# How do you evaluate sin^-1(cos((3pi)/4)) without a calculator?

Aug 3, 2018

${\sin}^{-} 1 \left(\cos \left(\frac{3 \pi}{4}\right)\right) = - \frac{\pi}{4}$

#### Explanation:

Let ,

$X = {\sin}^{-} 1 \left(\cos \left(\frac{3 \pi}{4}\right)\right)$

$X = {\sin}^{-} 1 \left(\cos \left(\pi - \frac{\pi}{4}\right)\right)$

$X = {\sin}^{-} 1 \left(- \cos \left(\frac{\pi}{4}\right)\right) \ldots . \to \left[\because {2}^{n d} Q u a \mathrm{dr} a n t\right]$

$X = {\sin}^{-} 1 \left(- \frac{1}{\sqrt{2}}\right) \ldots . \to \left[\because {\sin}^{-} 1 \left(- x\right) = - {\sin}^{-} 1 x\right]$

$X = - {\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right)$

$X = - \frac{\pi}{4}$

Aug 4, 2018

$- \frac{\pi}{4}$

#### Explanation:

$\cos \theta = \sin \left(\frac{\pi}{2} - \theta\right) , \theta \in \left(- \infty , \infty\right) r a \mathrm{di} a n$

${\sin}^{- 1} \cos \left(\frac{3}{4} \pi\right) = {\sin}^{- 1} \sin \left(\frac{\pi}{2} - \frac{3}{4} \pi\right)$

$= {\sin}^{- 1} \sin \left(- \frac{\pi}{4}\right)$

$= - \frac{\pi}{4}$.

For related information:

${\sin}^{- 1} \cos \left(- \frac{5}{4} \pi\right)$

$= {\sin}^{- 1} \sin \left(\frac{\pi}{2} + \frac{5}{4} \pi\right)$

$= {\sin}^{- 1} \sin \left(9 \frac{\pi}{4}\right)$

$\ne 9 \frac{\pi}{4}$, as it is $\notin \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

The calculators' answer is $- {45}^{o}$

There is need for the piecewise bijective definition

${\left(\sin\right)}^{- 1} x = k \pi + {\left(- 1\right)}^{k} {\sin}^{- 1} x , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$.

Here, in this example,

${\left(\sin\right)}^{- 1} \sin \left(\frac{9}{4} \pi\right) = \frac{9}{4} \pi$,

in accordance with ${f}^{- 1} f \left(x\right) = x$