How do you find the exact value of ${sin}^{- 1} (cos (\frac{π}{4}))$ $Sin^-1(cos(pi/4))$ ?

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Answer 1 · 2018-05-11T17:48:22

$Arc sin (cos (\frac{π}{4})) = \frac{π}{4}$ $text{Arc}text{sin}(cos (pi/4) ) = pi/4$

$arcsin(cos (pi/4)) = pi/4 + pi k, quad$ integer $k$

Oh my, $pi/4$ again. What a surprise.

Question writers, questions like this are your big chance to get away from the two cliche triangles of trig.

I prefer the notation $text{Arc}text{sin}(x)$ for the principal value and $arcsin(x)$ for all values. Let's run the problem both ways.

Of course,

$sin(pi/4) = cos (pi/4) = 1/sqrt{2}$

$text{Arc}text{sin}(cos (pi/4) ) = text{Arc}text{sin}( 1/sqrt{2} ) = pi/4$

$x = arcsin(cos (pi/4))$ is equivalent to

$sin(x) = cos(pi/4)$

I always remember $cos x =cos a$ has solutions $x=pm a + 2pi k quad$ integer $k$ .

$cos(pi/2 - x) = cos(pi/4)$

$pi/2 - x = pm pi/4 + 2pi k$

$x = pi/2 pm pi/4 + 2pi k$

$x = { pi/4, {3pi}/4 } + 2pi k$

We can rewrite that as

$x = pi/4 + pi k$

$arcsin(cos (pi/4)) = pi/4 + pi k, quad$ integer $k$

How do you find the exact value of Sin^-1(cos(pi/4))sin−1(cos(π4))Sin^-1(cos(pi/4))?