Question

@PeterPan help? http://prntscr.com/3leiae

Answer 1

Recall: \(\cos^2 \theta+\sin^2\theta=1\implies \sin^2\theta=1-\cos^2\theta\)
\[ \frac{\sin^2\theta}{1+\cos\theta}=\frac{1-\cos^2\theta}{1+\cos\theta}=\frac{(1-\cos\theta)(1+\cos\theta)}{1+\cos\theta}\]

The second step is a difference of squares

Answer 2

Work a little more magic with the identities...

Ever heard of the Pythagorean one?
\[\Large \cos^2(\theta) +\sin^2(\theta) = 1\]

Answer 3

So just cancel out both of the (1+cosθ)?

Answer 4

and that will make it just (1−cosθ)?

Answer 5

No actually there is an identity in the next math called called the Pythagorean identities that says it is \[\sin^2\theta+\cos^2\theta=1\]

Answer 6

he taught you something from a different math class

Answer 7

Ok what do i do now then?

Answer 8

What do you mean? I used the same identity...

Answer 9

so i was right then @kirbykirby ?

Answer 10

Ok so you just multiply and tat ends up giving you \[-(\cos(2x)-1)/2(\cos(x)+1)\]

Answer 11

Yes @tester97

Answer 12

mmmk thanks

Answer 13

go post on bae