How do you find the exact value of ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1 (sqrt2/2)$ ?

Question

How do you find the exact value of ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1 (sqrt2/2)$ ?

1 Answer

Impact of this question

10375 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-14T14:35:13

See below

Explanation:

Let $θ = {cos}^{- 1} (\frac{\sqrt{2}}{2})$ $theta = cos^-1(sqrt2/2)$

$cos θ = \frac{\sqrt{2}}{2}$ $costheta=sqrt2/2$

So in an imaginary right-angled triangle, the length of the $hyp$ $"hyp"$ is $2$ $2$ and the length of the $adj$ $"adj"$ is $\sqrt{2}$ $sqrt2$ . This means that the length of the $opp$ $"opp"$ is $\sqrt{2^{2} - {(\sqrt{2})}^{2}} = \sqrt{2}$ $sqrt(2^2-(sqrt2)^2)=sqrt2$ .

Since the $adj$ $"adj"$ and $opp$ $"opp"$ are equal lengths, our triangle is isosceles. This means that it also has two angles of equal lengths. Since one angle is $\frac{π}{2}$ $pi/2$ , the other two must be $\frac{π}{4}$ $pi/4$ .

So if we say that $θ = \frac{π}{4}$ $theta=pi/4$ , then $cos (\frac{π}{4}) = \frac{adj}{hyp} = \frac{\sqrt{2}}{2}$ $cos(pi/4)="adj"/"hyp"=sqrt2/2$
$thereforepi/4=cos^-1(sqrt2/2)$

Science

Math

Humanities

... and beyond

How do you find the exact value of ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1 (sqrt2/2)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the exact value of cos^-1 (sqrt2/2)cos−1(√22)cos^-1 (sqrt2/2)?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the exact value of ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1 (sqrt2/2)$ ?