How do you simplify $sin (arccos (x))$ $sin(arccos(x))$ ?

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Answer 1 · 2016-10-21T22:07:41

$sin (arccos (x)) = \sqrt{1 - x^{2}}$ $sin(arccos(x)) = sqrt(1-x^2)$

From Pythagoras, we have:

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2 theta + cos^2 theta = 1$

If $x \in [- 1, 1]$ $x in [-1, 1]$ and $θ = arccos (x)$ $theta = arccos(x)$ then:

$θ \in [0, π]$ $theta in [0, pi]$

$sin (θ) \geq 0$ $sin(theta) >= 0$

Hence:

$sin (arccos (x)) = sin (θ) = \sqrt{1 - {cos}^{2} θ} = \sqrt{1 - x^{2}}$ $sin(arccos(x)) = sin(theta) = sqrt(1 - cos^2 theta) = sqrt(1-x^2)$

Note we can use the non-negative square root since we have already established that $sin (arccos (x)) \geq 0$ $sin(arccos(x)) >= 0$

How do you simplify sin(arccos(x))sin(arccos(x))?