Can anyone please help me with precalc? identities..

[dcgames]

I've been trying to solve/prove these identities for like an hour now and can't seem to do it.

Csc4x - Cot4x = 2Csc2x - 1

(Cot x) / (Csc x - 1) = (Csc x + 1) / (Cot x)

Sin(x + y)Sin(x - y) = Sin2x - Sin2y

Can anyone help?

Thanks in advance..

 

[dcgames]

anyone?

 

[halik]

uhh isnt that somethign youre supposed to memorize ?

I finished calc 3 last semester and have no clue...

 

[Kyteland]

Um that 3rd one isn't an identity

try an example: x=35 y=25
Sin(x + y)Sin(x - y) = Sin(2x) - Sin(2y)
Sin(35 + 25)Sin(35 - 25) = Sin(2*35) - Sin(2*25)
Sin(60)Sin(10) = Sin(70) - Sin(50)
0.15038 != 0.1736

Edit: Are you sure you typed these up right? The first isn't an identity either. Only the second one is.

 

[EarthwormJim]

Originally posted by: dcgames
I've been trying to solve/prove these identities for like an hour now and can't seem to do it.

Csc4x - Cot4x = 2Csc2x - 1

(Cot x) / (Csc x - 1) = (Csc x + 1) / (Cot x)

Sin(x + y)Sin(x - y) = Sin2x - Sin2y

Can anyone help?

Thanks in advance..

You sure about trying to prove this identity?
Sin(x+y)Sin(x-y)=Sin2x-Sin2y

Because I get this:
Sin(x+y)Sin(x-y)=Sin²x-Sin²y

Here's the proof:
sin(x + y) sin(x ? y)
= (sin x cos y + cos x sin y)(sin x cos y ? cos x sin y
= sin²x cos²y ? sin²ycos²x
=sin²x(1-sin²y) - sin²y(1-sin²x)
=sin²x-sin²xin²y-sin²y+sin²xsin²y
=sin²x-sin²y

It's been too long for me to remember the other ones though (heh actually more like 6 months, just finished pre-calc )

 

[cmp1223]

still need help, i'll try them, im really good at these

 

[DrPizza]

Originally posted by: dcgames
I've been trying to solve/prove these identities for like an hour now and can't seem to do it.

Csc4x - Cot4x = 2Csc2x - 1

(Cot x) / (Csc x - 1) = (Csc x + 1) / (Cot x)

Sin(x + y)Sin(x - y) = Sin2x - Sin2y

Can anyone help?

Thanks in advance..

I believe your first one should be Csc^4(x)-cot^4(x)=2csc^2(x)-1
The left hand side is a difference of two squares, can easily be factorable, and one of the factors = 1.
Very easy identity.

For the 2nd identity, slightly trickier... try multiplying the numerator and the denominator on the left hand side by the conjugate of the denominator. (one of the tricks I teach on the first day) Surprise, it's an identity - a little cancelling andyou're done. Another easy one.

3rd identity, as someone else stated above, should be squared terms on the right hand side, not 2x as you have it typed. A longer identity, but still should still be fairly easy.

Wow, your pre-calc teacher must make it easy... The identities I assign take 7 or 8 steps... if they're lucky!


Also, to see if something actually *is* an identity, you can always grab a graphing calculator and graph y = left side and y=right side... see if the graphs overlap. If they do, nothing has been proven yet. If they don't, then you're not going to be able to prove it's an identity (cause it's not!). For your 3rd one, to graph, substitute some value in for y. (not a "nice" number though)