What is #cos (2 arcsin (3/5))#?

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Answer 1 · 2015-07-21T22:51:47

#7/25#

First consider that : #epsilon=arcsin(3/5)#

#epsilon# simply represents an angle.

This means that we are looking for #color(red)cos(2epsilon)!#

If #epsilon=arcsin(3/5)# then,

#=>sin(epsilon)=3/5#

To find #cos(2epsilon)# We use the identity : #cos(2epsilon)=1-2sin^2(epsilon)#

#=>cos(2epsilon)=1-2*(3/5)^2=(25-18)/25=color(blue)(7/25)#

Answer 2 · 2015-07-22T05:12:34

We have:

#y = cos(2arcsin(3/5))#

I will do something similar to Antoine's method, but expand on it.
Let #arcsin(3/5) = theta#

#y = cos(2theta)#

#theta = arcsin(3/5)#
#sintheta = 3/5#

Using the identity #cos(theta+theta) = cos^2theta - sin^2theta#, we then have:

#cos(2theta) = (1-sin^2theta) - sin^2theta = 1-2sin^2theta#
(I didn't remember the result, so I just derived it)

#= 1-2{sin[arcsin(3/5)]}^2#

#= 1-2(3/5)^2#

#= 25/25 - 2(9/25)#

#= 25/25 - 18/25 = color(blue)(7/25)#