Michael B. answered • 11/18/12

I can provide your 'A-HA' moment

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:

tan^{2}x - sin^{2}x = (tan^{2}x)(sin^{2}x)

**Proof Step 2**: Replace tan with sin/cos

(sin^{2}x/cos^{2}x) - sin^{2}x = (sin^{2}x/cos^{2}x)(sin^{2}x)

**Proof Step 3**: Obtain a common denominator on left, simplify right

(sin^{2}x - sin^{2}x cos^{2}x) / cos^{2}x = sin^{4}x / cos^{2}x

**Proof Step 4**: Cancel cos^{2}x from both denominators

sin^{2}x - sin^{2}x cos^{2}x = sin^{4}x

**Proof Step 5**: Factor out sin^{2}x from left

(sin^{2}x)(1 - cos^{2}x) = sin^{4}x

**Proof Step 6**: Use trig identity 1-cos^{2}x = sin^{2}x

(sin^{2}x)(sin^{2}x) = sin^{4}x

**Proof Step 7**: Simplify, DONE.

sin^{4}x = sin^{4}x

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).

11/18/12

Michael B.

Angel's answer is generally correct, if a bit confusing. I would have said it this way: think of (sin^{2}x - sin^{2}x cos^{2}x) as (a - ab) where a=sin^{2}x and b=cos^{2}x. 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=cos^{2}x, we know that:

sin^{2}x - sin^{2}x cos^{2}x = sin^{2}x (1 - cos^{2}x)

11/18/12

Fallou B.

04/01/15

Michael B.

^{2}x/cos

^{2}x and sin

^{2}x. We need the second term, sin

^{2}x to have a common denominator of cos

^{2}x, so we multiply both the numerator (sin

^{2}x) and denominator (an implicit 1) by cos

^{2}x, resulting in:

^{2}x sin

^{2}x cos

^{2}x

^{2}x cos

^{2}x

^{2}x - sin

^{2}x cos

^{2}x

^{2}x

05/14/15

Stefan C.

this is the best explanation on the internet!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!!!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.03/25/19

Kumar E.

How does sin^2x - sin^2x * cos^2x = sin^4x?????12/12/19

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^4x12/12/19

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.

11/18/12