How do you evaluate $cos[arcsin ((5sqrt 29)/29)]$ ?

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Answer 1 · 2015-05-15T22:59:28

If $theta = arcsin((5sqrt(29))/29)$ , then that means

$sin theta = (5sqrt(29))/29 = 5/sqrt(29)$

From Pythagoras theorem we know $sin^2 theta + cos^2 theta = 1$

So

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

$=sqrt(1-(5/sqrt(29))^2)$

$=sqrt(1-(5^2)/sqrt(29)^2)$

$=sqrt(1-25/29)$

$=sqrt((29-25)/29)$

$=sqrt(4/29)$

$=sqrt(4)/sqrt(29)$

$=2/sqrt(29)$

$=(2sqrt(29))/29$

Basically we are looking at a right angled triangle with sides $2$ , $5$ and $sqrt(29)$ .

How do you evaluate cos[arcsin ((5sqrt 29)/29)]cos[arcsin ((5sqrt 29)/29)]?