y)=sin⁡xcos⁡y×(sin⁡xcos⁡y−cos⁡xsin⁡y)+cos⁡xsin⁡y(sin⁡xcos⁡y−cos⁡xsin⁡y)\Rightarrow \sin (x + y)\sin (x - y) = \sin x\cos y \times \left( {\sin x\cos y - \cos x\sin y} \right) + \cos x\sin y\left( {\sin x\cos y - \cos x\sin y} \right)Multiply values outside the bracket to the terms inside bracket
⇒sin⁡(x+y)sin⁡(x−y)=sin⁡2xcos⁡2y−sin⁡xcos⁡xsin⁡ycos⁡y+sin⁡xcos⁡xsin⁡ycos⁡y−cos⁡2xsin⁡2y\Rightarrow \sin (x + y)\sin (x - y) = {\sin ^2}x{\cos ^2}y - \sin x\cos x\sin y\cos y + \sin x\cos x\sin y\cos y - {\cos ^2}x{\sin ^2}y
Cancel terms having same magnitude but opposite sign
⇒sin⁡(x+y)sin⁡(x−y)=sin⁡2xcos⁡2y−cos⁡2xsin⁡2y\Rightarrow \sin (x + y)\sin (x - y) = {\sin ^2}x{\cos ^2}y - {\cos ^2}x{\sin ^2}y
Now we know that sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 , then we can write sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x
Similarly, sin⁡2y=1−cos⁡2y{\sin ^2}y = 1 - {\cos ^2}y
Substitute these values in right hand side of the equation
⇒sin⁡(x+y)sin⁡(x−y)=(1−cos⁡2x)cos⁡2y−cos⁡2x(1−cos⁡2y)\Rightarrow \sin (x + y)\sin (x - y) = (1 - {\cos ^2}x){\cos ^2}y - {\cos ^2}x(1 - {\cos ^2}y)
Multiply values outside the bracket to the terms inside bracket
⇒sin⁡(x+y)sin⁡(x−y)=cos⁡2y−cos⁡2xcos⁡2y−cos⁡2x+cos⁡2xcos⁡2y\Rightarrow \sin (x + y)\sin (x - y) = {\cos ^2}y - {\cos ^2}x{\cos ^2}y - {\cos ^2}x + {\cos ^2}x{\cos ^2}y
Cancel terms having same magnitude but opposite sign
⇒sin⁡(x+y)sin⁡(x−y)=cos⁡2y−cos⁡2x\Rightarrow \sin (x + y)\sin (x - y) = {\cos ^2}y - {\cos ^2}x
This is equal to right hand side of the equation
So, left hand side of the equation is equal to right hand side of the equation
Hence proved

Note:
Many students make mistake of applying the property (a−b(a+b)=a2−b2(a - b(a + b) = {a^2} - {b^2} in this question which is wrong as in both the brackets the terms in place of ‘a’ and ‘b’ are different. Keep in mind we can only apply this property if we have two same terms in brackets, one with addition and one with subtraction. Also, students tend to change the equation into a form of sine function which is wrong here as it is required in the question for the values to be cosine function form.