$$\cos (x)=\left (\frac{1}{3}\right)$$for $0°\lt\ x\lt90°$
Find the exact value of $\sin(x)$. Give your answer as a surd.
I found the value with my calculator $$\sin (x)=\left(\frac{\sqrt 8}{3}\right)$$
Is there a way of doing this by hand?
$\endgroup$ 45 Answers
$\begingroup$$\sin^2 x + \cos^2 x = 1$
$\sin^2 x = 1 - \cos^2 x$
$\sin x = \pm \sqrt{1 - \cos^2 x}$
$\sin x = \pm \sqrt{1 - \left( \frac 13\right)^2}$
$\sin x = \pm \sqrt{1 - \frac 19}$
$\sin x = \pm \sqrt{\frac 89}$
$\sin x = \pm \frac {\sqrt 8}3$
$\endgroup$ $\begingroup$We use
$$\sin^2 (x)=1-\cos^2 (x) $$
$$=1-\frac19=\frac89$$
thus $$\sin (x)=\pm \sqrt {\frac{8}{9}} $$
$\endgroup$ $\begingroup$Since $$\cos^2(x) + \sin^2(x) = 1$$
it follows that:
$$\sin(x) = \pm\sqrt{1-\cos^2(x)} = \pm\sqrt{1-1/9} = \pm\sqrt{8/9}= \pm \sqrt{8}/3$$
$\endgroup$ $\begingroup$From $\sin^2 x +\cos^2 x=1$ we have:
$\sin x=\pm \sqrt{1-\cos^2 x} =\pm\sqrt{1-\frac{1}{9}}=\pm\sqrt{\frac{8}{9}}=\pm \frac{2}{3}\sqrt{2}$
$\endgroup$ $\begingroup$$\sin x = \pm\sqrt{1-\cos^2 x} = \pm\sqrt{1 - (\frac{1}{3})^2} = \pm\frac{\sqrt{8}}{3}$.
One would need to know the quadrant in which $x$ lies in to determine the "+" or "-" on the final answer.
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