1. ## Proving

2. It says, prove:
2csc^2(2A) -csc^2(A) = -2cot(2A)csc(2A)
where A = theta

As usual, develop the LHS to see if it matches the RHS.

LHS =
= 2csc^2(2A) -csc^2(A)
= [2 /sin^2(2A)] -[1 /sin^2(A)]
= [2 / (2sinAcosA)^2] -[1 /sin^2(A)]
= [1 / 2sin^2(A)cos^2(A)] -[1 /sin^2(A)]

Combine the two fractions into one fraction only,

= [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
= [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
= [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
= [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]
= [-cot(2A)]*[2 /sin(2A)]
= [-cot(2A)]*[2csc(2A)]
= -2cot(2A)csc(2A)
= RHS

Proven.

3. WOW!Exactly 10 steps lol.

[1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
= [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
= [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

There is something wrong here..did u just take out -1
from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

anyway thks alot..

Umm...wad abt this??is there something wrong here?

[-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
= [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

where did the 1/(1/2) come from??

4. Originally Posted by maybeline9216
WOW!Exactly 10 steps lol.

[1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
= [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
= [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

There is something wrong here..did u just take out -1
from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

anyway thks alot..

5. Originally Posted by maybeline9216
WOW!Exactly 10 steps lol.

[1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
= [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
= [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

There is something wrong here..did u just take out -1
from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

anyway thks alot..

$\displaystyle \begin{array}{l} 1 - 2\cos ^2 2A \\ = - 2\cos ^2 2A + 1 \\ = - (2\cos ^2 2A - 1) \\ \end{array}$

did u just take out -1
from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)]
Note that [-(2cos^2(A) +1) is exactly the same as [1 -2cos^2(A)]...

Umm...wad abt this??is there something wrong here?

[-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
= [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

where did the 1/(1/2) come from??

6. [1 -2cos^2(A)] / [2sin^2(A)cos^2(A)]
= [-(2cos^2(A) -1)] / [(2sinAcosA)(sinAcosA)]
= [-cos(2A)] / [sin(2A)*(1/2)sin(2A)]

There is something wrong here..did u just take out -1
from [1 -2cos^2(A)] ? then its supposed to be [-(2cos^2(A) +1)] isn't it??

1 -2cos^2(2A)
Multiply that by (-1)/(-1) ......(which is really 1 only),
= (-1)/(-1) *[1 -2cos^2(A)]
= [-1 +2cos^2(A)] / (-1)
= -[2cos^2(A) -1]
= -(2cos^2(A) -1)

With practice, you'd master those changing of signs eventually.

-------------------------
Umm...wad abt this??is there something wrong here?

[-cos(2A)] / [sin(2A)*(1/2)sin(2A)]
= [-cos(2A) /sin(2A)]*[1 / (1/2)sin(2A)]

where did the 1/(1/2) come from??

[-cos(2A)] is also [(-cos(2A))*(1)]
The (-cos(2A)) is made to be the numerator over sin(2A), so the (1) remained.
Then, that (1) is made to be the numerator over the (1/2)sin(2A)

More practice, or, in time you'd master all these.

7. i do know how to chg the sign...Erm...its because i was so confused by so many brackets, presenting it in latex form is much easier to see..