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1. Prove that:

(1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

I get too many cosines.

Laura xxxx
2. (Original post by Laura_M)
Prove that:

(1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

I get too many cosines.

Laura xxxx

1-tan²θ/1+tan²θ

1+tan²θ = sec²θ

=> = (1-tan²θ ) x cos²θ

=cos²θ - tan²θcos²θ

since tan²θ = sin²θ/cos²θ

=> = cos²θ - cos²θsin²θ/cos²θ

= cos²θ - sin²θ = cos2θ

3. (Original post by Laura_M)
Prove that:

(1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

I get too many cosines.

Laura xxxx
cos^2 x +sin^2 x=1
1+tan^2 x =sec^2x

this gives 1-tan^2 x/sec^2 x
= 1/sec^2 x- (tan^2 x)/sec^2 x
=cos^2 x- cos^2 x (tan^2 x)
=cos^2 x -sin^2 x
=cos 2x

be more specific
5. (Original post by Chiral)

be more specific
people? you mean me?
be more specific
6. Oooh, pwnage! :O

But to be fair, Chiral's got a point.

[And, also, to be fair, you (Laura) are not the only one to make an ambiguous thread. So 'people' in general is true.
But still, nice pwning. Sorta.]
7. (Original post by KAISER_MOLE)
1-tan²θ/1+tan²θ

1+tan²θ = sec²θ

=> = (1-tan²θ ) x cos²θ

=cos²θ - tan²θcos²θ

since tan²θ = sin²θ/cos²θ

=> = cos²θ - cos²θsin²θ/cos²θ

= cos²θ - sin²θ = cos2θ

How did you get from the 3rd line to the 4th line?
8. (Original post by mockel)
Oooh, pwnage! :O

But to be fair, Chiral's got a point.

[And, also, to be fair, you (Laura) are not the only one to make an ambiguous thread. So 'people' in general is true.
But still, nice pwning. Sorta.]

Lol
9. (Original post by Laura_M)
How did you get from the 3rd line to the 4th line?
He expanded the brackets:

cos²x(1-tan²x) = cos²x - cos²x.tan²x

etc...
10. (Original post by mockel)
Oooh, pwnage!
What's 'pwnage' ?
11. (Original post by Gaz031)
What's 'pwnage' ?
Ownage. (please say you know what that is.... )
12. Thank you v. much.

The second defenition gives a nice history to the word "pwn" or similar.
"Pwnage" is just a 'branch' of "pwn" (as you can see from the first defenition).
14. (Original post by Laura_M)
Prove that:

(1-tan^2θ)/(1+tan^2θ) ≡ cos2θ

I get too many cosines.

Laura xxxx
Alternative solution:

LHS = (1-tan^2θ)/(1+tan^2θ) = [1 - (sin²θ/cos²θ)]/[1 + (sin²θ/cos²θ)]

=> [cos²θ - sin²θ]/[cos²θ + sin²θ]

=> (cos2θ)/1 = cos2θ = RHS.

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