Precalc HELP

mousey.rr

United States103 Posts

December 13 2007 01:00 GMT

Okay I'm stumped on these proofs. I'm supposed to use trig, pythagorean, and algebraic identities to show that the left hand side is equal to the right hand side but I seem to hit a brick wall every time I start. Is anyone good at these?

Proove that the left hand side = right hand side:

1) (sinx-siny)/(sinx+siny) = (tan(x-y)/2)(cot(x+y)/2)

2) sin^2(x/2) = (cscx - cotx)/(2cscx)

3) tanx/2 + cotx/2 = 2cscx

Can anyone at least lead a math newb in the right direction? Any help is appreciated.

YoUr_KiLLeR

United States3420 Posts

December 13 2007 01:07 GMT

pretty much just look through a list of all the identities and try them out. after making one substitution, see if you can make any other ones, etc. theres also probably several ways to start out, so if your initial substitution didnt work, try another one.

just gotta practice and eventually youll start to more easily see what leads to what before doing them.

micronesia

United States24755 Posts

December 13 2007 01:17 GMT

Amazing... I've done tons of math courses since precalc and I haven't needed any of that stuff :-p

thoraxe

United States1449 Posts

December 13 2007 01:43 GMT

Is this college or high school? I'm in high school and I just come up with a complete blank.

mousey.rr

United States103 Posts

December 13 2007 02:04 GMT

college but I also did it in highschool, its the one thing in precalc i just never understood.

ShOoTiNg_SpElLs

Korea (South)690 Posts

December 13 2007 02:25 GMT

1) (sinx-siny)/(sinx+siny) = (tan(x-y)/2)(cot(x+y)/2)
Start from LHS.
Using the sum to product identity (sinx +/- siny) = 2(sin(x+/-y)/2)*(cos(x-/+y)/2), this one falls apart.
(+/- -/+ just refers to the order, all first ones correlate to each other, all second ones to each other)

2) sin^2(x/2) = (cscx - cotx)/(2cscx)
I would start from RHS.
Expand the right side, and you'll end up with (1-cosx)/2, which turns out to be sin^2(x/2).

3) tanx/2 + cotx/2 = 2cscx
Is the RHS supposed to be csc2x? Also, is the angle x/2 or (tanx)/2 and (cotx)/2 ?

skyglow1

New Zealand3962 Posts

December 13 2007 02:33 GMT

Ya if number 3 is (tanx)/2 and (cotx)/2 the thing should be csc2x on the right hand side.

For number 2 in case you don't know why with (1-cosx)/2 is sin^2(x/2):
cos(2x) = 1-2sin^2(x) (standard double angle cos formula)
so cos(x) = 1-2sin^2(x/2) (having angle on both sides)
so 1-cosx = 1-(1-2sin^2(x/2)) = 2sin^2(x/2)
And dividing by 2 gives sin^2(x/2).

mousey.rr

United States103 Posts

December 13 2007 02:47 GMT

thanks for the help, number 3 is tan (x/2) + cot (x/2) = 2 csc(x) sorry for confusion

Raithed

China7078 Posts

December 13 2007 03:06 GMT

i dont remember this crap but all calc has it, even trig has it.

ShOoTiNg_SpElLs

Korea (South)690 Posts

December 13 2007 03:09 GMT

#10

Then I'd start from LHS again.
We get from half angle identities on tan and cot,
cscx-cotx and cscx+cotx and you get the RHS.
To derive the half angle identities on tan and cot, we have
tan(x/2) = sqrt [(1-cosx)/(1+cosx)], and if you multiply top and bottom by sqrt(1-cosx), the result follows.

BuGzlToOnl

United States5918 Posts

December 13 2007 04:29 GMT

#11

On December 13 2007 10:17 micronesia wrote:
Amazing... I've done tons of math courses since precalc and I haven't needed any of that stuff :-p

hahahahahaha

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