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I remember reading somewhere that we can simplify $\cos(\arcsin x)$ and $\sin(\arccos x)$ in terms of a polynomial by making the substitution $m=\arcsin x$ or $m=\arccos x$ (respectively), then constructing a right angle triangle with appropriate ratios. Does anyone know what I'm talking about? I don't really remember how to do this, so if someone can show me I would really appreciate it!

Thanks.

CivilSigma
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  • You probably mean expressions like $cos^{-1}(sin(x))$, no? – Mose Wintner Aug 07 '15 at 23:14
  • in this case use the identity ${\cos{(x)}}^{2}+{\sin{(x)}}^{2} = 1$ – Oussama Boussif Aug 07 '15 at 23:21
  • $$\arcsin(\cos x)=\arcsin(\sin(\pi/2-x))=\pi/2-x,$$ if that is what you meant. – Jack D'Aurizio Aug 08 '15 at 01:52
  • You may expand the functions to a polynomial by well known methods: http://mathworld.wolfram.com/MaclaurinSeries.html Here you will find the cos and sin expansions. Use cos expansion and replace x with sin... – Moti Aug 08 '15 at 02:36
  • I don't think this question should be closed. It happens that there is no interesting answer. But how could the OP have known this a priori? For example, if he had asked about $\cos(m \cos^{-1}(x))$, he would have received interesting lectures on Tchebychev polynomials. – Stephen Montgomery-Smith Aug 08 '15 at 02:56
  • Thank you all for your input, it turns out I was thinking of cos( arcsinx ) and sin ( arccos x), how would we simplify those? – CivilSigma Aug 08 '15 at 03:52
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    I think here it is just what you were looking for (see the table): https://en.wikipedia.org/wiki/Inverse_trigonometric_functions – Quaerendo Feb 08 '20 at 14:58
  • yes it is thanks ! – CivilSigma Feb 09 '20 at 17:54

1 Answers1

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$$\cos(\arcsin x) = \sqrt{1 - \sin(\arcsin x)^2} = \sqrt{1-x^2}$$

This is the quick and dirty solution, I'll leave you to figure out the intricacies of the sign and the respective domains. Of course, you can simplify the other expression the same way.

Myridium
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Deepak
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