If sin (x)= 3/8, where x is in quadrant 2, how would you find cos (x)?

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If sin (x)= 3/8, where x is in quadrant 2, how would you find cos (x)?

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Answer 1 · 2017-08-06T18:24:56

$cos x = - \frac{\sqrt{55}}{8}$ $cosx=-sqrt55/8$

Explanation:

$∙ x {sin}^{2} x + {cos}^{2} x = 1$ $•color(white)(x)sin^2x+cos^2x=1$

$\Rightarrow cos x = \pm \sqrt{1 - {sin}^{2} x}$ $rArrcosx=+-sqrt(1-sin^2x)$

$x is in the second quadrant hence cos x < 0$ $"x is in the second quadrant hence "cosx<0$

$cos x = - \sqrt{1 - {(\frac{3}{8})}^{2}}$ $cosx=-sqrt(1-(3/8)^2)$

$cos x = - \sqrt{1 - \frac{9}{64}} = - \sqrt{\frac{55}{64}}$ $color(white)(cosx)=-sqrt(1-9/64)=-sqrt(55/64$

$cos x = - \frac{\sqrt{55}}{8}$ $color(white)(cosx)=-sqrt55/8$

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If sin (x)= 3/8, where x is in quadrant 2, how would you find cos (x)?

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Explanation:

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