Can someone help me answer this question? 'find the exact value of Cos theta, given sin theta = 5/9 and cos theta < 0

Jun 27, 2018

I got $0.831$

Explanation:

Have a look:

Jun 27, 2018

$\cos \left(\theta\right) = - 0.83147941928309808568527749607034$

= -0.8315 (4dp)

Explanation:

Find $\theta$ (Note: there are 2 possible angles for $\sin \left(\theta\right) = \frac{5}{9}$)
$S {\in}^{-} 1 \left(\frac{5}{9}\right) = 33.748988595888587817404332744458 \left(\mathrm{de} g r e e s\right)$

The sin will also have the value $\left(\frac{5}{9}\right)$ at

$180 - 33.748988595888587817404332744458$

$= 146.25101140411141218259566725554 \left(\mathrm{de} g r e e s\right)$

If you are not sure which of these angles will have cos < 1, try both.
(note that the numbers for both angles are actually the same but one is positive and the other is negative)

$\cos \left(33.748988595888587817404332744458\right)$
$= 0.83147941928309808568527749607034$

$\cos \left(146.25101140411141218259566725554\right)$
$= - 0.83147941928309808568527749607034$

(so $\theta$ is the 146.25101.... value)

Jun 27, 2018

$\cos \theta = - \frac{\sqrt{56}}{9}$

Explanation:

$\text{using the "color(blue)"trigonometric identity}$

•color(white)(x)sin^2theta+cos^2theta=1

$\Rightarrow \cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

$\text{given "sintheta=5/9" and } \cos \theta < 0$

$\text{then "theta" is in the second quadrant}$

$\cos \theta = - \sqrt{1 - {\left(\frac{5}{9}\right)}^{2}}$

$\textcolor{w h i t e}{\cos \theta} = - \sqrt{1 - \frac{25}{81}}$

$\textcolor{w h i t e}{\cos \theta} = - \sqrt{\frac{56}{81}} = - \frac{\sqrt{56}}{9}$