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)#.