I'm sure many people could.
I think I failed this part of trig.

Did I even take trig?
i can. I'm in Calc 3 right now. But do your homework. UG is not an answersheet
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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.
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And I shall inherit the earth.
you totally ripped off my trig thread...

jk
sinx = ((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))
(then cancel like terms)
= sinx

should work
Dear god, not on a keyboard. Btw, I just wrote a grade 12 trig test today.
Quote by kes1e
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.
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.
I am the meek.

And I shall inherit the earth.
((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))=searchbar
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.
sin(x)/n=?

n cancels

si(x)=6

ya rly
I'm glad I don't have to take math anymore
Quote by Holy Magenta
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.
Quote by Zaphikh
I hope instead of you copying it, you read through it and understood how 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.
Quote by Holy Magenta
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.

I dropped math, for a damn good reason. 12 months ago. I saw calculus!
I am me. Live with it.
Quote by Zaphikh
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
Remind me never to take trig.

God I hate math so much.
Quote by In The Mist
((cosx +sinx)/cosx)/((cosx+sinx)/(sinx*cosx))=searchbar

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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.
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Last edited by StreetLight3989 at Nov 2, 2009,
Quote by Holy Magenta
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.
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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.
I couldn't do the trig identeties part of my class... but on the test I copied off the guy beside me .
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I could've done this last year when I was in trig, but it's all gone now...
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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.
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What's with all these math threads lately?