#1
OK, im trying to revise and i have no idea how to prove this :-

secx - cosx =sinxtanx

and

sin2x + 2cos2x = 2 - sin2x

Any help would be greatly appreciated.
Quote by Tombe
Shut your noise, I was 18 when I had mine. About 30 seconds later the girl was sick on me. I'm not sure if the two events were related.
#2
For the second one, cos^2x = 1 - sin^2x

so it would be

sin^2x + 2 (1 - sin^2x) =

sin^x + 2 - 2sin^2x =

2 - sin^2x

QED
#6
secx - cosx =sinxtanx

1/cosx - cosx = sinx*sinx/cosx
1/cosx-cosx = sin(squared)/cosx
1/cosx-cosx = 1-cos(squard)x/cosx

...are you sure its not sec(squared) x?
for some reason i cant get this to work
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Last edited by el_strato at Mar 15, 2007,
#7
For the first one:

sec(x) - cos(x) = sin(x)tan(x)

1/cos(x) - cos(x) = sin^2(x)/cos(x) I rewrote sec(x) and distributed the sin(x)

(1 - cos^2(x))/cos(x) = sin^2(x)/cos(x) I combined the two terms on the left into one

1 - cos^2(x) = sin^2(x) I multiplied by cos(x)

cos^2(x) + sin^2(x) = 1, therefore
sin^2(x) = 1 - cos^2(x) Basic trigonometric identity

Hope that helps!
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#8
Slightly different way of doing one:
Edit: alright, maybe not different, but at least easier to read .

sec(x)-cos(x)
= (1/cos(x))-cos(x)            because sec(x) = 1/cos(x)
= (1-cos^2(x))/cos(x)    <-    multiply cos(x) by cos(x)/cos(x), make it into one fraction. 
= sin^2(x)/cos(x)              because sin(x)=sqrt(1-cos^2(x))
= sin(x)[sin(x)/cos(x)]        break apart the sin^2(x)
= sin(x)tan(x)                 because tan = [sin(x)/cos(x)] 


Yay math
#10
Quote by guywithguitar
Slightly different way of doing one:
Edit: alright, maybe not different, but at least easier to read .

sec(x)-cos(x)
= (1/cos(x))-cos(x) because sec(x) = 1/cos(x)
= (1-cos^2(x))/cos(x) <- multiply cos(x) by cos(x)/cos(x), make it into one fraction.
= sin^2(x)/cos(x) because sin(x)=sqrt(1-cos^2(x))
= sin(x)[sin(x)/cos(x)] break apart the sin^2(x)
= sin(x)tan(x) because tan = [sin(x)/cos(x)]


Yay math
Thanks for that, although i dont get this line:-

= (1-cos^2(x))/cos(x) <- multiply cos(x) by cos(x)/cos(x), make it into one fraction.

EDIT : nm, got it now thanks.
Quote by Tombe
Shut your noise, I was 18 when I had mine. About 30 seconds later the girl was sick on me. I'm not sure if the two events were related.
Last edited by adamstartin at Mar 15, 2007,
#11
Quote by umop-3p!sdn
For the second one, cos^2x = 1 - sin^2x

so it would be

sin^2x + 2 (1 - sin^2x) =

sin^x + 2 - 2sin^2x =

2 - sin^2x

QED
Cheers for this one too, im glad there are some decent mathematicians in this hellhole of the pit.
Quote by Tombe
Shut your noise, I was 18 when I had mine. About 30 seconds later the girl was sick on me. I'm not sure if the two events were related.
#13
Quote by adamstartin
Cheers for this one too, im glad there are some decent mathematicians in this hellhole of the pit.


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