Trigonometric Difference of Two Squares
True or False :
sin ( x + y ) × sin ( x − y ) = sin 2 x − sin 2 y
Inspiration behind the problem: It is well known that ( x + y ) ( x − y ) = x 2 − y 2 . Does this still work if we apply a trigonometric function, or is that a common misconception ?
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3 solutions
s i n 2 x − s i n 2 y = ( 1 − c o s 2 x ) / 2 − ( 1 − c o s 2 y ) / 2 1 / 2 ( c o s 2 y − c o s 2 x ) = 1 / 2 ( 2 s i n ( ( 2 x + 2 y ) / 2 ) s i n ( ( 2 x − 2 y ) / 2 ) ) = s i n ( x + y ) s i n ( x − y )
Bonus Question: Is it true that cos ( x + y ) × cos ( x − y ) = cos 2 x − cos 2 y ? Or is it a common misconception?
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Good one .... cos(x+y)cos(x-y)= c o s 2 x − s i n 2 y .It can be proved in a similar way. While c o s 2 x − c o s 2 y = s i n 2 y − s i n 2 x = s i n ( x + y ) s i n ( y − x )
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Seek the beauty:
cos ( x + y ) × cos ( x − y ) = cos 2 x − sin 2 y = cos 2 y − sin 2 x
It won't matter if we interchange x , y :)
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@Nihar Mahajan – That is a nice identity to share with others!
@Nihar Mahajan – cos is an even function and addition is commutative ! :p
In a simple way put x=0,y=0. So sin(0+0)×sin(0-0)=0. Sin^2(0)-sin^2(0)=0 .so two terms are equale #####
I do not understand your solution. At the most, you have shown that it is true at a specific value of x = 0 , y = 0 . This does not mean that we have an identity.
For example, the statement x 2 = y 2 is true when x = 0 , y = 0 . However, it is not an identity.
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First of all thank you for sending your comment , i agree with you, but I was checking the two expressions by calculations , I know that I should make two expressions identify by using rules, the question was true or false not to prove OK, but if we put for example x=60,y=30,sin(60+30)×sin(60-30)=.5,sin^2 (60)-sin^2 (30)=.5,so calculatoions also check identify of two expressions, you understand what I mean. Good bye
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Yes, I understand that you verified that equality holds under certain conditions. However, that doesn't imply that we have an identity.
For example, if we tried it with the scenario of x x = 1 , we cannot say "Just because it works for x = 1 , 2 , 3 , 4 , hence we have an identity". Because, if we check x = 0 (and only this case), then we do not have a true statement. Thus, x x = 1 is not an identity, even though it works for the cases that we tested,
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@Calvin Lin – That is perfectly correct what you are saying.
Remembering the trigonometric sum and difference formulas , we know that sin ( x + y ) = sin x cos y + cos x sin y , sin ( x − y ) = sin x cos y − cos x sin y .
Hence we have, \[\begin{array}{} \sin (x+y) \times \sin (x-y) & = \left( \sin x \cos y + \cos x \sin y \right) \times \left( \sin x \cos y - \cos x \sin y \right) \\ & = \sin^2 x \cos^2 y - \cos^2 x \sin^2 y \\ & = \sin^2 x ( 1-\sin^2 y) - ( 1- \sin^2 x ) \sin^2 y \\ & = \sin^2 x - \sin^2 x \sin^2 y -\sin^2 y + \sin^2 x \sin^2 y \\ & =\sin^2 x -\sin^2 y . \end{array} \]
Hence the given identity i.e. sin ( x + y ) × sin ( x − y ) = sin 2 x − sin 2 y is true. □