Trig problems

V

vsw

New member
Joined
Apr 11, 2005
Messages
13
ok so i have this take home test and ive finished 2/3 of it but this last problems are really challenging, maybe im just looking at it at the wrong angle, if you guys could point me in the right direction id be very grateful:

I need to "prove the following identities" or just make both sides equal to each other.

>2 means raised to the power of two or squared


an example of a right answer i got was as follows:

sin x (csc x - sin x) = cos>2 x

sin x (1/sin x) - (sin x) = cos>2 x

1 - sin>2 x = cos>2 x because of the Pythagonean Idenitities





1) 1-tan>2 x
----------------- = 1- 2sin>2 x
1+ tan>2 x

2) sec>2 x - csc>2 x = tan>2 x - cot>2 x

3) cotx - tanx = 2cot2x
(does this above one equal [cot (x/2) - tan (x/2)= cot 2x] cuz thats what i got and im not sure if thats right)

4) sin2x
----------- = tan x
1+cos 2x


5) 1 - sin x
------------ = (sec x - tan x)>2
1 + sin x



Thanks alot for any help you can give me im extremely puzzled on these.
 
The correct way to type sin>2 x is sin^2(x)

You have the right idea.
Change everything to sines and cosines on one side.

1) tan^2(x) to sin^2(x)/cos^2(x)
Change 1 to cos^2(x)/cos^2(x)
Simplify
Change cos^2(x) into 1-sin^2(x)
Come back if that isn't enough.

2) A little trickier. Do the LHS change.
Add the fractions.
Multiply by sin^2(x)+cos^2(x)
(remember that = 1)
Split up the fractions.
 
i got number one, thanks alot, number two is confusing me, you have to subtract them and where do u Multiply by sin^2(x)+cos^2(x), cuz im getting:

sin^2 x cos^2x sin^2x cos^2x
----------------- - --------------- = ---------- - ----------
cos^2x(sin^2x) cos^2x(sin^2x) cos^2x sin^2x


idk what im doing from here on, it just confuses me.
 
wait!!! ok i see what your saying now


i ended up with:


cos^2x - 1 = sin^2x - 1
------------ -----------
sin^2x cos^2x

am i on the right track?
 
Sorry, had to break.
3) You are close.
1/tan(x)-tan(x) =
(1-tan^2(x))/tan(x) =
2(1-tan^2(x))/2tan(x) =
2/{2tan(x)/(1-tan^2(x))}
The double angle formula is
tan(2x)=2tan(x)/(1-tan^2(x))
so that = 2/tan(2x)

4) Using the double angle formulas again
2sin(x)cos(x)
---------------------- =
1+(cos^2(x)-sin^2(x))

2sin(x)cos(x)/2cos^2(x) =
sin(x)/cos(x)

5) (1-sin(x))^2
--------------------
(1+sin(x))(1-sin(x))
Try it from there.
 
ok you lose me in 2,3, and 4

i posted above my problem with 2

i got lost around here in 3:

"2(1-tan^2(x))/2tan(x) =
2/{2tan(x)/(1-tan^2(x))} "
im confused on what goes where

and in four i got to

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

and i know cos^2 (x)- sin^2(x) is equal to cos 2(x)
but im so lost... sorry im picking this up as fast as i can and i seem to lack certain fundamentals
 
What I had in mind was
2) Do the LHS change.
1/cos^2(x) - 1/cos^2(x)
Add the fractions.
(sin^2(x)-cos^2(x)/((sin^2(x)cos^2(x))
Multiply by sin^2(x)+cos^2(x)
(remember that = 1)
(sin^4(x)-cos^4(x))/(sin^2(x)cos^2(x))
Split up the fractions
sin^2(x)/cos^2(x) - cos^2(x)/sin^2(x)
 
im still confused even though its spelled out for me whats the LHS

where are u getting two 1/cos^2 (x)


what fractions am i combining
split up what fractions?

and multiply what by sin^2(x)+cos^2(x)

my head is spinning and i feel really dumb haha im sorry
 
ok i figured out 4

now im stuck on only 3 and 5

for help on four, the 2cot2(x) throws me off, can i convert that to something to make this easier or do i need to make the other side of the problem equal 2cot2(x)
 
i got lost around here in 3:

"2(1-tan^2(x))/2tan(x) =
2/{2tan(x)/(1-tan^2(x))} "
Between those two equations I was just aranging it so that the denominator looked like the double angle formula. It is like
2*b/c = 2/(c/b)
b = 1-tan^2(x) and
c = 2tan(x)
Then I replaced it with the double angle

(Oh, LHS is code for the left hand side.)

Re two 1/cos. I goofed. Is should be
2) sec>2 x - csc>2 x =
1/cos^2(x) - 1/sin^2(x)
Those are the two fractions.
sin^2/(cos^2(x)*sin^2(x)) - cos^2(x)/(cos^2(x)*sin^2(x)) =
(sin^2(x)-cos^(x))/(cos^2(x)*sin^2(x)
It is the numerator you are multiplying by 1, disguised as sin^2(x)+cos^2(x)
 
ok so i understand what u did in that step for three but i dont understand three at all.

i know:

1/tan(x) - tanx= 1/2tan (2x)

but im a little confused on what to do from there, am i supposed to apply the double angle foruma to the LHS or what? Once again im sorry im so slow.
 
is 2cot (2x) basicly two times the reciprocal of the double angle tan? or does that make any sense at all?
 
so basicly right now im lookin at this:


(1-tan^2(x)) / (tan (x)) = (1- tan^2(x)) / (2tan(x))
 
Lets take then one at a time untill we are together.
3) 1/tan(x)-tan(x) =
Put them over a common denominator.
(1-tan^2(x))/tan(x) =
Multiply top and bottom by 2
2(1-tan^2(x))/2tan(x) =
do the 2b/c thingy
2/{2tan(x)/(1-tan^2(x))}
replace {2tan(x)/(1-tan^2(x))}
by tan(2x) using the double angle formula backwards
= 2/tan(2x)
cot(x)=1/tan(x) so we have
= 2cot(2x)
 
You must have gone beddy-bye.
5) (1 - sin(x))/(1 + sin(x)) =
multiply top and bottom by 1-sin(x)
(1-sin(x))^2/(1-sin^2(x) =
Substitute for 1-sin^2(x)
(1-sin(x))^2/cos^2(x) =
Bring cos^2(x) into the numerator ()^2
{1/cos(x)-sin(x)/cos(x)}^2
 
yeah, i did, it was pushin 12 here and i had to read a story for english, thanks so much for your help, it really pushed me in the right direction on these, and patiently explaining trig to a rookie must be hard haha and i thank you for that was well.
 
You must log in or register to reply here.
Top