Trigonometric Relation Formulas

[TheDestroyer]

I know some of them, such as :

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

sin (2x) = 2(cos x)(sin x)

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

sin (a+b) = sin a cos b + cos a sin b

cos (a+b) = cos a cos b - sin a sin b

I need the other formulas such as

sin x/2
cos x/2
sin x^2
cos x^2
sin x^3
cos x^3
sin (x^(1/2))
cos (x^(1/2))

And any others, everyyything about them,

Anyone can help? or guide me to a link?

 

[PrudensOptimus]

$$Tan[x +- y] = \frac{Tan(x) +- Tan (y)}{1 -+ Tan(x)Tan(y)}$$

$$Sin^{2}(x) + Cos^{2}(x) = 1$$

$$\frac{Sin A}{A} = \frac{Sin B}{B}$$

$$A^{2} = B^{2} + C^{2} - 2BC Cos A$$

$$Cos A = \frac{A^{2} - B^{2} - C^{2}}{-2BC}$$

$$Sin(\frac{x}{2}) = +-\sqrt{\frac{1 - Cos(x)}{2}}$$

$$Cos(\frac{x}{2}) = +-\sqrt{\frac{1 + Cos(x)}{2}}$$

$$Sin^{2}(\frac{x}{2}) = \frac{1 - Cos(x)}{2}$$

$$Cos^{2}(\frac{x}{2}) = \frac{1 + Cos(x)}{2}$$

$$Sin^{2}(x) = \frac{1 - Cos(2x)}{2}$$

$$Cos^{2}(x) = \frac{1 + Cos(2x)}{2}$$

 

[TheDestroyer]

Thanks But ...

Thank you, but I still need sin (x^1/2), cos (x^1/2), sin (x^2), cos (x^2), sin (x^3), cos (x^3), sin (x^1/3), cos (x^1/3)

The roots and poweres are for the angles not for the function

:)

 

[PrudensOptimus]

Sin[sqrt(x)] ? ... I'm going to leave that for the expert of this forum to answer lol. But I will cogitate on it.


Edit: After playing around with a right triangle a bit, I got this from a right triangle with sides a,b,c, with c being the hyp, and a being the opp side with angle x.

$$sin(x) = \frac{a}{c}$$

$$x = sin^{-1}\frac{a}{c}$$

$$\sqrt(x) = \sqrt{sin^{-1}\frac{a}{c}} = \phi$$

$$\begin{equation*}<br /> \begin{split}<br /> sin \phi = +- \sqrt{\frac{1 - cos(2\phi)}{2}}<br /> &= +- \sqrt{\frac{1 - cos(2\sqrt{sin^{-1}(\frac{a}{c})})}{2}}<br /> \end{split}<br /> \end{equation*}$$

 

[TheDestroyer]

:)

Thank you for making a try, anyone else can help also?

 

[gnome]

google search "trigonometric identities" will give you thousands of sites

 

[PrudensOptimus]

Originally posted by gnome
google search "trigonometric identities" will give you thousands of sites

but not the ones he want.

 

[chroot]

How about

http://functions.wolfram.com

It's the base of relationships used by the Mathematica software and has every identity, I believe, known to man.

- Warren

 

[master_coda]

I doubt you'll find any identities for those functions. At least nothing that isn't even messier than the original function. I can think of some ugly expansions, like:

$$\sin\left(x^2\right)=\sum_{n=0}^\infty(-1)^n\frac{x^{2+4n}}{(2n+1)!}$$

Of course, I could be wrong. I don't know how to prove that there isn't a simple identity.

 

[StephenPrivitera]

how about sin(arctanx)? can this be simplified?
how about sin(arccosx)?
I can get this:
sinx=cos(pi/2-x)
y=sinx=cos(pi/2-x)
x=arcsiny=pi/2-arccosy
so arccosy=pi/2-arcsiny
sin(arccosy)=sin(pi/2-arcsiny)=cos(arcsiny)

 

[master_coda]

$$\begin{align*}<br /> \sin(\arctan(x))&=\frac{x}{\sqrt{1+x^2}} \\<br /> \sin(\arccos(x))&=\sqrt{1-x^2}<br /> \end{align*}$$

 

[master_coda]

Now that I think about it we also have:

$$\begin{align*}<br /> \cos(\arcsin(x))&=\sqrt{1-x^2} \\<br /> \cos(\arctan(x))&=\frac{1}{\sqrt{1+x^2}} \\<br /> \tan(\arcsin(x))&=\frac{x}{\sqrt{1-x^2}} \\<br /> \tan(\arccos(x))&=\frac{\sqrt{1-x^2}}{x}<br /> \end{align*}$$

Using those identities, I can spot a few more identities, like

$$\sin(\arccos(x))=\cos(\arcsin(x))$$

which was already mentioned, as well as

$$\begin{align*}<br /> \sin(\arctan(x))&=x\cos(\arctan(x)) \\<br /> \tan(\arcsin(x))&=\frac{1}{\tan(\arccos(x))}<br /> \end{align*}$$

 

[PrudensOptimus]

Originally posted by master_coda
$$\begin{align*}<br /> \sin(\arctan(x))&=\frac{x}{\sqrt{1+x^2}} \\<br /> \sin(\arccos(x))&=\sqrt{1-x^2}<br /> \end{align*}$$

$$= dSin^{-1}(x)/dx$$

 

[master_coda]

I thought

$$\frac{d}{dx}(\arcsin(x))=\frac{1}{\sqrt{1-x^2}}$$