Yes, it simplifies to give you zero. Now to make our lives easier, it would be better to turn the equation to:
#cos(45+a)cos(45-a) = sin(45-a)sin(45+a)#
This way we can resolve both sides without things getting mixed and muddled together.
You should be comfortable with your trigonometric identities so by looking at this equation you are able to tell the four identities we need to know on both LHS and RHS.
#sin(a+b) = sinacosb + sinbcosa#
#sin(a-b) = sinacosb - sinbcosa#
#cos(a+b) = cosacosb - sinasinb#
#cos(a-b) = cosacosb + sinasinb#
LHS (lefthandside):
#cos(45+a) = cos45cosa - sin45sina#
#cos(45-a) = cos45cosa + sin45sina#
#cos45# put in the calculator will give you #1/sqrt2# which is the same value given for #sin45#
#cos(45+a) = 1/sqrt2cosa - 1/sqrt2sina#
#cos(45+a) = 1/sqrt2(cosa - sina)#
#cos(45-a) = 1/sqrt2cosa + 1/sqrt2sina#
#cos(45-a) = 1/sqrt2(cosa + sina)#
#cos(45+a)cos(45-a) = 1/sqrt2(cosa + sina) times 1/sqrt2(cosa - sina) #
= #1/sqrt2(cosa + sina) times 1/sqrt2(cosa - sina)#
= #1/2(cosa+sina)(cosa-sina)#
= #1/2(cos^2a-sin^2a)#
Now that's the Left Hand Side Done.
RHS (righthandside):
follow the same steps as the LHS
#sin(45+a) = sin45cosa + sinacos45#
#sin(45+a) = 1/sqrt2(cosa+sina)#
#sin(45-a) = sin45cosa - sinacos45#
#sin(45-a) = 1/sqrt2(cosa-sina)#
#sin(45-a)sin(45+a) = 1/sqrt2(cosa-sina) times 1/sqrt2(cosa+sina)#
= #1/sqrt2(cosa-sina) times 1/sqrt2(cosa+sina)#
= #1/2(cosa-sina)(cosa+sina)#
= #1/2(cos^2a - sin^2a)#
This is the same simplified value given for the LHS
If the LHS gives the same value as the RHS, this will simplify to give you zero.
