How do you do simplify $sin(cos^(-1)(1/5))$ ?

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Answer 1 · 2015-11-17T23:58:58

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

Consider a right angled triangle with hypotenuse $5$ and one leg $1$ . Then the other leg will have length $sqrt(5^2-1^2) = sqrt(24) = 2sqrt(6)$ .

So if $cos(theta) = 1/5$ then $sin(theta) = (2sqrt(6))/5$

So:

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

Another way of calculating this is to start with:

$cos^2(theta) + sin^2(theta) = 1$

Let $theta = cos^(-1)(1/5)$

Then $cos(theta) = 1/5$ and we find:

$sin(theta) = sqrt(1-cos^2(theta)) = sqrt(1-(1/5)^2)) = sqrt(24/25) = (2sqrt(6))/5$

How do you do simplify sin(cos^(-1)(1/5))sin(cos^(-1)(1/5))?