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Tricky Trig proof

Is this proof valid?
I can't seem to prove LHS=RHS

sin⁡2xcos⁡2y−cos⁡2xsin⁡2y=sin⁡2x−sin⁡2y{{\sin }^{2}}x{{\cos }^{2}}y-{{\cos }^{2}}x{{\sin }^{2}}y={{\sin }^{2}}x-{{\sin }^{2}}y
Maybe consider some trig identities??
Cos(A+B) may be a good start point.

I just need to be careful about how i post thats all :P
you can choose values of x and y ...put them into both sides & see if you get the same number on both sides.
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Avatar for xlaser31
xlaser31
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6
Original post by JD2015AS
Maybe consider some trig identities??
Cos(A+B) may be a good start point.

I just need to be careful about how i post thats all :P


Its supposed to be C2 level. Advanced C3 Trig identities should not be used.
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Avatar for xlaser31
xlaser31
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Original post by the bear
you can choose values of x and y ...put them into both sides & see if you get the same number on both sides.


Yes it works!
Next question is how to prove it?
Oh i didn't know -_-
Maybe divide by Cos^2 (x or y) ?
Original post by xlaser31
Its supposed to be C2 level. Advanced C3 Trig identities should not be used.
(edited 7 years ago)
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Avatar for Zacken
Zacken
22
Original post by xlaser31
Is this proof valid?
I can't seem to prove LHS=RHS

sin⁡2xcos⁡2y−cos⁡2xsin⁡2y=sin⁡2x−sin⁡2y{{\sin }^{2}}x{{\cos }^{2}}y-{{\cos }^{2}}x{{\sin }^{2}}y={{\sin }^{2}}x-{{\sin }^{2}}y


Convert all the cos's to sin's: this is obvious, you have an expression filled with squares of sines and cosines and you want to get to something filled with squares of sines. This cries out to change all squares of cosines to squares of sines.

So

sin⁡2x(1−sin⁡2y)−sin⁡2y(1−sin⁡2x)=sin⁡2x−sin⁡2xsin⁡2y−sin⁡2y+sin⁡2ysin⁡2x\sin^2 x(1 - \sin^2 y) - \sin^2 y(1-\sin^2 x) = \sin^2 x - \sin^2 x \sin^2 y - \sin^2 y + \sin^2 y \sin^2 x
(edited 7 years ago)
Original post by xlaser31
Is this proof valid?
I can't seem to prove LHS=RHS

sin⁡2xcos⁡2y−cos⁡2xsin⁡2y=sin⁡2x−sin⁡2y{{\sin }^{2}}x{{\cos }^{2}}y-{{\cos }^{2}}x{{\sin }^{2}}y={{\sin }^{2}}x-{{\sin }^{2}}y


I haven't tried it as I don't have access to pen and paper, but use of a^2 - b^2 = (a+b)(a-b) may help.

Edit: not even that. Using sin^2 x + cos^2 x = 1.
(edited 7 years ago)
Original post by xlaser31
Yes it works!
Next question is how to prove it?


a fun way is to add cos2x - cos2y to both sides...

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