Question

What is the exact value of sin(cos^-1 (√2/2)) + tan^-1 (sin(π/2))

Answer 1

`\qquad \qquad \textit{Inverse Trigonometric Identities} \\\\ \begin{array}{cccl} Function&Domain&Range\\[-0.5em] \hrulefill&\hrulefill&\hrulefill\\ y=cos^(-1)(\theta)&-1 ~\le~ \theta ~\le~ 1& 0 ~\le~ y ~\le~ \pi \\\\ y=tan^(-1)(\theta)&-\infty ~\le~ \theta ~\le~ +\infty &-(\pi)/(2) ~\le~ y ~\le~ (\pi)/(2) \end{array} \\\\[-0.35em] ~\dotfill`

`cos^(-1)\left( -\cfrac{√(2)}{2} \right)\implies \theta \hspace{5em}\stackrel{\textit{so we can say}}{cos(\theta )=-\cfrac{√(2)}{2}} \\\\\\ \theta =cos^(-1)\left( -\cfrac{√(2)}{2} \right)\implies \stackrel{ \textit{on the II Quadrant} }{\theta =\cfrac{3\pi }{4}} \\\\[-0.35em] ~\dotfill\\\\ sin\left[ cos^(-1)\left( -\cfrac{√(2)}{2} \right) \right]\implies sin\left( \cfrac{3\pi }{4} \right)\implies \boxed{\cfrac{√(2)}{2}}`

now let's find the angle for the inverse tangent

`sin\left( \cfrac{\pi }{2} \right)\implies 1\hspace{5em}\stackrel{\textit{so we can say}}{tan^(-1)\left[ sin\left( (\pi )/(2) \right) \right]}\implies tan^(-1)(1) \stackrel{ \textit{on the I Quadrant} }{\implies\boxed{\cfrac{\pi }{4}}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin\left[ cos^(-1)\left( -(√(2))/(2) \right) \right]~~ + ~~tan^(-1)\left[ sin\left( (\pi )/(2) \right) \right]\implies \cfrac{√(2)}{2}~~ + ~~\cfrac{\pi }{4} \implies \boxed{\cfrac{2√(2)+\pi }{4}}`

for the sine function we end up in the II Quadrant because the inverse cosine function range is constrained to the I and II Quadrants only, so our angle comes from that range.

Likewise, our angle from the inverse tangent comes from the I Quadrant, because inverse tangent range is only I and IV Quadrants.