Lauren L.

# 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

By: Certified High School AP Calculus and Physics Teacher 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.

Report

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).

Report

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)

Report

11/18/12

Fallou B.

Wait im not getting how you found a common denominator

Report

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
Report

05/14/15

Stefan C.

this is the best explanation on the internet!
Report

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!!!
Report

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.
Report

03/25/19

Kumar E.

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

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
Report

1d

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.
Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.