How do you simplify ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1(sqrt2/2)$ ?

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How do you simplify ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1(sqrt2/2)$ ?

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Answer 1 · 2017-08-03T03:51:23

${cos}^{- 1} (\frac{\sqrt{2}}{2}) = \frac{π}{4}$ $cos^-1(sqrt2/2)=pi/4$

Explanation:

This value is actually a standard value in disguise. Surds are often removed from the denominator of a fraction by multiplying top and bottom by the surd in question. We can do exactly the same thing to put the surd back in the denominator:

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

Maybe you recognise the value now.

What value of theta gives this value? It is one that should be memorised if you have exams on trigonometry.

${cos}^{- 1} (\frac{\sqrt{2}}{2}) = {cos}^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$ $cos^-1(sqrt2/2)=cos^-1(1/sqrt2)=pi/4$

Writing this another way, we have:

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

$θ = \frac{π}{4}$ $theta=pi/4$

Answer 2 · 2017-08-03T04:05:06

$\frac{π}{4}; \frac{7 π}{4}$ $pi/4; (7pi)/4$

Explanation:

Trig table and unit circle give -->
$cos x = \frac{\sqrt{2}}{2}$ $cos x = sqrt2/2$ --> arc $x = \pm \frac{π}{4}$ $x = +- pi/4$ ,
or, using co-terminal arcs:
$x = \frac{π}{4}$ $x = pi/4$ and $x = \frac{7 π}{4}$ $x = (7pi)/4$

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How do you simplify ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1(sqrt2/2)$ ?

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