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

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

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Answer 1 · 2016-07-03T06:57:22

Multiple values possible but assuming angles $A$ $A$ and $B$ $B$ in first quadrant, $sin (arctan (\frac{1}{4}) + arccos (\frac{3}{4})) = \frac{3 + 4 \sqrt{7}}{4 \sqrt{17}}$ $sin(arctan(1/4)+arccos(3/4))=(3+4sqrt7)/(4sqrt17)$

Explanation:

Let $A = arctan (\frac{1}{4})$ $A=arctan(1/4)$ and $B = arccos (\frac{3}{4})$ $B=arccos(3/4)$ and thus

$tan A = \frac{1}{4}$ $tanA=1/4$ and $cos B = \frac{3}{4}$ $cosB=3/4$ and $sin B = \sqrt{1 - {(\frac{3}{4})}^{2}} = \pm \frac{\sqrt{7}}{4}$ $sinB=sqrt(1-(3/4)^2)=+-sqrt7/4$

$tan A = \frac{1}{4}$ $tanA=1/4$ leads to $cot A = 4$ $cotA=4$ and

$csc A = \pm \sqrt{1 + {cot}^{2} A} = \pm \sqrt{1 + 4^{2}} = \pm \sqrt{17}$ $cscA=+-sqrt(1+cot^2A)=+-sqrt(1+4^2)=+-sqrt17$ and $sin A = \pm \frac{1}{\sqrt{17}}$ $sinA=+-1/sqrt17$ and $cos A = \pm \frac{4}{\sqrt{17}}$ $cosA=+-4/sqrt17$ - note that they will have same sign.

Hence $sin (arctan (\frac{1}{4}) + arccos (\frac{3}{4})) = sin (A + B)$ $sin(arctan(1/4)+arccos(3/4))=sin(A+B)$

= $sin A cos B + cos A sin B$ $sinAcosB+cosAsinB$

Hence assuming all angles in first quadrant,

$sin (A + B) = \frac{1}{\sqrt{17}} \times \frac{3}{4} + \frac{4}{\sqrt{17}} \frac{\sqrt{7}}{4} = \frac{3 + 4 \sqrt{7}}{4 \sqrt{17}}$ $sin(A+B)=1/sqrt17xx3/4+4/sqrt17sqrt7/4=(3+4sqrt7)/(4sqrt17)$

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

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