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

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

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Answer 1 · 2018-04-07T18:18:49

$\frac{\sqrt{2}}{2}$ $sqrt2/2$

Explanation:

First, recall that $cos (x - \frac{π}{2}) = sin x$ $cos(x-pi/2)=sinx$ , so here, we're truly being asked to find

$sin ({cos}^{- 1} (\frac{\sqrt{2}}{2}))$ $sin(cos^-1(sqrt2/2))$

Now, determine ${cos}^{- 1} (\frac{\sqrt{2}}{2})$ $cos^-1(sqrt2/2)$ .

$x = {cos}^{- 1} (\frac{\sqrt{2}}{2}) \Leftrightarrow cos x = \frac{\sqrt{2}}{2}$ $x=cos^-1(sqrt2/2) hArr cosx=sqrt2/2$ , from the definition of an inverse function.

Keeping in mind that the domain of the inverse cosine is $[- 1, 1],$ $[-1, 1],$ the only solution to the above equation is

$x = \frac{π}{4}$ $x=pi/4$

Thus, we get

$sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin(pi/4)=sqrt2/2$

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

1 Answer

Explanation:

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