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

Question

5195 views around the world

You can reuse this answer
Creative Commons License

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,

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

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