How do you find the exact value #cos(x+y)# if #tanx=5/3,siny=1/3#?
1 Answer
Explanation:
Use the cosine angle addition formula:
#cos(x+y)=cosxcosy-sinxsiny#
We need to determine
Determining
#tan^2x+1=sec^2x" "" "" "# use#tanx=5/3#
#25/9+1=sec^2x#
#secx=sqrt(34/9)=sqrt34/3#
Then:
#color(blue)(cosx=1/secx=3/sqrt34#
Now using:
#sin^2x+cos^2x=1" "" "" "# where#cosx=3/sqrt34#
#sin^2x+9/34=1#
#color(blue)(sinx=sqrt(25/34)=5/sqrt34#
Determining
#sin^2y+cos^2y=1" "" "" "# use#siny=1/3#
#1/9+cos^2y=1#
#color(blue)(cosy=sqrt(8/9)=(2sqrt2)/3#
Returning to
#cos(x+y)=cosxcosy-sinxsiny#
#color(white)(cos(x+y))=3/sqrt34((2sqrt2)/3)-5/sqrt34(1/3)#
#color(white)(cos(x+y))=(6sqrt2-5)/(3sqrt34)#
Note these are all assuming that