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

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Answer 1 · 2016-07-03T06:57:22

Multiple values possible but assuming angles #A# and #B# in first quadrant, #sin(arctan(1/4)+arccos(3/4))=(3+4sqrt7)/(4sqrt17)#

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

#tanA=1/4# and #cosB=3/4# and #sinB=sqrt(1-(3/4)^2)=+-sqrt7/4#

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

#cscA=+-sqrt(1+cot^2A)=+-sqrt(1+4^2)=+-sqrt17# and #sinA=+-1/sqrt17# and #cosA=+-4/sqrt17# - note that they will have same sign.

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

= #sinAcosB+cosAsinB#

Hence assuming all angles in first quadrant,

#sin(A+B)=1/sqrt17xx3/4+4/sqrt17sqrt7/4=(3+4sqrt7)/(4sqrt17)#