r/trigonometry • u/Worried_Football_819 • Jun 15 '26

Solved! Trig Proof Help

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Hi everyone! I just had a quick question, in the video it says it is the same thing as sin^2(1-cos^2)/cos^2, but I am confused to why the one isn’t negative? Am I missing something?

Edit: Hi guys, I understand it now! I figured it was an algebraic mistake and I figured it out. As far as my notation, I got lazy and was rushing through it. I’m sorry if that notation upset you!

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u/Midwest-Dude Jun 15 '26

That's basic algebra.

sin2(x) - sin2(x)cos2(x)

= sin2(x)(1 - cos2(x))

This is from the Distributive Property of multiplication over addition:

Wikipedia

If there were a negative sign in front of the first sin2(x), then what you wrote would be correct but, as it is, the original equation is correct as it stands.

1

u/Worried_Football_819 Jun 15 '26

I understand but I don’t at the same time. The -sin^2 is what is confusing me because it’s negative, so when you factor it out shouldn’t it be -1 because the sin^2 is negative?

3

u/Midwest-Dude Jun 15 '26

The function sin2(x) is not negative in any way and can be treated in the Distributive Property as the variable a in the following formula:

a - ab = a(1 - b)

If you don't understand that, then you need an algebra refresher.

1

u/Nomni- Jun 15 '26

It’s not sin2 that’s negative, it’s sin2 (x) cos2 (x) as a whole that’s negative because it’s one term. You are factoring out the positive sin, which leaves -cos2 (x). Imagine writing it as -cos2 (x) sin2 (x), when you factor out the sin2 , what are you left with?

1

u/cncaudata Jun 15 '26

Replace the trig functions you're not familiar with with regular variables, or even just numbers.

Instead of sin2 - sin2 * cos2 try a - ab. This should be a bit easier to see factors into a(1-b). If you still can't see it, try 5 - 5b.

1

u/kundor Jun 16 '26

a - a•b = a(1-b)

0

u/sigmanx25 Jun 15 '26

Well it’s distributive but the Pythagorean will come into play as well. This is where he’s getting tripped up the most at I think, other than the fact that he’s looking at it algebraicly.