Question 736384: tan[arcsin1/4- arccos1/4]
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! tan[arcsin1/4- arccos1/4]
I will assume you want an exact answer.
let O=opposite side
let A=adjacent side
let H=hypotenuse
..
let s=arcsin1/4
sin s=1/4=O/H
O=1, H=4
A=√(H^2-O^2)=√(4^2-1^2)=√(16-1)=√15
tan s=O/A=1/√15
..
let t=arccos1/4
cos t=1/4=A/H
A=1, H=4
O=√(H^2-A^2)=√(4^2-1^2)=√(16-1)=√15
tan t=O/A=√15/1=√15
..
Identity: tan(s-t)=(tan s-tan t)/(1+tan s tan t)
=(1/√15-√15)/(1+1/√15*√15)
=(-14/√15)/2
=-7/√15
tan[arcsin1/4- arccos1/4]=-7/√15
..
Check with calculator:
arcsin1/4≈14.48º
arccos1/4≈75.52º
14.48-75.52=-61.04º
tan(-61.04º)≈-1.807..
exact value as calculated=-7/√15≈-1.807..
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