What is #Cos(arcsin(5/13) - arccos(7/25))#?

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Answer 1 · 2018-07-28T05:16:51

#\frac{204}{325}#

#\cos(\sin^{-1}(5/13)-\cos^{-1}(7/25))#

#=\cos(\sin^{-1}(5/13))\cos(\cos^{-1}(7/25))+\sin(\sin^{-1}(5/13))\sin(\cos^{-1}(7/25))#

#=\cos(\cos^{-1}(12/13))(7/25)+(5/13)\sin(\sin^{-1}(24/25))#

#=(12/13)(7/25)+(5/13)(24/25)#

#=\frac{84+120}{13\cdot 25}#

#=\frac{204}{325}#

Answer 2 · 2018-07-28T06:06:43

The answer is #=218/325#

Let #theta_1=arcsin(5/13)#

Then,

#sintheta_1=5/13#

And

#costheta_1=sqrt(1-sin^2theta_1)#

#=sqrt(1-25/169)#

#=sqrt(144/169)#

#=14/13#

Let #theta_2=arccos(7/25)#

#costheta_2=7/25#

#sintheta_2=sqrt(1-cos^2theta_2)#

#=sqrt(1-49/625)#

#=sqrt(576/625)#

#=24/25#

Therefore,

#cos(arcsin(5/13)-arccos(7/25))#

#=cos(theta_1-theta_2)#

#=costheta_1costheta_2+sintheta_1sintheta_2#

#=14/13*7/25+5/13*24/25#

#=218/325#