How do you evaluate $sin (arccos (\frac{2}{8}))$ $sin (arccos (2/8))$ ?

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Answer 1 · 2015-06-23T08:39:08

$sin (arccos (\frac{2}{8}))$ $sin(arccos(2/8))$ = $\frac{\sqrt{15}}{4}$ $sqrt(15)/4$

Let

$sin ({cos}^{- 1} (\frac{2}{8})) = sin (x)$ $sin(cos^(-1)(2/8))=sin(x)$ $(1)$ $" " "color(red)((1))$

From $(1)$ $color(red)((1))$ ;

$x = {cos}^{- 1} (\frac{2}{8})$ $x=cos^(-1)(2/8)$

$cos x = \frac{2}{8}$ $cosx=2/8$

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

$sin x = \sqrt{1 - {(\frac{2}{8})}^{2}}$ $sinx=sqrt(1-(2/8)^2$

$sin x = \sqrt{1 - \frac{4}{64}}$ $sinx=sqrt(1-4/64$

$sin x = \frac{\sqrt{64 - 4}}{\sqrt{64}}$ $sinx=sqrt(64-4)/sqrt64$

$sin x = \frac{\sqrt{60}}{\sqrt{64}}$ $sinx=sqrt60/sqrt64$

$sin x = \frac{\sqrt{15 \times 4}}{8}$ $sinx=sqrt(15xx4)/8$

$sin x = \frac{2 \sqrt{15}}{8}$ $sinx=(2sqrt(15))/8$

$sin x = \frac{\sqrt{15}}{4}$ $sinx=sqrt(15)/4$

The above sinx value substitute in $(1)$ $" " "color(red)((1))$

$sin (arccos (\frac{2}{8}))$ $sin(arccos(2/8))$ = $\frac{\sqrt{15}}{4}$ $sqrt(15)/4$

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