anonymous
  • anonymous
Prove that (1 + secx)(1 - cosx) = sinxtanx??
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
write everything in terms of sin and cos to see things better
anonymous
  • anonymous
so 1 - cosx + secx - cosxsecx = secx - cosx. Now what??
anonymous
  • anonymous
sec=cos/sin

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anonymous
  • anonymous
no, sec is 1/cos
anonymous
  • anonymous
hhaha
amistre64
  • amistre64
i was thinking more of: \[1-cos+sec-cos.sec=sin.tan\] \[1-cos+sec-cos/cos=sin.sin/cos\] \[1-cos+sec-1=sin^2/cos\] \[-cos+1/cos=sin^2/cos\] for starters
anonymous
  • anonymous
but we don't have to do that. we have to work from one side, and expand that to get the other side. It has to be proven.
amistre64
  • amistre64
this is working one side; I havent changed the left at all except to see it better
amistre64
  • amistre64
and by left i mean right lol
amistre64
  • amistre64
\[-cos+1/cos=sin^2/cos\] \[(-cos^2+1)/cos=sin^2/cos\] \[(1-cos^2)/cos=sin^2/cos\] \[sin^2/cos=sin^2/cos\]
anonymous
  • anonymous
oh silly me it is a product!! i thought it was a fraction...
anonymous
  • anonymous
but we are not supposed to even supposed to do ANYTHING with the other side. It's like we're not supposed to know the other side.
anonymous
  • anonymous
^that's right :P
amistre64
  • amistre64
your requirements are absurd
amistre64
  • amistre64
your not spose to know tha tan = sin/cos? your not spose to know that sin.sin = sin^2? your not spose to know what; any trig relations and identities? but somehow transform them into each other?
anonymous
  • anonymous
not that, yaar. Imagine they give you a question, saying expand ....., leaving it terms of sin/cos/tan. SO we just do that. we expand. but how??
anonymous
  • anonymous
\[(1+\frac{1}{a})(1-a)=\frac{1}{a}-a=\frac{1-a^2}{a}\] now replace a by cosine and you should get it
amistre64
  • amistre64
since the right side isnt changed; just revert it back into its origanl form and and leave it there; but the process is still the same ...
anonymous
  • anonymous
it is much easier to do the algebra without the sines and cosines. then put them in at the end.
anonymous
  • anonymous
but how does that give us sinxtanx??
anonymous
  • anonymous
but what amistre said is important. it in not a matter of leaving one side untouched. if you can change one side, you can change it back. if you have to hand it in in a certain form, fine, but if you can alter one side you can presumably go backwards
amistre64
  • amistre64
what? how does sin.sin/cos = sin tan?
amistre64
  • amistre64
by definition of what sin and tan ARE lol
anonymous
  • anonymous
ok we did the algebra part, now comes the trig part. replace a by cosine and get \[\frac{1-\cos^2(x)}{\cos(x)}\] now come the only trig step, writing \[1-\cos^2(x)=\sin^2(x)\] that is the one step that is trig
amistre64
  • amistre64
can know that 1-cos^2 = sin^2 sat; its forbidden knowledge ;)
anonymous
  • anonymous
I GOT IT. I love you guys :D
anonymous
  • anonymous
so you get \[\frac{1-\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}=\frac{\sin(x)}{\cos(x)}\times \sin(x)\]
amistre64
  • amistre64
:) good luck
anonymous
  • anonymous
and we love you!
anonymous
  • anonymous
okayy, another question. given that tanx = p, find an expression, in terms of p, for cosec^2 x. GO.
amistre64
  • amistre64
gone!! ;)
anonymous
  • anonymous
??
anonymous
  • anonymous
oh i see what they want
anonymous
  • anonymous
so is it 1 + 1/p^2 ??
anonymous
  • anonymous
i don't think so let me draw a triangle. you should draw one too
anonymous
  • anonymous
|dw:1327848673010:dw|
anonymous
  • anonymous
there is my triangle, the triangle where tan(x) = p
anonymous
  • anonymous
now you need the hypotenuse which you find by pythagoras
anonymous
  • anonymous
|dw:1327848731602:dw|
anonymous
  • anonymous
hypotenuse is \[\sqrt{p^2+1}\] you are looking for \[\csc(x)\] and that is \[\frac{\sqrt{p^2+1}}{p}\]
anonymous
  • anonymous
nonononon. See, cosec^2x = 1 + cot^2x = 1 + 1/tan^2x = 1 + 1/p^2, right??
anonymous
  • anonymous
oh sorry you want cosecant squared, so it is the square of what i wrote, which is \[\frac{p^2+1}{p^2}=1+\frac{1}{p^2}\] yes you have it!
anonymous
  • anonymous
i forgot about the square, sorry
anonymous
  • anonymous
hahahahah, studd. Since you love me, you have to do me a favour. You have to join my group.
anonymous
  • anonymous
group of what?
anonymous
  • anonymous
my openstudy group. I created it B) It's called, wait for it, Herp-Derp.

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