Question 958763
how do you find the exact value of the expression 
cos(sin^-1 1/3 - tan^-1 1/2)
sinx=1/3
cosx=√(1-sin^2(x))=√1-1/9=√(8/9)=√8/3
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tany=1/2
hypotenuse of  reference right triangle in quadrant I=√1+2^2=√5
cosy=2/√5
siny=1/√5
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cos(sin^-1 1/3 - tan^-1 1/2)=cos(x+y)=cosx*cosy-sinx*siny
=√8/3*2/√5-1/3*1/√5=2√8/3√5-1/3√5=2√8/3√5-1/3√5=(2√8-1)/3√5
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Check: w/calculator
sinx=1/3
x≈19.471˚
tany=1/2
y≈46.036
x+y=46.036˚
cos(sin^-1 1/3 - tan^-1 1/2)=cos(x+y)≈cos(46.036˚)≈0.6942
exact value=(2√8-1)/3√5≈0.6942