Question

simplfy
sin(arccos0 + arccos1/2)

Answer 1

cos (pi/2) =?

Answer 2

Still need help?

Answer 3

simplify
[0, 180º]
sin(arccos 0 + arccos 1/2)
..
arccos 0=90º

arccos 1/2=60º
..
sin(arccos 0 + arccos 1/2)
sin(90º+ 60º)
=sin 150º
=1/2

Answer 4

did tht help?

Answer 5

To approach it from a slightly different direction: Let \(x=\arccos0\) (so that \(\cos x=0\)) and \(y=\arccos\dfrac{1}{2}\) (so that \(\cos y=\dfrac{1}{2}\)).
\[\begin{align*}\sin(x+y)&=\sin x\cos y+\cos x\sin y\\[1ex]
&=\frac{1}{2}\sin x
\end{align*}\]
Since \(\cos x=0\), we have that \(x=\dfrac{k\pi}{2}\) for nonzero integer \(k\). Keeping in mind that \(0\le\arccos z\le\pi\), we omit all values of \(k\) except \(k=1\), so \(x=\dfrac{\pi}{2}\).

Finally, what's \(\dfrac{1}{2}\sin\dfrac{\pi}{2}\)?