# Question 18d08

Aug 8, 2017

$\cos A = \pm \frac{2 \sqrt{2}}{3}$

#### Explanation:

•color(white)(x)sin^2x+cos^2x=1

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

$\Rightarrow \cos A = \pm \sqrt{1 - {\left(\frac{1}{3}\right)}^{2}}$

$\textcolor{w h i t e}{\Rightarrow \cos A} = \pm \sqrt{\frac{8}{9}}$

$\textcolor{w h i t e}{\Rightarrow \cos A} = \pm \frac{2 \sqrt{2}}{3}$

$\text{the sign of "cosA}$ will be dependent on which quadrant A is in.

Aug 8, 2017

$\cos A = \pm \frac{2 \sqrt{2}}{3}$

#### Explanation:

The sine of a right triangle is the ratio of the lengths opposite side to the hypotenuse, and the cosine is the ratio of the adjacent side to the hypotenuse.

From the given $\sin$, we know that the ratio gives us

• $\text{opposite} = 1$

• $\text{hypotenuse} = 3$

We can use the Pythagorean theorem to find the length of the adjacent side:

"adjacent" = sqrt(3^2 - 1^2) = color(red)(ul(+-2sqrt2

And since

$\cos = \text{adjacent"/"hypotenuse}$

We have

color(blue)(ulbar(|stackrel(" ")(" "cos = +-(2sqrt2)/3" ")|)#

depending on the quadrant it lies in.