How do you evaluate $Sin[Arccos(-1/5)]$ ?

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Answer 1 · 2015-06-29T18:33:07

$sin(arccos(-1/5)) = (2sqrt(6))/5$

Let $alpha = arccos(-1/5)$

By definition of $arccos$ , we have $0 <= alpha <= pi$

More specifically, since $cos(alpha) = -1/5 < 0$ ,

we have $pi/2 < alpha <= pi$

Since $alpha$ is in this quadrant, $sin(alpha) >= 0$

Now $cos^2(alpha) + sin^2(alpha) = 1$

So:

$sin(alpha) = sqrt(1 - cos^2(alpha))$

(must be positive square root since $sin(alpha) >= 0$ )

$=sqrt(1-(-1/5)^2))$

$=sqrt(1-1/25)$

$=sqrt(24/25)$

$=sqrt(6*4/25)$

$=sqrt(6)sqrt(4/25)$

$=(2sqrt(6))/5$

How do you evaluate Sin[Arccos(-1/5)]Sin[Arccos(-1/5)]?