1. ## Deriving identities

Okay, so I have a possibly sort of unique situation!

My instructor does not give any identities or formulas for the tests, and while my final is a long ways off I need to do this at least once now that I'm done with this chapter.

This is a whopper, but I know that once I can see the relationships between them I'll have an easy time with it.

The sets of identities I need to be able to derive from scratch:

-Cosine, Sine, and Tangent of a sum or difference (6 identities)
-Double angle identities of a cosine (3 identities), sine (1) and tangent (1)
-( I have the cofunction identities, they're easy enough)
-Product to sum identities (4)
-Sum to product identities (4)
and lastly the half angle identities (5)

In total that's 30 identities, and I have 6 memorized so 24 identities I have to be able to derive to be able to use on my final exam. This is an online class ( tests are proctored ) and I am allowed to use my TI-89. ( That 'solve(' function is such a time saver )

I have all of the identities written down in my notes ( which are in notepad, so they're ugly. I have dysgraphia and taking written notes is an exercise in futility for me ) I've attached the text document with all my identities for this section

Where do I start? Which ones do I memorize and which ones do I derive?

2. Originally Posted by Wolvenmoon
Okay, so I have a possibly sort of unique situation!

My instructor does not give any identities or formulas for the tests, and while my final is a long ways off I need to do this at least once now that I'm done with this chapter.

This is a whopper, but I know that once I can see the relationships between them I'll have an easy time with it.

The sets of identities I need to be able to derive from scratch:

-Cosine, Sine, and Tangent of a sum or difference (6 identities)
-Double angle identities of a cosine (3 identities), sine (1) and tangent (1)
-( I have the cofunction identities, they're easy enough)
-Product to sum identities (4)
-Sum to product identities (4)
and lastly the half angle identities (5)

In total that's 30 identities, and I have 6 memorized so 24 identities I have to be able to derive to be able to use on my final exam. This is an online class ( tests are proctored ) and I am allowed to use my TI-89. ( That 'solve(' function is such a time saver )

I have all of the identities written down in my notes ( which are in notepad, so they're ugly. I have dysgraphia and taking written notes is an exercise in futility for me ) I've attached the text document with all my identities for this section

Where do I start? Which ones do I memorize and which ones do I derive?
Everything can be derived from cos(A + B) = ..... and sin(A + B) = .....

However, derivations take time and may themselves require memorisation ....

3. this may help, the WEB have plenty of it only IF you try browsing,
this one is the best: Trigonometric Identities -- from Wolfram MathWorld
also this List of trigonometric identities - Wikipedia, the free encyclopedia

Double-, triple-, and half-angle formulae
These can be shown by using either the sum and difference identities or the multiple-angle formulae.
Double-angle formulae[16][17]Triple-angle formulae[18][14]Half-angle formulae[19][20]

The tangent half-angle formulae are as follows. Let
Then we have

4. Okay, so I'm working on memorizing these by working the identities. Right now I'm on the double angle identities. What I derived are:

(Latex never ceases to humiliate me, so they're in calculator terms)

cos2A = cos^2(A) - sin^2(A)

sin2A = 2sin(A)*cos(A)

tan2A = (2tan(A))/(1-(tan^2(A))

There are two others for cosine,

Cos2A = 2*cos^2(A) - 1
cos2A = 1 - 2sin^2(A)

This would mean that:
cos^2(A) - sin^2(A) = 2*cos^2(A) - 1

Where I'm stuck is -sin^2(A) becomes cos^2(A) - 1, and how cos^2(A) becomes -sin^2(A)

This is something to do with the Pythagorean identity, but I'm not seeing it.

5. Originally Posted by Wolvenmoon
Okay, so I'm working on memorizing these by working the identities. Right now I'm on the double angle identities. What I derived are:

(Latex never ceases to humiliate me, so they're in calculator terms)

cos2A = cos^2(A) - sin^2(A)

sin2A = 2sin(A)*cos(A)

tan2A = (2tan(A))/(1-(tan^2(A))

There are two others for cosine,

Cos2A = 2*cos^2(A) - 1
cos2A = 1 - 2sin^2(A)

This would mean that:
cos^2(A) - sin^2(A) = 2*cos^2(A) - 1

Where I'm stuck is -sin^2(A) becomes cos^2(A) - 1, and how cos^2(A) becomes -sin^2(A)

This is something to do with the Pythagorean identity, but I'm not seeing it.
Remember that $sin^2(\theta)+cos^2(\theta)=1$

So you can solve this for $sin^2(\theta)=1-cos^2(\theta)$ and substitute. The negative of this is just $cos^2(\theta)-1$

6. I accept with information: cos2A = cos^2(A) - sin^2(A)

sin2A = 2sin(A)*cos(A)

tan2A = (2tan(A))/(1-(tan^2(A))

There are two others for cosine,

Cos2A = 2*cos^2(A) - 1
cos2A = 1 - 2sin^2(A)

This would mean that:
cos^2(A) - sin^2(A) = 2*cos^2(A) - 1
______________________________
Liposuction before and after photos | laser liposuction cost | Tumescent liposuction prices