How do you find the exact value of $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt(2))$ ?

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How do you find the exact value of $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt(2))$ ?

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Answer 1 · 2016-10-14T15:01:00

$\frac{3 π}{4}$ $(3pi)/4$

Explanation:

$arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt2)$

First, it would be helpful to rationalize $- \frac{1}{\sqrt{2}}$ $-1/sqrt2$ because unit circle values are usually rationalized.

$- \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$ $-1/sqrt2*sqrt2/sqrt2=-sqrt2/2$

Arccos is asking for the ANGLE with a cosine of the given value.

The range of arccos is between zero and $π$ $pi$ . So if you are finding an arccos of a positive value, the answer is between zero and $\frac{π}{2}$ $pi/2$ . If you are finding the arccos of a negative value, the answer is between $\frac{π}{2}$ $pi/2$ and $π$ $pi$ .

According to the unit circle, the angle in the second quadrant (between $\frac{π}{2}$ $pi/2$ and $π$ $pi$ ) with a cosine of $- \frac{\sqrt{2}}{2}$ $-sqrt2/2$ is $\frac{3 π}{4}$ $(3pi)/4$ .

Answer 2 · 2017-01-16T03:22:14

$\frac{3 π}{4}, \frac{5 π}{4}$ $(3pi)/4, (5pi)/4$

Explanation:

cos x = - 1/(sqrt2) = - sqrt2/2
On the trig unit circle, there are 2 arcs that have the same cos value:
$x = \frac{3 π}{4}$ $x = (3pi)/4$ and $x = \frac{5 π}{4}$ $x = (5pi)/4$
Answers for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{3 π}{4}, \frac{5 π}{4}$ $(3pi)/4, (5pi)/4$
Check with calculator:
$cos (\frac{3 π}{4}) = {cos 135}^{\circ} = - 0.707$ $cos ((3pi)/4) = cos 135^@ = - 0.707$
$cos (\frac{5 π}{4}) = {cos 225}^{\circ} = - 0.707$ $cos ((5pi)/4) = cos 225^@ = - 0.707$ .

Answer 3 · 2017-02-26T12:23:16

$\frac{3 π}{4}$ $(3pi)/4$

Explanation:

$arccos (- \frac{1}{\sqrt{2}})$ $color(blue)(arccos(-1/(sqrt2))$

First we should understand what the question is about.

It means that, we need to find an angle, when it is inside a cosine function gives $- \frac{1}{\sqrt{2}}$ $-1/sqrt2$

Let's rationalize it

$\to - \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$ $rarr-1/sqrt2*sqrt2/sqrt2$

$\Rightarrow - \frac{\sqrt{2}}{2}$ $rArr-sqrt2/2$

Now let's find out the angle using the unit circle

enter image source here

The angle is $\frac{3 π}{4}$ $color(green)((3pi)/(4)$

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How do you find the exact value of $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt(2))$ ?

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How do you find the exact value of $arccos (- \frac{1}{\sqrt{2}})$ $arccos(-1/sqrt(2))$ ?