#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.

secx - cosx =sinxtanx

and

sin2x + 2cos2x = 2 - sin2x

Any help would be greatly appreciated.

#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

so it would be

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

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

2 - sin^2x

QED

#3

...Uh, im going to assume its triginomotery

#4

Not really, you dont have to actually use trig, its just a matter of substitution.

#5

what is this?

it looks confusing

it looks confusing

#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

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

*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!

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!

#8

Slightly different way of doing one:

Edit: alright, maybe not different, but at least easier to read .

Yay math

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

#9

wheres a popapez when we need one?

#10

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

= (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.

*Last edited by adamstartin at Mar 15, 2007,*

#11

Cheers for this one too, im glad there are some decent mathematicians in this hellhole of the pit.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

#12

i did identities last year and i hated them

#13

Cheers for this one too, im glad there are some decent mathematicians in this hellhole of the pit.

college theoretical calculus .

Yeah, I'm cool.

#14

Im in 8th grade

but double advanced >.>

silly college kids xD

but double advanced >.>

silly college kids xD

#15

i just got a 33% on my math test yesterday

#16

sry. cant help u. way to confusing.