You can put this solution on YOUR website! Evaluate sin(arccos 4/5 + arctan 2) Let A = arccos 4/5 and B = arctan 2.
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let o=opposite side
let a=adjacent side
let h=hypotenuse
..
cos(A)=4/5=a/h
a=4, h=5
o=√(h^2-a^2)=√(25-16)=√9=3
sin(A)=o/h=3/5
..
tan(B)=2=o/a
o=2, a=1
h=√a^2+o^2)=√(1+4)=√5
cos(B)=a/h=1/√5
sin(B)=o/h=2/√5
..
sin(arccos 4/5 + arctan 2)=sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
=(3/5)*(1/√5)+(4/5)*(2/√5)=3/5√5+8/5√5=11/(5√5)
..
Calculator check:
cosA=4/5
A=36.87º
tanB=2
B=63.43
A+B=100.30º
sin(A+B)=sin(100.30º)=0.9839..
11/(5√5)=0.9839..
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