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tan^2x-sin^2x=tan^2xsin^2x

If you are familiar with trig. identiy, you can prove from the left to the right in a much simpler way:

tan^2x-sin^2x

= tan^2x (1 - cos^2x), since tan^2x cos^2x = sin^2x

= tan^2 sin^2x

When trying to prove trig identities, it is often helpful to convert TAN functions into SIN/COS functions:

Proof Step 1: Start with the original equation to prove:
tan2x - sin2x = (tan2x)(sin2x)

Proof Step 2: Replace tan with sin/cos
(sin2x/cos2x) - sin2x = (sin2x/cos2x)(sin2x)

Proof Step 3: Obtain a common denominator on left, simplify right
(sin2x - sin2x cos2x) / cos2x = sin4x / cos2x

Proof Step 4: Cancel cos2x from both denominators
sin2x - sin2x cos2x = sin4x

Proof Step 5: Factor out sin2x from left
(sin2x)(1 - cos2x) = sin4x

Proof Step 6: Use trig identity 1-cos2x = sin2x
(sin2x)(sin2x) = sin4x

Proof Step 7: Simplify, DONE.
sin4x = sin4x

Could you remind me how you factored out sin^2x from the left in proof step 5? Other than that, everything else was very helpful, it makes so much more sense.

11/18/2012

look at (sin^2(x)-sin^2(x)cos^2(x)) as (2-2x). Both terms have a two therefore can be factored out, 2(1-x).

11/18/2012

Angel's answer is generally correct, if a bit confusing.  I would have said it this way: think of (sin2x - sin2x cos2x) as (a - ab) where a=sin2x and b=cos2x.  Now, when 'factoring out' we are looking for an expression that is common in both terms 'a' and 'ab'.  The common term is simply 'a', so we can factor it out.  Factoring it out is the opposite of the 'distribution' property - that is, it is distribution in reverse.  So for each term, we can ask "a times what is equal to the term?"  So for the first term, a, we say "a times what is equal to a?" and the answer is 1.  For the second term, "a times what is equal to ab?" and the answer is b.  So: (a - ab) = a(1 - b).

Now since we said a=sin2x and b=cos2x, we know that:

sin2x - sin2x cos2x = sin2x (1 - cos2x)

11/18/2012

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