How do you evaluate sin (arctan (1/4) + arccos (3/4))sin(arctan(14)+arccos(34))?

1 Answer
Jul 3, 2016

Multiple values possible but assuming angles AA and BB in first quadrant, sin(arctan(1/4)+arccos(3/4))=(3+4sqrt7)/(4sqrt17)sin(arctan(14)+arccos(34))=3+47417

Explanation:

Let A=arctan(1/4)A=arctan(14) and B=arccos(3/4)B=arccos(34) and thus

tanA=1/4tanA=14 and cosB=3/4cosB=34 and sinB=sqrt(1-(3/4)^2)=+-sqrt7/4sinB=1(34)2=±74

tanA=1/4tanA=14 leads to cotA=4cotA=4 and

cscA=+-sqrt(1+cot^2A)=+-sqrt(1+4^2)=+-sqrt17cscA=±1+cot2A=±1+42=±17 and sinA=+-1/sqrt17sinA=±117 and cosA=+-4/sqrt17cosA=±417 - note that they will have same sign.

Hence sin(arctan(1/4)+arccos(3/4))=sin(A+B)sin(arctan(14)+arccos(34))=sin(A+B)

= sinAcosB+cosAsinBsinAcosB+cosAsinB

Hence assuming all angles in first quadrant,

sin(A+B)=1/sqrt17xx3/4+4/sqrt17sqrt7/4=(3+4sqrt7)/(4sqrt17)sin(A+B)=117×34+41774=3+47417