Trigonometric Relation Formulas

AI Thread Summary
The discussion centers on various trigonometric identities and formulas, including double angle and sum formulas. Participants request additional identities for functions like sin(x^2), cos(x^2), and their roots, indicating a need for more comprehensive resources. Some members suggest using online references, such as Wolfram's functions site, for extensive trigonometric identities. The conversation also touches on the simplification of expressions involving inverse trigonometric functions, leading to the derivation of several related identities. Overall, the thread highlights a collaborative effort to explore and expand knowledge on trigonometric relations.
TheDestroyer
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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?
 
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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}
 
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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

:)
 
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.

<br /> sin(x) = \frac{a}{c}

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

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

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

Thank you for making a try, anyone else can help also?
 
google search "trigonometric identities" will give you thousands of sites
 
Originally posted by gnome
google search "trigonometric identities" will give you thousands of sites


but not the ones he want.
 
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
 
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:

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

Of course, I could be wrong. I don't know how to prove that there isn't a simple identity.
 
  • #10
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)
 
  • #11
<br /> \begin{align*}<br /> \sin(\arctan(x))&amp;=\frac{x}{\sqrt{1+x^2}} \\<br /> \sin(\arccos(x))&amp;=\sqrt{1-x^2}<br /> \end{align*}<br />
 
  • #12
Now that I think about it we also have:

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

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

<br /> \sin(\arccos(x))=\cos(\arcsin(x))<br />

which was already mentioned, as well as

<br /> \begin{align*}<br /> \sin(\arctan(x))&amp;=x\cos(\arctan(x)) \\<br /> \tan(\arcsin(x))&amp;=\frac{1}{\tan(\arccos(x))}<br /> \end{align*}<br />
 
  • #13
Originally posted by master_coda
<br /> \begin{align*}<br /> \sin(\arctan(x))&amp;=\frac{x}{\sqrt{1+x^2}} \\<br /> \sin(\arccos(x))&amp;=\sqrt{1-x^2}<br /> \end{align*}<br />

= dSin^{-1}(x)/dx
 
  • #14
I thought

<br /> \frac{d}{dx}(\arcsin(x))=\frac{1}{\sqrt{1-x^2}}<br />
 
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