How do you find the exact value of $sin [arcsin (\frac{1}{2}) + arccos (0)]$ $sin[arcsin(1/2) + arccos (0) ]$ ?

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Answer 1 · 2018-04-29T18:41:50

$\frac{\sqrt{3}}{2}$ $sqrt3/2$ .

Recall the following definitions of $a r c sin and a r c cos$ $arc sin and arc cos$ functions :

$a r c sin x = θ, | x | \leq 1 \Leftrightarrow sin θ = x, | θ | \leq \frac{π}{2}$ $arc sinx=theta, |x|le1 iff sintheta=x, |theta|lepi/2$ .

$a r c cos x = θ, | x | \leq 1 \Leftrightarrow cos θ = x, θ \in [0, π]$ $arc cosx=theta, |x|le1 iff costheta=x, theta in [0,pi]$ .

Now, $sin(pi/6)=1/2, and, |pi/6| lepi/2 :. arcsin(1/2)=pi/6$ .

Similarly, $arc cos0=pi/2$ .

$:. sin[arcsin(1/2)+arc cos0]=sin(pi/6+pi/2)$ ,

$=cos(pi/6)$ ,

$=sqrt3/2$ .

How do you find the exact value of sin[arcsin(1/2) + arccos (0) ] sin[arcsin(12)+arccos(0)]sin[arcsin(1/2) + arccos (0) ] ?