Simplify: (sin Θ − cos Θ)2 - (sin Θ - cos Θ)2

Question

anonymous · Answer

0
2
sin^2sigma
Cos^2sigma

anonymous · Answer

@alyssa_xo

anonymous · Answer

oh and those twos on the () are ^2

anonymous · Answer

I tried to do it the way you showed me but I kept doing something wrong

alyssa_xo · Answer

you're in bio btw, you can switch subjects by hovering over http://puu.sh/i1Vm4/6075c5fc3e.jpg and either clicking your new subject or clicking find more subjects

anonymous · Answer

oh shoot

alyssa_xo · Answer

nobody cares lol

alyssa_xo · Answer

just a heads up when you ask your next question so you can get the appropriate help

anonymous · Answer

haha ill remember that!

alyssa_xo · Answer

ok so let's see
$(a-b)^2=a^2-2ab+b^2\(a-b)^2-(a-b)^2=(a^2-2ab+b^2)-(a^2-2ab+b^2)$

this is probably where you went wrong, did you distribute the minus sign correctly?

anonymous · Answer

nope it doesnt look like it haha

alyssa_xo · Answer

note $(a^2-2ab+b^2)-(a^2-2ab+b^2)=\(a^2-2ab+b^2)-a^2-(-2ab)-b^2$

anonymous · Answer

okay, that makes sense

alyssa_xo · Answer

and so we have $a^2-2ab+b^2-a^2+2ab-b^2$

alyssa_xo · Answer

looks like 0 to me

alyssa_xo · Answer

wow, lmao a 6 year old could have done this

anonymous · Answer

Oh wow thanks for your help! And for being mean lol

alyssa_xo · Answer

lol didn't mean to be mean, just to point out that we did unneeded work

alyssa_xo · Answer

look at the question, it's of the form $a^2-a^2$

anything minus itself = 0

alyssa_xo · Answer

so we didn't have to expand the stuff

anonymous · Answer

ahhhhh good point lol

anonymous · Answer

thanks!