#1

Can anyone prove sinx=(1+tanx)/(secx+cscx) ?

#2

I'm sure many people could.

#3

I think I failed this part of trig.

Did I even take trig?

Did I even take trig?

#4

i can. I'm in Calc 3 right now. But do your homework. UG is not an answersheet

#5

Well let's see, tan = sin/cos, sec = 1/cos and csc = 1/sin, sooo...

sinx = (1+sinx/cosx)/(1/cosx+1/sinx)

That's as far as I can get you without thinking about this too hard. Hope it helps.

sinx = (1+sinx/cosx)/(1/cosx+1/sinx)

That's as far as I can get you without thinking about this too hard. Hope it helps.

#6

you totally ripped off my trig thread...

jk

jk

#7

sinx = ((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))

(then cancel like terms)

= sinx

should work

(then cancel like terms)

= sinx

should work

#8

Dear god, not on a keyboard. Btw, I just wrote a grade 12 trig test today.

#9

sinx = ((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))

(then cancel like terms)

= sinx

should work

I got to that point but I'm not sure how you can cancel everything out afterwards.

#10

sinx = (1+sinx/cosx)/(1/cosx+1/sinx)

Multiply both sides by (1/cosx + 1/sinx)

sinx(1/cosx + 1/sinx) = 1+sinx/cosx

Distribute the sinx

(sinx/cosx) + (sinx/sinx) = 1 + sinx/cosx

And sinx/sinx = 1, so

tanx + 1 = 1 + tanx

I know I said I wouldn't finish it but it was bugging me and I wanted to see if I still remember trig.

Multiply both sides by (1/cosx + 1/sinx)

sinx(1/cosx + 1/sinx) = 1+sinx/cosx

Distribute the sinx

(sinx/cosx) + (sinx/sinx) = 1 + sinx/cosx

And sinx/sinx = 1, so

tanx + 1 = 1 + tanx

I know I said I wouldn't finish it but it was bugging me and I wanted to see if I still remember trig.

#11

((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))=searchbar

#12

sin(x) = [1 + tan(x)] / [sec(x) + csc(x)]

There you go. I hope you never build any bridges I have to drive over.

- sin(x) = [1 + (sin(x) / cos(x))] / [ (1/cos(x)) + (1/(sin(x))]
- sin(x) = [(cos(x) + sin(x))/cos(x)] / [(sin(x) + cos(x)) / sin(x)cos(x)]
- sin(x) = [(cos(x) + sin(x))/cos(x)] x [(sin(x)cos(x)) / (sin(x) + cos(x))] so...
- sin(x) = [
~~(cos(x) + sin(x))~~/~~cos(x)~~] x [(sin(x)~~cos(x)~~) / (~~sin(x) + cos(x))~~] - sin(x) = sin(x)

There you go. I hope you never build any bridges I have to drive over.

#13

sin(x)/n=?

n cancels

si(x)=6

ya rly

n cancels

si(x)=6

ya rly

#14

I'm glad I don't have to take math anymore

#15

Thanks! I got it.

#16

Thanks! I got it.

I hope instead of you copying it, you read through it and

*understood*how the answer was derived. Don't slack in your studies.

#17

I hope instead of you copying it, you read through it andunderstoodhow the answer was derived. Don't slack in your studies.

I did. I actually had your final step already written but I didn't notice you could cancel things out so I kept going and making it more confusing. Thanks again.

#18

I did. I actually had your final step already written but I didn't notice you could cancel things out so I kept going and making it more confusing. Thanks again.

#19

I dropped math, for a damn good reason. 12 months ago. I saw calculus!

#20

sin(x) = [1 + tan(x)] / [sec(x) + csc(x)]

- sin(x) = [1 + (sin(x) / cos(x))] / [ (1/cos(x)) + (1/(sin(x))]
- sin(x) = [(cos(x) + sin(x))/cos(x)] / [(sin(x) + cos(x)) / sin(x)cos(x)]
- sin(x) = [(cos(x) + sin(x))/cos(x)] x [(sin(x)cos(x)) / (sin(x) + cos(x))] so...
- sin(x) = [
~~(cos(x) + sin(x))~~/~~cos(x)~~] x [(sin(x)~~cos(x)~~) / (~~sin(x) + cos(x))~~]- sin(x) = sin(x)

There you go. I hope you never build any bridges I have to drive over.

I swear I used to know how to do all of this

#21

Remind me never to take trig.

God I hate math so much.

God I hate math so much.

#22

((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))=searchbar

Your sig is the most clever way I've ever lost the game. Congrats.

#23

I'm going to fail any higher level calc classes so hard. I'm in pre calc(basically trig) right now and EVERYTHING goes over my head. I hate it.

*Last edited by StreetLight3989 at Nov 2, 2009,*

#24

Can anyone prove sinx=(1+tanx)/(secx+cscx) ?

I always saw "sex" when I put this in my calculator. Probably another reason why I barely passed trig.

I used to be able to do stuff like this in my sleep. My, how the years have passed.

#25

I always saw "sex" when I put this in my calculator. Probably another reason why I barely passed trig.

I used to be able to do stuff like this in my sleep. My, how the years have passed.

Why did you put it in a calculator?

Identities can be done intuitively pretty easily..

Maybe

*THAT'S*why.

#26

I couldn't do the trig identeties part of my class... but on the test I copied off the guy beside me .

#27

I could've done this last year when I was in trig, but it's all gone now...

#28

Your sig is the most clever way I've ever lost the game. Congrats.

I don't get it...

#29

I don't get it...

The chords (G Am E)

#30

sinx = (1+sinx/cosx)/(1/cosx+1/sinx)

Multiply both sides by (1/cosx + 1/sinx)

sinx(1/cosx + 1/sinx) = 1+sinx/cosx

Distribute the sinx

(sinx/cosx) + (sinx/sinx) = 1 + sinx/cosx

And sinx/sinx = 1, so

tanx + 1 = 1 + tanx

I know I said I wouldn't finish it but it was bugging me and I wanted to see if I still remember trig.

for identities, you cant do stuff to "both" sides.

you have to pick a side and go from there. but TS, do your homework and pay attention in class, this is an easy problem.

#31

What's with all these math threads lately?

Has the math thread died?!

GOD! SAY IT AIN'T SO!

Has the math thread died?!

GOD! SAY IT AIN'T SO!