1. ## Simplifying (sin2x)(-sin2x)

The equation is (1-sin2x)(1+sin2x) = 1+sin2x-sin2x- (sin2x)(sin2x)
Does that equal = 1 - 2sin2x or 1 - (sin2x)^2 = 1- sin4x^2 ?? I have not done this before

2. ## Re: Simplifying (sin2x)(-sin2x)

Originally Posted by sakonpure6
The equation is (1-sin2x)(1+sin2x) = 1+sin2x-sin2x- (sin2x)(sin2x)
Does that equal = 1 - 2sin2x or 1 - (sin2x)^2 = 1- sin4x^2 ?? I have not done this before

Do you mean 1) $(\sin(2x))(-\sin(2x))$ or do you mean 2) $(\sin^2(x))(-\sin^2(x))$?

3. ## Re: Simplifying (sin2x)(-sin2x)

Sorry, here is the proper equation

$(1+\sin(2x)) (1-\sin(2x))$

4. ## Re: Simplifying (sin2x)(-sin2x)

It would be very strange to be given problems like this involving sine if you have never taken an algebra class. But if you have taken algebra then surely you should know that (a+ b)(a- b)= a^2+ ab- ab- b^2= a^2- b^2. So (1- sin(2x))(1+ sin(2x))= 1- sin^2(2x). No, the square does does NOT apply to the argument, 2x, of the sine, it applies to the sine itself.

5. ## Re: Simplifying (sin2x)(-sin2x)

Originally Posted by HallsofIvy
It would be very strange to be given problems like this involving sine if you have never taken an algebra class. But if you have taken algebra then surely you should know that (a+ b)(a- b)= a^2+ ab- ab- b^2= a^2- b^2. So (1- sin(2x))(1+ sin(2x))= 1- sin^2(2x). No, the square does does NOT apply to the argument, 2x, of the sine, it applies to the sine itself.
OMG I Love you!!!!!!! You just helped me solve an identity that took me forever!!!!!!!!!!!!!!!!!!!!!!!!!!

$(1-sin2x)/(cos2x) = (cos2x)/(1+sin2x)$

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### sin2x sin2x

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