if sin x=.76, what is cos?

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Guest Apr 30, 2014

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Best Answer

$\sin^2(x)+\cos^2(x)=1$

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

$\cos(x)=\sqrt{1-(0.76)^2}$

${\sqrt{{\mathtt{1}}{\mathtt{\,-\,}}{\left({\mathtt{0.76}}\right)}^{{\mathtt{2}}}}} = {\mathtt{0.649\: \!923\: \!072\: \!370\: \!876\: \!8}}$

$\cos(x) \approx 0.65$

Rom Apr 30, 2014

Rom · Answer 1 · 2014-04-30T09:44:04

$\sin^2(x)+\cos^2(x)=1$

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

$\cos(x)=\sqrt{1-(0.76)^2}$

${\sqrt{{\mathtt{1}}{\mathtt{\,-\,}}{\left({\mathtt{0.76}}\right)}^{{\mathtt{2}}}}} = {\mathtt{0.649\: \!923\: \!072\: \!370\: \!876\: \!8}}$

$\cos(x) \approx 0.65$

.

Rom · Answer 2 · 2014-04-30T09:44:04

$\sin^2(x)+\cos^2(x)=1$

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

$\cos(x)=\sqrt{1-(0.76)^2}$

${\sqrt{{\mathtt{1}}{\mathtt{\,-\,}}{\left({\mathtt{0.76}}\right)}^{{\mathtt{2}}}}} = {\mathtt{0.649\: \!923\: \!072\: \!370\: \!876\: \!8}}$

$\cos(x) \approx 0.65$

.