Question

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

Answer 1

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

simplify that

Answer 2

okay...

Answer 3

\(a=sin\theta\\
b=cos \theta\)
note that there's a -2ab and +2ab

Answer 4

is it 2a^2+2b^2

Answer 5

yep. let's factor the 2 out so we're left with \(2(a^2+b^2)\)

substitute in the original values
\(2(\sin^2 \theta + \cos^2 \theta)\)

do you remember your trig identities?

Answer 6

No I do not lol

Answer 7

But I understand so far

Answer 8

it's on here, somewhere

Answer 9

So what would I look for on that?

Answer 10

f it, no point in making you play hide and seek
\(sin^2x+cos^2x=1\)

Answer 11

I actually just found it haha

Answer 12

But thanks

Answer 13

haha

Answer 14

so since it equals 1, what next?

Answer 15

So is the answer just 2?

Answer 16

plug that back into our simplification.
\((a-b)^2=a^2-2ab+b^2\\(a+b)^2=a^2+2ab+b^2\\(a-b)^2+(a+b)^2=a^2\cancel{-2ab}+b^2+a^2\cancel{+2ab}+b^2\\2a^2+2b^2=2(a^2+b^2)=2(1)=2\)

Answer 17

yeah

Answer 18

Wow! Thank you Very Very much!!!!

Answer 19

np np