Question

there is one problem i can not solve at all it is sin^2/1+cos

Answer 1

simplify?

Answer 2

yes i need simplify it

Answer 3

do oyu know any trig identites where you get a (1 + cosx) or maybe one for sin^2x?

Answer 4

Think. sin(x+y)=sinxcosy+sinycosx
sin(x+x)=?
???
Profit.

Answer 5

the closest one is 1-sin^\[1-\sin ^2\theta\]

Answer 6

@eyust707 the one i think you are suggesting is 1-sin^2 and that equals cos^2

Answer 7

yes.... there is a version of that for sin^2x...

Answer 8

i have this crazy idea....you know that

\[\Huge \frac{\frac{a}{c}}{\frac{b}{c}} = \frac{a}{b}\]

right?

so dived the numerator and denominator by sin^2 x :DDDD

Answer 9

@lgbasallote you really confused me

Answer 10

okay..disregard what i said :P lol it's just a trick haha..i'll try to find something else

Answer 11

Remember, 1=sin^2+cos^2... Why don't you try putting that into the 1?

Answer 12

1-cos^2/1+cos is as far as i got but i think if i was to take cos out of the top i will be left with 1-cos/1+cos then idk

Answer 13

in terms of what are you supposed to simply it in?

Answer 14

lol... sin^2 x = 1 - cos ^2x

\[\huge \frac {1- cos^{2}x}{1-cos}\]
\[\huge \frac{(1+cos)(1-cos)}{1-cos}\]

why didn't we see it before :P

Answer 15

im just suppose to get it to the simpliest form

Answer 16

lol thank you its 1+ cos but that isnt the answer ugh

Answer 17

lol then cancel the 1 + cos haha

1- cos will be left..

Answer 18

im doing a puzzle with pieces and that isnt one of the answer

Answer 19

1- cos is not the answer??

Answer 20

the problems are printed on the pieces and we have to solve so im guessing that not the answer

Answer 21

his answer looks correct to me, can you list what options you have?

Answer 22

im trying to see if it can be simplifed again but i might have to email my teacher if it isnt too late

Answer 23

well do you have the answer? maybe i can see what they did from the answer...what i did was..

\[\huge \frac {sin^{2}x}{1+cosx}\]
\[\huge \frac{1-cos^2{x}}{1+cosx}\]
\[\huge \frac{(1+cosx)(1-cosx)}{1+cosx}\]

cancel 1+ cosx...

1-cosx should be the answer...i don't see how it's not :/

Answer 24

that is what i got but my teacher said it should be able to simplify more bt i dnt thinkso

Answer 25

nope 1-cosx is as simple as it gets

Answer 26

you should get
\[1-\cos(x)\] and that is it

Answer 27

ok

Answer 28

hmmm it can be simplified further actually...

\[\huge 1 - cosx\]
\[\huge sin^2{x} + cos^{2}x - cosx\]
\[\huge sin^2{x} + cosx(cosx - 1)\]
\[\large 1 - cos^2{x} + cosx(cosx-1)\]
\[\large 1 - cosx(cosx + cosx - 1)\]
\[\large 1 - cosx(2cosx - 1)\]

lol i don't think this is simple :P

Answer 29

that seems like entirely too much

Answer 30

so im going to stick with 1-cosA