Question #bd09c

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Answer 1 · 2018-01-21T23:26:12

#cos(x-y)=(5+24sqrt2)/39#

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#x# and #y# are both in the first quadrant where both sine and cosine are positive.

#sinx=1/3#

#sin^2x+cos^2x=1#

#cos^2x=1-sin^2x=1-(1/3)^2=1-1/9=8/9#

#cosx=(2sqrt2)/3#

#secy=13/12=1/cosy#

#cosy=12/13#

#sin^2y=1-cos^2y=1-(12/13)^2=1-144/169=25/169#

#siny=5/13#

#cos(x-y)=cosxcosy+sinxsiny#

#cos(x-y)=((2sqrt2)/3)(12/13)+(1/3)(5/13)=(24sqrt2)/39+5/39#

#cos(x-y)=(5+24sqrt2)/39#