Question

Simply: sin(x)+cot(x)cos(x)

Answer 1

I'm assuming this is being simplified to just one trigonometric function?

Answer 2

yes

Answer 3

\[\sin(x)+\cot(x)\cos(x)\] turn everything into sin and cos.\[\sin(x)+\frac{\cos(x)}{\sin(x)}*\frac{\cos(x)}{1}\]\[\sin(x)+\frac{\cos^2(x)}{\sin(x)}\] make a common denominator. \[\frac{\sin^2(x)}{\sin(x)}+\frac{\cos^2(x)}{\sin(x)}\] you know that \[\sin^2(x)+ \cos^2(x) = 1\]combine fractions. \[\frac{\sin^2(x)+\cos^2(x)}{\sin(x)}\]\[\frac{1}{\sin(x)}\]

Answer 4

You can simplify 1/sin(x) to csc(x). :)

Answer 5

yes!! :)
\[\frac{1}{\sin(x)} = \csc(x)\]

Answer 6

@Jhannybean Glad I'm not the only one who enjoys Fundamental Identities!

Answer 7

so \[\large \sin(x)+\cot(x)\cos(x)= \csc(x)\]

Answer 8

IKR?! theyre fun :D and thank you for the medal!!

Answer 9

Ok, got it! thanks so much! i appreciate it!

Answer 10

Just out of curiousity, how do you give medals?

Answer 11

nooo problemo :)

Answer 12

You just gave me one xD

Answer 13

Well, alrighty. I have no idea how I did that!

Answer 14

how do you people like fundamental identities?!!!!

Answer 15

you can vote/unvote for the best response

Answer 16

It's like solving a puzzle for me. I find it entertaining. :)

Answer 17

fundamental identities are better than graphing parabolas and stuff xD in my opinion

Answer 18

oh, ok. see, now i love graphing parabolas! :D

Answer 19

Definitely. Graphing gets old. Can't same I'm a huge fan of Double Angle and Half Angle Identities though.

Answer 20

I came onto OS to relearn how to graph and all that algebra 2 stuff. Haha. Those are important!

Answer 21

which math class are you guys in?

Answer 22

I actually just found OS about an hour ago. I'm in Pre-Cal right now.

Answer 23

haha thats funny because i just stumbled upon this website like a few minutes ago! I am in Pre-Calc too! i am NOT loving all the trig stuff though.. :(

Answer 24

OH! double angle
\[\sin(2x)=2\sin(x)\cos(x)\]\[\cos(2x)=\cos^2x-\sin^2x = 2\sin^2x-1 = 1-2\sin^2x\]

Half angle just remember you ALWAYS use cos (2x) and sin is negative. \[\sin^2(x)=\frac{1-\cos(2x)}{2}\]\[\sin^2(x)=\frac{1+\cos(2x)}{2}\]

Answer 25

Don't forget the Square root over the half angle. It'll be positive or negative depending on the quadrant you're given.

Answer 26

@cristian500 Trig isn't too bad. It's just very repetitive.

Answer 27

There is no square root over it? not according to the definition i learned. Ohhh you're right, depending on which quadrant, yes. Good one!!

Answer 28

@justanotherstudent that's true... i guess i just dont have a great teacher to teach me... and i have a test tomorrow and i am totally stressin it!

Answer 29

This stuff seems easy, but I still don't get it! haha

Answer 30

@cristian500 I also have a test tomorrow! Best of luck! It's not hard at all, the hardest part is just getting the hang of it. I'm sure you'll catch on with enough practice! If you need more help just let me know. :)

Answer 31

Let me know toooo :)

Haha i just got done with calc 2. WHEW.This all is an AWESOME review!!

Answer 32

@Jhannybean Calc 2? Ouch. Glad we could help you review though!

Answer 33

No way! Good luck to you as well!! but of course, you're already SMART! ;) so how about you give me your luck! haha yeah, i will definitely contact you if i need help! thanks!! hahha @Jhannybean i will be sure to let you know too! :D

Answer 34

Nooo Calc 2 is SIPLE once you get your trig down. it's ALLL about trig, lol.

Answer 35

Haha you'll do fine!!

Answer 36

yes, @Jhannybean see, we both benefited from this! :D

Answer 37

Calc 2?? is that like AP Calc? @Jhannybean

Answer 38

Well, if you need anything just send me a message! I'll be heading off now. Good luck with your test!

Answer 39

thanks to both of you! yes, i will be sure to contact both of you in the future! :D

Answer 40

thanks @Jhannybean ! i appreciate it sooo much! i will message you if i ever need help again! haha

Answer 41

sure!

Answer 42

bye! :D