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prove:
1/sinx -sinx = cosx/tanx
1-sinx/cosx = cosx/(1 + sinx)
1/tanx + tanx = 1/(sinxcosx)
(1-2sin^2x)/(cosx+sinx) = cosx-sinx
explanations for each one would be greatly appreciated.. and a few hints to get confident with these thanks for help!
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1/sinx - (sinx)^2/sinx
= 1 - (sinx)^2 / sinx
= (cosx)^2 / sinx
that is equal to cosx/tanx
tanx = sinx/cosx
cosx/(sinx/cosx)
= (cosx)^2 / sinx
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#3.)
1/tanx + tanx = 1/(sinxcosx)
c/s + s/c = 1/sc
cc/sc + ss/sc = 1/sc
cc + ss = 1 Yes, that is pythagoreans theorm.
igloo myrtilles fourmis
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#4.)
(1-2sin^2x)/(cosx+sinx) = cosx-sinx
(1 - 2s^2)/(c+s) = c-s
mult both sides by c+s:
1 - 2s^2 = (c+s)(c-s) (do you remember diff between 2 squares?)
multiply the right side out to see the difference between 2 squares.
1 - 2s^2 = c^2 - s^2 + sc - cs ( the sc - cs are is 0, they are the same)
s is sine c is cos so sc and cs are just sin*cos or cos*sin
1 - 2s^2 = c^2 - s^2
Add 2s^2 to both sides:
1 = c^2 + s^2 True!, pythagorean's theorm again.
igloo myrtilles fourmis
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#2.) 1-sinx/cosx = cosx/(1 + sinx)
This one cannot be proven because it is false, just try 22 degrees
on your calculator.
(see mathsyperson below me)
Last edited by John E. Franklin (2008-04-23 02:44:17)
igloo myrtilles fourmis
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The left side of #2 is (1-sinx)/cosx, as opposed to 1-tanx. With those brackets, the identity works.
(1-sinx)/cosx
[(1-sinx)(1+sinx)]/[cosx(1+sinx)]
(1-sin²x)/[cosx(1+sinx)]
cos²x/[cosx(1+sinx)]
cosx/(1+sinx)
Why did the vector cross the road?
It wanted to be normal.
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woot! cheers guys
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