Lauren L.

asked • 11/18/12

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

The directions say to prove the identity. I understand how these two functions work, but I don't understand how to go through the process of getting from the left side to the right side. 

2 Answers By Expert Tutors

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Robert J. answered • 11/18/12

Certified High School AP Calculus and Physics Teacher

Michael B. answered • 11/18/12

I can provide your 'A-HA' moment

Lauren L.

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.

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11/18/12

Angel M.

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).

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11/18/12

Michael B.

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)

 

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11/18/12

Fallou B.

Wait im not getting how you found a common denominator
 
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04/01/15

Michael B.

Fallou, there are two additive terms on the left: sin2x/cos2x and sin2x.  We need the second term, sin2x to have a common denominator of cos2x, so we multiply both the numerator (sin2x) and denominator (an implicit 1) by cos2x, resulting in:
 
sin2x       sin2x cos2x
------   -  --------------
cos2x           cos2x
 
which is then combined into:
 
  sin2x - sin2x cos2x
 ------------------------
            cos2x
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05/14/15

Stefan C.

this is the best explanation on the internet!
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02/22/19

Gabrielle G.

where does the cos^2x denominator go? I am at sin^4x/cos^2x=sin^4x/cos^2x but i cannot see how the cos^2x disappears. HELP!!!
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03/25/19

Michael B.

Gabrielle, I'm not quite sure what step you are looking at. To cancel the cos^2x, just multiply both sides by cos^2x. But remember that the original question was to prove that the identity is true. If you ever get to a step where you have the same expression on both sides of the equal sign, then you are done - you have completed the proof. Since you say you are at sin^4x/cos^2x = sin^4x/cos^2x (the same thing on both sides) you can say that you have completed the proof.
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03/25/19

Kumar E.

How does sin^2x - sin^2x * cos^2x = sin^4x?????
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1d

Michael B.

Kumar remember the trig identity: cos^2x = 1 - sin^2x. When you substitute this into your equation, you get: sin^2x - sin^2x * (1 - sin^2x) = sin^4x.... then distribute the sin^2x sin^2x - sin^2x + sin^4x = sin^4x..... then the sin^2x terms cancel sin^4x = sin^4x
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1d

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